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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , Exercise 3.1, 1., , Akhila went to a fair in her village. She wanted to enjoy rides on the Giant Wheel and play, Hoopla (a game in which you throw a rig on the items kept in the stall, and if the ring, covers any object completely you get it). The number of times she played Hoopla is half, the number of rides she had on the Giant Wheel. Each ride costs Rs 3, and a game of, Hoopla costs Rs 4. If she spent Rs 20 in the fair, represent this situation algebraically and, graphically., Sol:, The pair of equations formed is:, 1, y x, 2, i.e., x 2 y 0, ……..(1), 3x 4 y 20, , ………(2), Let us represent these equations graphically. For this, we need at least two solutions for, each equation. We give these solutions in Table, x, , y, , x, 2, , 0, , 2, , 0, , 1, , x, , y, , 20 3x, 4, , 0, , 2, , 4, , 5, , 0, , 2, , Recall from Class IX that there are infinitely many solutions of each linear equation. So, each of you choose any two values, which may not be the ones we have chosen. Can you, guess why we have chosen x O in the first equation and in the second equation? When, one of the variables is zero, the equation reduces to a linear equation is one variable, which, can be solved easily. For instance, putting x O in Equation (2), we get 4y = 20 i.e.,, 20, 20, is, y 5. Similarly, putting y O in Equation (2), we get 3x 20 i.e., x . But as, 3, 3, not an integer, it will not be easy to plot exactly on the graph paper. So, we choose y = 2, which gives x = 4, an integral value.
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , Plot the points A O, O , B 2,1 and P O,5 , Q 412 , corresponding to the draw the lines, AB and PQ, representing the equations x 2 y O and 3x 4 y 20, as shown in figure, In fig., observe that the two lines representing the two equations are intersecting at the, point (4,2),, , 2., , Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also,, three years from now, I shall be three times as old as you will be.” Is not this interesting?, Represent this situation algebraically and graphically., Sol:, Let the present age of Aftab and his daughter be x and y respectively. Seven years ago., Age of Ahab x 7, Age of his daughter y 7, According to the given condition., x 7 7 y 7, x 7 7 y 49, , x 7 y 42, , Three years hence, Age of Aftab x 3, Age of his daughter y 3, According to the given condition,, , x 3 3 y 3, x 3 3y 9
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, x 3y 6, , Thus, the given condition can be algebraically represented as, x 7 y 42, x 3y 6, x 7 y 42 x 42 7 y, Three solution of this equation can be written in a table as follows:, x, , 7, , 0, , 7, , y, , 5, , 6, , 7, , x 3y 6 x 6 3y, , Three solution of this equation can be written in a table as follows:, x, , 6, , 3, , 0, , y, , 0, , -1, , -2, , The graphical representation is as follows:, , Concept insight In order to represent a given situation mathematically, first see what we, need to find out in the problem. Here. Aftab and his daughters present age needs to be, found so, so the ages will be represented by variables z and y. The problem talks about, their ages seven years ago and three years from now. Here, the words ’seven years ago’, means we have to subtract 7 from their present ages. and ‘three years from now’ or three, years hence means we have to add 3 to their present ages. Remember in order to represent, the algebraic equations graphically the solution set of equations must be taken as whole, numbers only for the accuracy. Graph of the two linear equations will be represented by a, straight line., 3., , The path of a train A is given by the equation 3x + 4y − 12 = 0 and the path of another, train B is given by the equation 6x + 8y − 48 = 0. Represent this situation graphically.
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, Sol:, The paths of two trains are giver by the following pair of linear equations., 3x 4 y 12 0, ... 1, , 6 x 8 y 48 0, , ... 2 , , In order to represent the above pair of linear equations graphically. We need two points on, the line representing each equation. That is, we find two solutions of each equation as given, below:, We have,, 3x 4 y 12 0, Putting y 0, we get, , 3x 4 0 12 0, , 3x 12, 12, , x 4, 3, Putting x 0, we get, 3 0 4 y 12 0, , 4 y 12, 12, , y 3, 4, Thus, two solution of equation 3x 4 y 12 0 are 0,3 and 4, 0 , We have,, 6 x 8 y 48 0, Putting x 0, we get, 6 0 8 y 48 0, , , , 8 y 48, 48, , y, 8, , y6, Putting y 0, we get, , , , , 6 x 8 0 48 0, 6 x 48, 48, x, 8, 6, , Thus, two solution of equation 6 x 8 y 48 0 are 0, 6 and 8, 0
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , Clearly, two lines intersect at 1, 2 , Hence, x 1, y 2 is the solution of the given system of equations., 4., , Gloria is walking along the path joining (− 2, 3) and (2, − 2), while Suresh is walking, along the path joining (0, 5) and (4, 0). Represent this situation graphically., Sol:, It is given that Gloria is walking along the path Joining 2,3 and 2, 2 , while, Suresh is walking along the path joining 0,5 and 4, 0 ., , We observe that the lines are parallel and they do not intersect anywhere., 𝑎, , 𝑏, , 𝑐, , 5. On comparing the ratios 𝑎1 , 𝑏1 𝑎𝑛𝑑 𝑐1 and and without drawing them, find out whether the, 2, , 2, , 2, , lines representing the following pairs of linear equations intersect at a point, are parallel or, coincide:, (i) 5x− 4y + 8 = 0, (ii) 9x + 3y + 12 = 0 (iii) 6x − 3y + 10 = 0, 7x + 6y − 9 = 0, 18x + 6y + 24 = 0, 2x – y + 9 = 0
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , a1 6 3, ,, a2 2 1, b1 3 3, , ,, b2 1 1, And, , c1 10, , c2 9, , , , a1 b1 c1, , a2 b2 c2, , The lines are parallel, 6. Given the linear equation 2x + 3y - 8 = 0, write another linear equation in two variables such, that the geometrical representation of the pair so formed is:, (i) intersecting lines, (ii) parallel lines, (iii) coincident lines., Sol:, We have,, 2x 3 y 8 0, Let another equation of line is:, 4x 9 y 4 0, Here,, a1 2, b1 3, c1 8, a2 4, b2 9, c2 4, , Now,, , a1 2 1, ,, a2 4 2, , b1 3 1, ,, b2 9 3, And, , c1 8 2, , , c2 4 1, , , , a1 b1, , a2 b2, , 2 x 3 y 8 0 and 4 x 9 y 4 0 intersect each other at one point., , Hence, required equation of line is 4 x 9 y 4 0, We have,, 2x 3 y 8 0, Let another equation of line is:, 4x 6 y 4 0
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, Here,, a1 2, b1 3, c1 8, a2 4, b2 6, c2 4, , Now,, , a1 2 1, ,, a2 4 2, , b1 3 1, ,, b2 6 2, And, , c1 8 2, , , c2 4 1, , , , a1 b1 c1, , a2 b2 c2, , Lines are parallel to each other., Hence, required equation of line is 4 x 6 y 4 0., 7., , The cost of 2kg of apples and 1 kg of grapes on a day was found to be Rs 160. After a, month, the cost of 4kg of apples and 2kg of grapes is Rs 300. Represent the situation, algebraically and geometrically., Sol:, Let the cost of 1 kg of apples and 1 kg grapes be Rs x and Rs y., The given conditions can be algebraically represented as:, 2 x y 160, 4 x 2 y 300, 2 x y 160 y 160 2 x, Three solutions of this equation cab be written in a table as follows:, x, , 50, , 60, , 70, , y, , 60, , 40, , 20, , 300 4 x, 2, Three solutions of this equation cab be written in a table as follows:, 4 x 2 y 300 y , , x, , 70, , 80, , 75, , y, , 10, , -10, , 0, , The graphical representation is as follows:
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , Concept insight: cost of apples and grapes needs to be found so the cost of 1 kg apples and, 1kg grapes will be taken as the variables from the given condition of collective cost of, apples and grapes, a pair of linear equations in two variables will be obtained. Then In, order to represent the obtained equations graphically, take the values of variables as whole, numbers only. Since these values are Large so take the suitable scale., , Exercise 3.2, Solve the following systems of equations graphically:, 1. x y 3, 2 x 5 y 12, Sol:, We have, x y 3, 2 x 5 y 12, Now,, x y 3, When y 0, we have, , x3, When x 0, we have, y 3, Thus, we have the following table giving points on the line x y 3, x, , 0, , 3, , y, , 3, , 0, , Now,
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, 2 5 y 12, , 12 2 x, 5, When x 1, we have, y, , y, , 12 11, , 2, 5, When x 4, we have, y, , 12 1 4 , , 4, 5, Thus, we have the following table giving points on the line 2 x 5 y 12, x, , 1, , -4, , y, , 2, , 4, , Graph of the equation x y 3 and 2 x 5 y 12 :, , Clearly, two lines intersect at P 1, 2 ., Hence, x 1, y 2 is the solution of the given system of equations., , 2., , x 2y 5, 2 x 3 y 10, Sol:, We have, x 2y 5, , 2 x 3 y 10, Now,, , , , x 2y 5, x 5 2y
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, When y 0, we have, , x 5 2 0 5, When y 2, we have, x 5 2 2 1, Thus, we have the following table giving points on the line x 2 y 5, x, , 5, , 1, , y, , 0, , -2, , Now,, , 2 x 3 y 10, 2 x 10 3 y, , , , 10 3 y, 2, When y 0, we have, , , x, , 10, 5, 2, When y 0, we have, x, , 10, 5, 2, When y 2, we have, x, , 10 3 2, 2, 2, Thus, we have the following table giving points on the line 2 x 3 y 10, x, , x, , 5, , 2, , y, , 0, , 2, , Graph of the equation x 2 y 5 and 2 x 3 y 10 :
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, Clearly, two lines intersect at (5,0)., Hence, x 5, y 0 is the solution of the given system of equations., , 3x y 1 0, 2x 3y 8 0, , 3., , Sol:, We have,, 3x y 1 0, , 2x 3y 8 0, Now,, , 3x y 1 0, y 1 3x, , , , When x 0, we have, y 1, When x 1, we have, , y 1 3 1 2, Thus, we have the following table giving points on the line 3x y 1 0, x, , -1, , 0, , y, , 2, , -1, , Now,, , 2x 3y 8 0, , 2x 3y 8, 3y 8, , x, 2, When y 0, we have, , 3 0 8, 4, 2, When y 2, we have, x, , 3 2 8, 1, 2, Thus, we have the following table giving points on the line 2 x 3 y 8 0, x, , x, , -4, , -1, , y, , 0, , -2, , Graph of the equation are:
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , Clearly, two lines intersect at - 1, 2 ., Hence, x 1, y 2 is the solution of the given system of equations., , 2x y 3 0, 2x 3y 7 0, , 4., , Sol:, We have, 2x y 3 0, , 2x 3y 7 0, Now,, , 2x y 3 0, y 3 2x, , , , When x 0, we have, y 3, When x 1, we have, y 1, Thus, we have the following table giving points on the line 2 x y 3 0, x, , 0, , 1, , y, , 3, , 1, , Now,, , , , 2x 3y 7 0, 3y 2x 7, , 25 7, 1, 3, When x 5, we have, , , y
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , 25 7, 1, 3, When x 2, we have, y, , 2 2 7, 1, 3, Thus, we have the following table giving points on the line 2 x 3 y 7 0, y, , x, , 2, , 5, , y, , -1, , 1, , Graph of the given equation are, , Clearly, two lines intersect at 2, 1 ., Hence, x 2, y 1 is the solution of the given system of equations., , x y 6, x y 2, , 5., , Sol:, We have., x y 6, , x y 2, Now,, , , , x y 6, y 6 x, , When x 2, we have, y4, When x 3, we have
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, y 3, , Thus, we have the following table giving points on the line x y 6, x, , 2, , 3, , y, , 4, , 3, , Now,, , x y 2, y x2, , , , When x 0, we have, y 2, When x 2, we have, y0, Thus, we have the following table giving points on the line x y 6, x, , 0, , 2, , y, , -2, , 0, , Graph of the given equation are, , Clearly, two lines intersect at (4,2)., Hence, x 4, y 2 is the solution of the given system of equations., , 6., , x 2y 6, 3x 6 y 0, Sol:, We have., x 2y 6, , 3x 6 y 0
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, Now,, , x 2y 6, x 6 2y, , , , When y 2, we have, , x 6 2 2 2, When y 3, we have, , x 6 2 3 0, Thus, we have the following table giving points on the line x 2 y 6, x, , 2, , 0, , y, , -2, , -3, , Now,, 3x 6 y 0, 3x 6 y, x 2y, , , , , When y 0, we have, , x0, When y 1, we have, x2, Thus, we have the following table giving points on the line 3x 6 y 0, x, , 0, , 2, , y, , 0, , 1, , Graph of the given equation are, , Clearly, two lines are parallel to each other. So, the two lines have no common point, Hence, the given system of equations has no solution.
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , x y 4, 2x 3y 3, , 7., , Sol:, We have., x y 4, , 2x 3y 3, Now,, , x y 4, x 4 y, , , , When y 0, we have, , x4, When y 2, we have, x2, Thus, we have the following table giving points on the line x y 4, x, , 4, , 2, , y, , 0, , 2, , Now,, , 2x 3y 3, , 2x 3y 3, 3y 3, , x, 2, When y 1, we have, , x3, When y 1, we have, x0, Thus, we have the following table giving points on the line 2 x 3 y 3, x, , 3, , 0, , y, , 1, , -1, , Graph of the given equation are
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , Clearly, two lines intersect at (3, 1)., Hence, x 3, y 1 is the solution of the given system of equations., , 2x 3y 4, x y3 0, , 8., , Sol:, We have., 2x 3y 4, , x y3 0, Now,, , 2x 3y 4, , 2x 4 3y, 4 3y, , x, 2, When y 0, we have, , 4 3 2, 1, 2, When y 2, we have, x, , 4 3 2, 1, 2, Thus, we have the following table giving points on the line 2 x 3 y 4, x, , x, , -1, , 2, , y, , 2, , 0
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, Now,, , x y3 0, x y 3, , , , When y 3, we have, , x0, When y 4, we have, x 1, Thus, we have the following table giving points on the line x y 3 0, x, , 0, , 1, , y, , 3, , 4, , Graph of the given equation are, , , ., Clearly, two lines intersect at 12, Hence, x 1, y 2 is the solution of the given system of equations., , 2 x 3 y 13 0, 3x 2 y 12 0, , 9., , Sol:, We have,, 2 x 3 y 13 0, , 3x 2 y 12 0, Now,
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , 2 x 3 y 13 0, , 2 x 3 y 13, 3 y 13, , x, 2, When y 1, we have, , 3 1 13, 5, 2, When y 3, we have, x, , 3 3 13, 2, 2, Thus, we have the following table giving points on the line 2 x 3 y 13 0, x, , x, , -5, , -2, , y, , 1, , 3, , Now,, , 3x 2 y 12 0, , 3x 2 y 12, 2 y 12, , x, 3, When y 0, we have, , 2 0 12, 14, 3, When y 3, we have, x, , 2 3 12, 2, 3, Thus, we have the following table giving points on the line 3 y 2 y 12 0, x, , x, , -4, , -2, , y, , 0, , 3, , Graph of the given equations are:
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , Clearly, two lines intersect at (-2, 3), Hence, x 2, y 3 is the solution of the given system of equations., , 2x 3 y 5 0, 3x 2 y 12 0, , 10., , Sol:, We have,, 2x 3 y 5 0, , 3x 2 y 12 0, Now,, , 2x 3y 5 0, , 2 x 3 y 5, 3 y 5, , x, 2, When y 1, we have, , 3 1 5, 4, 2, When y 1, we have, x, , x, , 3 1 5, , 1, 2, Thus, we have the following table giving points on the line 2 x 3 y 5 0, x, , -4, , -1, , y, , 1, , -1
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, Now,, , 3x 2 y 12 0, , 3x 2 y 12, 2 y 12, , x, 3, When y 0, we have, , 2 0 12, 4, 3, When y 3, we have, x, , 2 3 12, 6, 3, Thus we have the following table giving points on the line 3x 2 y 12 0, x, , x, , 4, , 6, , y, , 0, , 3, , Graph of the given equations are:, , Clearly, two lines intersect at 2, 3 ., Hence, x 2, y 3 is the solution of the given system of equations., Show graphically that each one of the following systems of equations has infinitely many, solutions:, , 2x 3 y 6, 4 x 6 y 12, , 11., Sol:
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, We have,, 2x 3 y 6, , 4 x 6 y 12, Now,, , 2x 3y 6, , 2x 6 3y, 6 3y, , x, 2, When y 0, we have, , x3, When y 2, we have, , 6 3 2, 0, 2, Thus, we have the following table giving points on the line 2 x 3 y 6, x, , x, , 0, , 3, , y, , 2, , 0, , Now,, , 4 x 6 y 12, , 4 x 12 6 y, 12 6 y, , x, 4, When y 0, we have, , x3, When y 2, we have, , 12 6 2, 0, 3, Thus, we have the following table giving points on the line 4 x 6 y 12, x, , x, , 0, , 3, , y, , 2, , 0, , Graph of the given equations:
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , Thus, the graphs of the two equations are coincident., Hence, the system of equations has infinitely many solutions., , x 2y 5, 3x 6 y 15, , 12., , Sol:, We have,, x 2y 5, , 3x 6 y 15, Now,, , x 2y 5, x 2y 5, , , , When y 1, we have, , x 2 1 5 3, When y 0, we have, , x 2 0 5 5, Thus, we have the following table giving points on the line x 2 y 5, x, , 3, , 5, , y, , 1, , 0, , Now,, , 3x 6 y 15, , 3x 15 6 y, 15 6 y, , x, 3, When y 2, we have
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , x, , 15 6 2 , , 1, 3, When y 3, we have, , x, , 15 6 3, , 1, 3, Thus, we have the following table giving points on the line 3x 6 y 15, x, , 1, , -1, , y, , -2, , -3, , Graph of the given equations:, , 3x y 8, 6 x 2 y 16, , 13., , Sol:, We have,, 3x y 8, , 6 x 2 y 16, Now,, , 3x y 8, y 8 3x, , , , When x 2, we have, y 8, 3 2 2, When x 3, we have, y 8, 3 3 1, Thus we have the following table giving points on the line 3x y 8, x, , 2, , 3, , y, , 2, , -1
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, Now,, , 6 x 2 y 16, , 2 y 16 6 x, 16 6 x, , y, 2, When x 1, we have, , 16 6 1, 5, 2, When x 3, we have, y, , 16 6 3, 1, 2, Thus we have the following table giving points on the line 6 x 2 y 16, y, , x, , 1, , 3, , y, , 5, , -1, , Graph of the given equations:, , Thus, the graphs of the two equations are coincident., Hence, the system of equations has infinitely many solutions,, , x 2 y 11 0, 3x 6 y 33 0, , 14., , Sol:, We have,, x 2 y 11 0, , 3x 6 y 33 0, Now,
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , x 2 y 11 0, x 2 y 11, , , , When y 5, we have, , x 2 5 11 1, When x 4, we have, x 2 4 11 3, Thus we have the following table giving points on the line x 2 y 11 0, x, , -1, , -3, , y, , 5, , 4, , Now,, , 3x 6 y 33 0, , 3x 6 y 33, 6 y 33, , x, 1, 3, When y 6, we have, , 6 6 33, 1, 3, When y 5, we have, x, , 6 5 33, 1, 2, Thus we have the following table giving points on the line 3x 6 y 33 0, x, , x, , 1, , -1, , y, , 6, , 5, , Graph of the given equations:, , Thus, the graphs of the two equations are coincident,, Hence, the system of equations has infinitely many solutions,
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, Show graphically that each one of the following systems of equations is in-consistent (i.e.,, has no solution), , 3x 5 y 20, 6 x 10 y 40, , 15., , Sol:, We have,, 3x 5 y 20, , 6 x 10 y 40, Now, , , 3x 5 y 20, 5 y 20, , x, 3, When y 1, we have, x, , 5 1 20, , 5, 3, When y 4, we have, , x, , 5 4 20, , 0, 3, Thus we have the following table giving points on the line 3x 5 y 20, x, , 5, , 0, , y, , -1, , -4, , Now, , 6 x 10 y 40, , 6 x 40 10 y, 40 10 y, , x, 6, When y 4, we have, , 40 10 4, 0, 6, When y 1, we have, x, , 40 10 1, 5, 6, Thus we have the following table giving points on the line 6 x 10 y 40, x, , x, , 0, , -5
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, y, , 4, , 1, , Graph of the given equations:, , Clearly, there is no common point between these two lines, Hence, given system of equations is in-consistent., , x 2y 6, 3x 6 y 0, , 16., , Sol:, We have, x 2y 6, , 3x 6 y 0, Now,, , x 2y 6, x 6 2y, , , , When y 0, we have, , x 6 2 0 6, When y 2, we have, x 6 2 2 2, Thus, we have the following table giving points on the line x 2 y 6, x, , 6, , 2, , y, , 0, , -2, , Now,
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , 3x 6 y 0, , 3x 6 y, 6y, , x, 3, , x 2y, When y 0, we have, , x 2 0 0, When y 1, we have, x 2 1 2, Thus, we have the following table giving points on the line 3x 6 y 0, x, , 0, , 2, , y, , 0, , 1, , Graph of the given equations:, , We find the lines represented by equations x 2 y 6 and 3x 6 y 0 are parallel. So, the, two lines have no common point., Hence, the given system of equations is in-consistent., , 2y x 9, 6 y 3x 21, , 17., , Sol:, We have, 2y x 9, , 6 y 3x 21, Now,, , , , 2y x 9, 2y 9 x, x 2y 9, , When y 3, we have
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , x 2 3 9 3, When y 4, we have, x 2 4 9 1, Thus, we have the following table giving points on the line 2 x x 9, x, -3, -1, y, , 3, , 4, , Now,, , , , , 6 y 3 x 21, 6 y 21 3 x, 3x 6 y 21, , , , x, , 3 2 y 7, , 3, x 2y 7, , , , When y 2, we have, , x 2 2 7 3, When y 3, we have, x 2 3 7 1, Thus, we have the following table giving points on the line 6 y 3x 21., x, , -3, , -1, , y, , 2, , 3, , Graph of the given equations:, , We find the lines represented by equations 2 y x 9 and 6 y 3x 21 are parallel. So, the, two lines have no common point., Hence, the given system of equations is in-consistent.
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , 3x 4 y 1 0, 8, 2x y 5 0, 3, , 18., , Sol:, We have, 3x 4 y 1 0, , 8, 2x y 5 0, 3, Now,, , 3x 4 y 1 0, , 3x 1 4 y, 1 4 y, , x, 3, When y 2, we have, , 1 4 2, 3, 3, When y 1, we have, x, , x, , 1 4 1, , 1, 3, Thus, we have the following table giving points on the line 3x 4 y 1 0., x, , -1, , 3, , y, , -1, , 2, , Now,, 8, 2x y 5 0, 3, 6 x 8 y 15, , 0, 3, , 6 x 8 y 15 0, , 6 x 8 y 15, 8 y 15, , x, 6, When y 0, we have, , 8 0 15, 2.5, 6, When y 3, we have, x, , x, , 8 3 15, 1.5, 6
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , 8, Thus, we have the following table giving points on the line 2 x y 5 0., 3, x, -2.5, 1.5, y, , 0, , 3, , Graph of the given equations:, , 8, We find the lines represented by equations 3x 4 y 1 0 and 2 x y 5 0 are, 3, parallel. So, the two lines have no common point., Hence, the given system of equations is in-consistent., 19., , Determine graphically the vertices of the triangle, the equations of whose sides are given, below:, 2y x 8, 5 y x 14, y 2x 1, (i), yx, , (ii), , y0, 3x 3 y 10, , Sol:, We have, 2y x 8, 5 y x 14, y 2x 1, Now,, , , , 2y x 8, 2y 8 x, x 2y 8
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, When y 2, we have, , x 2 2 8 4, When y 4, we have, , x 2 4 8 0, Thus, we have the following table giving points on the line 2 y x 8., x, , -4, , 0, , y, , 2, , 4, , Now,, 5 y x 14, 5 y 14 x, x 5 y 14, , , , , When y 2, we have, , x 5 2 14 1, When y 3, we have, x 5 3 14 1, Thus, we have the following table giving points on the line 5 y x 14., x, , -4, , 1, , y, , 2, , 3, , We have, y 2x 1, , y 1 2x, y 1, , x, 2, When y 3, we have, , 3 1, 1, 2, When y 1, we have, x, , 1 1, 1, 2, Thus, we have the following table giving points on the line y 2 x 1., x, , x, , -1, , 1, , y, , 1, , 3, , Graph of the given equations:
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , From the graph of the lines represented by the given equations, we observe that the lines, taken in pairs intersect each other at points A 4, 2 , B 1,3 and C 2,5 , Hence, the vertices of the triangle are A 4, 2 , B 1,3 and C 2,5 ., The given system of equations is, yx, , y0, 3x 3 y 10, We have,, yx, When x 1, we have, y 1, When x 2, we have, y 2, Thus, we have the following table points on the line y x, x, , 1, , -2, , y, , 7/3, , 4/3, , Graph of the given equation:
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , From the graph of the lines represented by the given equations, we observe that the lines, 10 , 5 5, taken in pairs intersect each other at points A 0, 0 , B , 0 and C , , 3 , 3 3, , 10 , 5 5, Hence, the required vertices of the triangle are A 0, 0 , B , 0 and C , ., 0 , 3 3, 20., , Determine, graphically whether the system of equations x − 2y = 2, 4x − 2y = 5 is, consistent or in-consistent., Sol:, We have, x 2y 2, , 4x 2 y 5, Now, , x 2y 2, x 2 2y, , , , When y 0, we have, , x 2 2 0 2, When y 1, we have, x 2 2 1 0, Thus, we have the following table giving points on the line x 2 y 2, x, , 2, , 0, , y, , 0, , -1, , Now,
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , 4x 2 y 5, , 4x 5 2 y, 5 2y, , x, 4, When y 0, we have, , 5 2 0 5, , 4, 4, When y 1, we have, x, , 5 2 1 7, , 4, 4, Thus, we have the following table giving points on the line 4 x 2 y 5, x, , x, , 5/4, , 7/4, , y, , 0, , 1, , Graph of the given equations:, , Clearly, the two lines intersect at (i!)., Hence, the system of equations is consistent., 21., , Determine, by drawing graphs, whether the following system of linear equations has a, unique solution or not:, (i) 2x − 3y = 6, x + y = l, (ii) 2y = 4x − 6, 2x = y + 3, Sol:, We have, 2x 3y 6, , x y 1, Now
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , 2x 3y 6, 2x 6 3y, , , , When y 0, we have, , 6 3y, 2, When y 2, we have, x, , x, , 6 3 2 , , 0, 2, Thus, we have the following table giving points on the line 2 x 3 y 6, x, , 3, , 0, , y, , 0, , -2, , Now,, , x y 1, x 1 y, , , , When y 1, we have, , x 1 1 0, When y 0, we have, x 1 0 1, Thus, we have the following table giving points on the line x y 1, x, , 0, , 1, , y, , 1, , 0, , Graph of the given equations:, , We have,
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , 2 y 4x 6, 2x y 3, Now,, , 2 y 4x 6, 2 y 6 4x, 4x 2 y 6, 2y 6, x, 4, , , , , , When y 1, we have, x, , 2 1 6, , 1, 4, When y 5, we have, , 25 6, 4, 4, Thus, we have the following table giving points on the line 2 y 4 x 6, x, , x, , 1, , 4, , y, , -1, , 5, , Now,, 2x y 3, y3, , x, 2, When y 1, we have, , 1 3, 2, 2, When y 3, we have, x, , 33, 3, 2, Thus, we have the following table giving points on the line 2 x y 3, x, , x, , 2, , 3, , y, , 1, , 3, , Graph of the given equations:
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , We find the graphs of the two equations are coincident,, Hence, the system of equations has infinity many solutions, , 22., , Solve graphically each of the following systems of linear equations. Also find the, coordinates of the points where the lines meet axis of y., 2x 5 y 4 0, (i), 2x y 8 0, (ii), , 3x 2 y 12, 5x 2 y 4, , (iii), , 2 x y 11 0, x y 1 0, , (iv), , (v), (vi), , 𝑥 + 2𝑦 − 7 = 0, 2𝑥 − 𝑦 − 4 = 0, 3x y 5 0, , 2x y 5 0, 2x y 5 0, x y 3 0, , Sol:, We have, 2x 5 y 4 0, , 2x y 8 0
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, Now,, , 2x 5 y 4 0, , 2x 5 y 4, 5y 4, , x, 2, When y 2, we have, , 5 2 4, 3, 2, When y 4, we have, x, , 5 4 4, 8, 2, Thus, we have the following table giving points on the line 2 x 5 y 4 0, x, , x, , 3, , 8, , y, , 2, , 4, , Now,, , 2x y 8 0, , 2x 8 y, 8 y, , x, 2, When y 4, we have, , 84, 2, 2, When y 2, we have, x, , 82, 3, 2, Thus, we have the following table giving points on the line 2 x 5 y 4 0, x, , x, , 3, , 8, , y, , 2, , 4, , Graph of the given equations:
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , Clearly, two intersect at P 3, 2 ., Hence, x 2, y 3 is the solution of the given system of equations., We also observe that the lines represented by 2 X 5 y 4 0 and 2x y 8 0 meet y-, , 4, axis at A 0, and B 0,8 respectively., 5, We have,, 3x 2 y 12, , 5x 2 y 4, Now,, , 3x 2 y 12, , 3x 12 2 y, 12 2 y, , x, 3, When y 3, we have, , 12 2 3, 2, 3, When y 3, we have, x, , x, , 12 2 3, , 6, 3, Thus, we have the following table giving points on the line 3x 2 y 12, x, , 2, , 6, , y, , 3, , -3, , Now,
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , 5x 2 y 4, , 5x 4 2 y, 4 2y, , x, 5, When y 3, we have, , 4 23, 2, 5, When y 7, we have, x, , x, , 4 2 7 , , 2, 5, Thus, we have the following table giving points on the line 5 x 2 y 4, x, , 2, , -2, , y, , 3, , -7, , Graph of the given equation, , Clearly, two intersect at p 2,3 ., Hence, x 2, y 3 is the solution of the given system of equations., We also observe that the lines represented by 3x 2 y 12 and 5 x 2 y 4 meet y-axis at, , A 0, 6 and B 0, 2 respectively., We have,, 2 x y 11 0, , x y 1 0, Now,
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , 2 x y 11 0, y 11 2 x, , , , When x 4, we have, y 11 2 4 3, When x 5, we have, y 11 2 5 1, Thus, we have the following table giving points on the line 2 x y 11 0, x, , 4, , 5, , y, , 3, , 1, , Now,, x y 1 0, x 1 y, y x 1, , , , , When x 2, we have, y 2 1 1, When x 3, we have, y 3 1 2, Thus, we have the following table giving points on the line x y 1 0, x, , 2, , 3, , y, , 1, , 2, , Graph of the given equation, We have,, 2 x y 11 0, , x y 1 0, Now,, , 2 x y 11 0, y 11 2 x, , , , When x 4, we have, y 11 2 4 3, When x 5, we have, y 11 2 5 1, Thus, we have the following table giving points on the line 2 x y 11 0, x, , 4, , 5, , y, , 3, , 1, , Now,
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, x y 1 0, x 1 y, y x 1, , , , , When x 2, we have, y 2 1 1, When x 3, we have, y 3 1 2, Thus, we have the following table giving points on the line x y 1 0, x, , 2, , 3, , y, , 1, , 2, , Graph of the given equations:, , Clearly, two intersect at P 4,3 ., Hence, x 4, y 3 is the solution of the given system of equations., We also observe that the lines represented by 2 x y 11 0 and x y 1 0 meet y-axis, at, A 0,11 and B 0, 1 respectively., We have, x 2 y 7 0, Now,, 2x y 4 0, x 2y 7 0, x 7 2y, When, , y 1, x 5, y 2, x 3
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, x, , 5, , 3, , y, , 1, , 2, , Also,, , 2x y 4 0, y 2x 4, , x, , 2, , 0, , y, , 0, , -4, , From the graph, the solution is A 3, 2 ., Also, the coordinates of the points where the lines meet the y-axis are, B 0,3.5 and C 0, 4 ., We have, 3x y 5 0, , 2x y 5 0, Now,, , 3x y 5 0, y 5 3x, , , , When x 1, we have, y 5, 3 1 2, When x 2, we have, y 5, 3 2 1, Thus, we have the following table giving points on the line 3x y 5 0, x, , 1, , 2
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, y, , 2, , -1, , Now,, 2x y 5 0, 2x 5 y, y 2x 5, , , , , When x 0, we have, y 5, When x 2, we have, y 2 2 5 1, Thus, we have the following table giving points on the line 2 x y 5 0, x, , 0, , 2, , y, , -5, , -1, , Graph of the given equations:, , Clearly, two intersect at P 2, 1 ., Hence, x 2, y 1 is the solution of the given system of equations., We also observe that the lines represented by 3x y 5 0 and 2 x y 5 0 meet y-axis, at A 0,5 and 8 0, 5 respectively., We have,, 2x y 5 0, , x y 3 0, Now,, , , , 2x y 5 0, 2x 5 y, y 2x 5
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, When x 1, we have, y 2 1 5 3, When x 2, we have, y 2 2 5 1, Thus, we have the following table giving points on the line 2 x y 5 0, x, , 1, , 2, , y, , -3, , -1, , Now,, x y 3 0, x 3 y, y x3, , , , , When x 3, we have, y 33 0, When x 4, we have, y 4 3 1, Thus, we have the following table giving points on the line x y 3 0, x, , 3, , 4, , y, , 0, , 1, , Graph of the given equations:, , Clearly, two intersect at P 2, 1 ., Hence, x 2, y 1 is the solution of the given system of equations?, We also observe that the lines represented by 2 x y 5 0 and x y 3 0 meet y-axis, at A 0, 5 and 8 0, 3 respectively.
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, 23., , Determine graphically the coordinates of the vertices of a triangle, the equations of whose, sides are:, yx, (i), y 2x, yx6, yx, (ii), 3y x, x y 8, Sol:, The system of the given equations is,, yx, y 2x, yx6, Now,, yx, When x 0, we have, , y0, When x 1, we have, , y 1, Thus, we have the following table:, x, 0, -1, y, , 0, , -2, , We have, y 2x, When x 0, we have, , y 2 0 0, When x 1, we have, y 2 1 2, Thus, we have the following table:, x, , 0, , -1, , y, , 0, , -2, , We have, yx6, , y 6 x, When x 2, we have, , y 62 4
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, When x 4, we have, , y 64 2, Thus, we have the following table:, x, 2, 4, y, , 4, , 2, , Graph of the given system of equations:, , From the graph of the three equations, we find that the three lines taken in pairs intersect, each other at points A 0, 0 , B 2, 4 and C 3,3 ., Hence, the vertices of the required triangle are 0, 0 , 2, 4 and, The system of the given equations is,, yx, 3y x, x y 8, Now,, yx, , , , x y, , When y 0, we have, x0, When y 3, we have, x 3, Thus, we have the following table., x, , 0, , -3, , y, , 0, , -3, , 3,3 .
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, We have, 3y x, , x 3y, When y 0, we have, x 3 0 0, When y 1, we have, , y 3 1 3, Thus, we have the following table:, x, , 0, , -3, , y, , 0, , -1, , We have, x y 8, , x 8 y, When y 4, we have, x 84 4, When y 5, we have, x 85 3, Thus, we have the following table:, x, , 4, , 5, , y, , 4, , 3, , Graph of the given system of equations:
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, From the graph of the three equations, we find that the three lines taken in pairs intersect, each other at points A 0, 0 , B 4, 4 and C 6, 2 ., Hence, the vertices of the required triangle are 0, 0 , 44 and, 24., , 6, 2 ., , Solve the following system of linear equations graphically and shade the region between, the two lines and x-axis:, 2 x 3 y 12, (i), x y 1, (ii), , 3x 2 y 4 0, 2x 3y 7 0, , (iii), , 3x 2 y 11 0, 2 x 3 y 10 0, , Sol:, The system of given equations is, 2 x 3 y 12, x y 1, Now,, , 2 x 3 y 12, , 2 x 12 3 y, 12 3 2, , x, 3, 2, When y 2, we have, , 12 3 2, 3, 2, When y 4, we have, x, , 12 3 4, 0, 2, Thus, we have the following table:, x, 0, 3, x, , y, , 4, , 2, , We have,, x y 1, , x 1 y, When y 0, we have, x 1, When y 1, we have, x 11 2
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, Thus, we have the following table:, x, , 1, , 2, , y, , 0, , 1, , Graph of the given system of equations:, , Clearly, the two lines intersect at P 3, 2 ., Hence, x 3, y 2 is the solution of the given system of equations. The system of the, given equations is,, 3x 2 y 4 0, 2x 3y 7 0, Now,, , 3x 2 y 4 0, , 3x 4 2 y, 4 2y, , x, 3, When y 5, we have, , 4 25, 2, 3, When y 8, we have, x, , 4 28, 4, 3, Thus, we have the following table:, x, -2, -4, x, , y, , We have,, , 5, , 8
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , 2x 3y 7 0, , 2x 3y 7, 3y 7, , x, 2, When y 1, we have, , 3 1 7, 5, 2, When y 1, we have, x, , x, , 3 1 7, , 2, 2, Thus, we have the following table:, x, 5, 2, y, , 1, , -1, , Graph of the given system of equations:, , Clearly, the two lines intersect at P 2, 1 ., Hence, x 2, y 1 is the solution of the given system of equations., The system of the given equations is,, 3x 2 y 11 0, 2 x 3 y 10 0, Now,, , 3x 2 y 11 0, , 3x 11 2 y, 11 2 y, , x, 3, When y 1, we have
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , 11 2 1, 3, 3, When y 4, we have, x, , 11 2 4, 1, 3, Thus, we have the following table:, x, 3, 1, x, , y, , 1, , 4, , We have,, 2 x 3 y 10 0, , 2 x 3 y 10, 3 y 10, , x, 2, When y 0, we have, , 3 0 10, 5, 2, When y 2, we have, x, , 3 2 10, 2, 2, Thus, we have the following table:, x, -5, -2, x, , y, , 0, , 2, , Graph of the given system of equations:, , Clearly, the two lines intersect at P 1, 4 ., Hence, x 1, y 4 is the solution of the given system of equations
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, 25., , Draw the graphs of the following equations on the same graph paper:, 2 x 3 y 12, x y 1, Sol:, The system of the given equations is, 2 x 3 y 12, x y 1, Now,, , 2 x 3 y 12, , 2 x 12 3 y, 12 3 y, , x, 2, When y 0, we have, , 12 3 0, 6, 2, When y 2, we have, x, , 12 3 2, 3, 2, Thus, we have the following table:, x, 6, 3, x, , y, , 0, , 2, , We have, x y 1, , x 1 y, When y 0, we have, x 1, When y 1, we have, x 1 1 0, Thus, we have the following table:, x, , 1, , 0, , y, , 0, , -1, , Graph of the given system of equations:
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , Clearly, the two lines intersect at A 3, 2 ., We also observe that the lines represented by the equations 2 x 3 y 12 and x y 1, meet y-axis at B 0, 1 and C 0, 4 ., Hence, the vertices of the required triangle are A 3, 2 , B 0, 1 and C 0, 4 ., 26., , Draw the graphs of x − y + 1 = 0 and 3x + 2y − 12 = 0. Determine the coordinates of the, vertices of the triangle formed by these lines and x- axis and shade the triangular area., Calculate the area bounded by these lines and x-axis., Sol:, The given system of equations is, x y 1 0, 3x 2 y 12 0, Now,, x y 1 0, x y 1, , , , When y 3, we have, x 3 1 2, When y 1, we have, x 1 1 2, Thus, we have the following table:, x, , 2, , -2, , y, , 3, , -1, , We have
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , 3x 2 y 12 0, , 3x 12 2 y, 12 2 y, , x, 3, When y 6, we have, , 12 2 6, 0, 3, When y 3, we have, x, , 12 2 3, 2, 3, Thus, we have the following table:, x, 0, 2, x, , y, , 6, , 3, , Graph of the given system of equations:, , Clearly, the two lines intersect at A 2,3 ., We also observe that the lines represented by the equations, , x y 1 0 and 3x 2 y 12 0 meet x-axis at B 1, 0 and C 4, 0 respectively., Thus, x 2, y 3 is the solution of the given system of equations., Draw AD perpendicular from A on x-axis., Clearly, we have, AD y coordinate of point A 2,3, , , AD 3 and, BC 4 1 4 1 5
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, 27., , Solve graphically the system of linear equations:, 4x 3y 4 0, 4 x 3 y 20 0, Find the area bounded by these lines and x-axis., Sol:, The given system of equation is, 4x 3y 4 0, 4 x 3 y 20 0, Now,, , 4x 3y 4 0, , 4x 3y 4, 3y 4, , x, 4, When y 0, we have, , 3 0 4, 1, 4, When y 4, we have, x, , x, , 3 4 4, 2, 4, , Thus, we have the following table:, x, , 2, , -1, , y, , 4, , 0, , We have, 4 x 3 y 20 0, , 4 x 20 3 y, 20 3 y, , x, 4, When y 0, we have, , 20 3 0, 5, 4, When y 4, we have, x, , 20 3 4, 2, 4, Thus, we have the following table:, x, 5, 2, x, , y, , 0, , 4
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, Graph of the given system of equation:, , Clearly, the two lines intersect at A 2, 4 . Hence x 2, y 4 is the solution of the given, system of equations., We also observe that the lines represented by the equations, , 4 x 3 y 4 0 and 4 x 3 y 20 0 meet x-axis at B 1, 0 and C 5, 0 respectively., Thus, x 2, y 4 is the solution of the given system of equations., Draw AD perpendicular from A on x-axis., Clearly, we have, AD y coordinate of point A 2, 4 , , , AD 4 and, BC 5 1 5 1 6, , Area of the shaded region = Area of ABC, 1, Area of the shaded region Base Height , 2, 1, BC AD , 2, 1, 6 4, 2, 6 2, 12 sq. units, Area of shaded region 12 sq. units
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, 28., , Solve the following system of linear equations graphically:, 3x y 11 0, x y 1 0, Shade the region bounded by these lines and y -axis. Also, find the area of the region, bounded by these lines and y-axis., Sol:, The given system of equation is, 3x y 11 0, x y 1 0, Now,, 3x y 11 0, y 11 3x, , , , When x 0, we have, , y 11 3 0 11, When x 3 we have, y 11 3 3 2, Thus, we have the following table:, x, 0, 3, y, , 11, , 2, , We have, x y 1 0, , x 1 y, , y x 1, When x 0, we have, , y 0 1 1, When x 3, we have, y 3 1 2, Thus, we have the following table:, x, 0, 3, y, , -1, , 2, , Graph of the given system of equations:
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , Clearly, the two lines intersect at A 3, 2 . Hence x 3, y 2 is the solution of the given, system of equations., We so observe that the lines represented by the equations 3x y 11 0 and x y 1 0, meet y-axis at B 0,11 and C 0, 1 respectively., Thus, x 3, y 2 is the solution of the given system of equations., Draw AD perpendicular from A on y-axis., Clearly, we have, AD x coordinate of point A 3, 2 , , , AD 3 and, BC 11 1 11 1 12, , Area of the shaded region = Area of ABC, 1, Area of the shaded region Base Height , 2, 1, BC AD , 2, 1, 12 3, 2, 63, 18 sq. units, Area of the shaded region 18 sq. units, , 29., , Solve graphically each of the following systems of linear equations. Also, find the, coordinates of the points where the lines meet the axis of x in each system:, 2x y 2, (i), 4x y 8
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, (ii), , 2x y 2, 4x y 8, , (iii), , x 2y 5, 2 x 3 y 4, , (iv), , 2x 3 y 8, x 2 y 3, , Sol:, The given system of equation is, 2x y 2, 4x y 8, Now,, , 2x y 2, , 2x y 2, y2, , x, 2, When y 0, we have, , 02, 1, 2, When y 2, we have, x, , 22, 2, 2, Thus, we have the following table:, x, 1, 2, x, , y, , 0, , 2, , We have,, 4x y 8, , 4x y 8, y 8, , x, 4, When y 0, we have, , 08, 2, 4, When y 4 we have, x, , 4 8, 1, 4, Thus, we have the following table:, x
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, x, , 2, , 1, , y, , 0, , -4, , Graph of the given system of equations:, , Clearly, the two lines intersect at A 2, 2 . Hence x 2, y 2 is the solution of the given, system of equations., We so observe that the lines represented by the equations 2 x y 6 and x 2 y 2 meet, x-axis at B 3, 0 and C 2, 0 respectively., The system of the given equations is, 2x y 6, x 2 y 2, Now,, , 2x y 6, 6 y, , x, 2, When y 0, we have, , 60, 3, 2, When y 2, we have, x, , 62, 2, 2, Thus, we have the following table:, x, 3, 2, x, , y, , We have,, , 0, , 2
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, x 2 y 2, y 2y 2, , , , When y 0, we have, x 2 0 2 2, When y 1, we have, x 2 1 2 0, Thus, we have the following table:, x, , -2, , 0, , y, , 0, , 1, , Graph of the given system of equations:, , Clearly the two lines intersect at A 3, 4 . Hence x 3, y 4 is the solution of the given, system of equations., We so observe that the lines represented by the equations 2 x y 2 and 4 x y 8 meet, x-axis at B 1, 0 and C 2, 0 respectively, The system of the given equations is, x 2y 5, 2 x 3 y 4, Now,, , , x 2y 5, x 5 2y, , When y 2, we have, x 5 2 2 1, When y 3, we have, , x 5 2 3 1
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, Thus, we have the following table:, x, , 1, , -1, , y, , 2, , 3, , We have,, 2 x 3 y 4, , 2x 3y 4, 3y 4, , x, 2, When y 0, we have, , 3 0 4, 2, 2, When y 2, we have, x, , 3 2 4, 1, 2, Thus, we have the following table:, x, -2, 1, x, , y, , 0, , 2, , Graph of the given system of equations:
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , The given system of equation is, 2x 3 y 8, x 2 y 3, Now,, , 2x 3y 8, , 2x 8 3y, 8 3y, , x, 2, When y 2, we have, , 8 3 4, 1, 2, When y 4, we have, x, , 8 3 4, 2, 2, Thus, we have the following table:, x, 1, -2, x, , y, , 2, , 4, , We have,, x 2 y 3, , x 2y 3, When y 0, we have, x 2 0 3 3, When y 1, we have, x 2 1 3 1, Thus, we have the following table:
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, x, , -3, , -1, , y, , 0, , 1, , Graph of the given system of equations:, , Clearly, the two lines intersect at A 1, 2 . Hence x 1, y 2 is the solution of the given, system of equations., We also observe that the lines represented by the equations 2 x 3 y 8 and x 2 y 3, meet x-axis at B 4, 0 and C 3, 0 respectively., 30., , Draw the graphs of the following equations:, 2x 3y 6 0, 2 x 3 y 18 0, y2 0, Find the vertices of the triangle so obtained. Also, find the area of the triangle., Sol:, The given system of equation is, 2x 3y 6 0, 2 x 3 y 18 0, y2 0, Now,, , 2x 3y 6 0, , 2x 3y 6, 3y 6, , x, 2, When y 0, we have, , x, , 3 0 6, 3, 2
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, When y 2, we have, , 3 2 6, 0, 2, Thus, we have the following table:, x, -3, 0, x, , y, , 0, , 2, , We have,, 2 x 3 y 18 0, , 2 x 18 3 y, 18 3 y, , x, 2, When y 2, we have, , 18 3 2, 6, 2, When y 6, we have, x, , 18 3 6, 0, 2, Thus, we have the following table:, x, 6, 0, x, , y, , 2, , 6, , We have, y20, , y 2, Graph of the given system of equations:
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , From the graph of the three equations, we find that the three lines taken in pairs intersect, each other at points A 3, 4 , B 0, 2 and C 6, 2 ., Hence, the vertices of the required triangle are 3, 4 , 0, 2 and, , 6, 2 ., , From graph, we have, AD 4 2 2, , BC 6 0 6, Area of ABC , , 1, Base Height , 2, , 1, BC AD, 2, 1, 6 2, 2, 6 sq. units, Area of ABC 6 sq.units, 31., , Solve the following system of equations graphically:, 2x 3y 6 0, 2 x 3 y 18 0, Also, find the area of the region bounded by these two lines and y-axis., Sol:, The given system of equation is, 2x 3y 6 0, 2 x 3 y 18 0, Now,
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , 2x 3y 6 0, , 2x 6 3y, , 3y 2x 6, 2x 6, , y, 3, When x 0, we have, 20 6, 2, 3, When x 3, we have, y, , y, , 2 3 6, , 0, 3, Thus, we have the following table:, x, 0, -3, y, , 2, , 6, , Graph of the given system of equations:, , Clearly, the two lines intersect at A 3, 4 . Hence, x 3, y 4 is the solution of the given, system of equations., We also observe that the lines represented by the equations, 2 x 3 y 6 0 and 2 x 3 y 18 0 meet y-axis at B 0, 2 and C 0, 6 respectively., Thus, x 3, y 4 is the solution of the given system of equations., Draw AD perpendicular from A on y-axis., Clearly, we have,, AD x coordinate of point A 3, 4 , , , AD 3 and, BC 6 2 4, Area of the shaded region = Area of ABC
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, Area of the shaded region , , 1, Base Height , 2, , 1, BC AD , 2, 1, 43, 2, 23, 6 sq. units, Area of the region bounded by these two lines and y-axis is 6 sq. units., , , 32., , Solve the following system of linear equations graphically:, 4 x 5 y 20 0, 3x 5 y 15 0, Determine the vertices of the triangle formed by the lines representing the above equation, and the y-axis., Sol:, The given system of equation is, 4 x 5 y 20 0, 3x 5 y 15 0, Now,, , 4 x 5 y 20 0, , 4 x 5 y 20, 5 y 20, , x, 5, 4, When y 0, we have, , 5 0 20, 5, 4, When y 4, we have, x, , x, , 5 4 20, , 0, 4, Thus, we have the following table:, x, 5, 0, y, , 0, , -4, , We have,, 3x 5 y 15 0, , 3x 15 5 y, 15 5 y, , x, 3
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, When y 0, we have, , 15 5 3, 0, 3, When y 3, we have, x, , 15 5 3, 0, 3, Thus, we have the following table:, x, , x, , 5, , 0, , y, , 0, , 3, , Graph of the given system of equations:, , Clearly, the two lines intersect at 4 5, 0 . Hence, x 5, y 0 is the solution of the given, system of equations., We also find that the two lines represented by the equations, 4 x 5 y 20 0 and 3x 5 y 15 0 meet y-axis at B 0, 4 and C 0,3 respectively,, The vertices of the required triangle are 5, 0 ,, , 33., , 0, 4 , , and, , 0,3 ., , Draw the graphs of the equations 5x − y = 5 and 3x − y = 3. Determine the co-ordinates of, the vertices of the triangle formed by these lines and y-axis. Calculate the area of the, triangle so formed., Sol:, 5x y 5 y 5x 5, Three solutions of this equation can be written in a table as follows:, x, , 0, , 1, , 2, , y, , -5, , 0, , 5
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, 3x y 3 y 3x 3, x, , 0, , 1, , 2, , y, , -3, , 0, , 3, , The graphical representation of the two lines will be as follows:, , It can be observed that the required triangle is ABC., The coordinates of its vertices are A 1, 0 , B 0, 3 , C 0, 5 ., Concept insight: In order to find the coordinates of the vertices of the triangle so formed., Find the points where the two lines intersects the y-axis and also where the two lines, intersect each other. Here, note that the coordinates of the intersection of lines with y-axis, is taken and not with x-axis, this is became the question says to find the triangle formed by, the two lines and the y-axis., 34., , Form the pair of linear equations in the following problems, and find their solution, graphically:, (i) 10 students of class X took part in Mathematics quiz. If the number of girls is 4 more, than the number of boys, find the number of boys and girls who took part in the quiz., (ii) 5 pencils and 7 pens together cost Rs 50, whereas 7 pencils and 5 pens together cost Rs, 46. Find the cost of one pencil and a pen., (iii)Champa went to a ‘sale’ to purchase some pants and skirts. When her friends asked her, how many of each she had bought, she answered, “The number of skirts is two less than, twice the number of pants purchased. Also, the number of skirts is four less than four, times the number of pants purchased.” Help her friends to find how many pants and, skirts Champa bought., Sol:, (i) Let the number of girls and boys in the class be x and y respectively., According to the given conditions, we have:
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, x y 10, x y 4, x y 10 x 10 y, , Three solutions of this equation can be written in a table as follows:, x, , 4, , 5, , 6, , y, , 6, , 5, , 4, , x y 4 x 4 y, , Three solutions of this equation can be written in a table as follows:, x, , 5, , 4, , 3, , y, , 1, , 0, , -1, , The graphical representation is as follows:, , From the graph, it can be observed that the two lines intersect each other at the point 7, 3 ., So. x 7 and y 3., Thus, the number of girls and boys in the class are 7 and 3 respectively., (ii) Let the cost of one pencil and one pen be Rs x and Rs y respectively., According to the given conditions, we have:, 5 x 7 y 50, 7 x 5 y 46, 50 7 y, 5 x 7 y 50 x , 5, Three solutions of this equation can be written in a table as follows:
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, x, , 3, , 10, , -4, , y, , 5, , 0, , 10, , 46 5 y, 7, Three solutions of this equation can be written in a table as follows:, 7 x 5 y 46 x , , x, , 8, , 3, , -2, , y, , -2, , 5, , 12, , The graphical representation is as follows:, , From the graph. It can be observed that the two lines intersect each other at the point 3, 5 ., So. x 3 and y 5., Therefore, the cost of one pencil and one pen are Rs 3 and Rs 5 respectively., (iii) Let us denote the number of pants by x and the number of skirts by y. Then the, equations formed are:, y 2 x ? 2...... 1 and y 4 x ? 4...... 2 , Let us draw the graphs of Equations (1) and (2) by finding two solutions for each of the, equations., They are given in Table, They are giving table
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, x, , 2, , 0, , y 2 x ?2, , 2, , -2, , x, , 0, , 1, , y 2 x ?2, , -4, , 0, , Plot the point and draw the lines passing through them to represent the equation, as shown, in fig., The t lines intersect at the point 1, 0 . So. x 1, y 0 is the required solution of the pair of, linear equations, i.e, the number of pants she purchased island she did not buy any skirt, Concept insight: Read the question carefully and examine what are the unknowns., Represent the given conditions with the help of equations by taking the unknowns, quantities as variables. Also carefully state the variables as whole solution is based on it on, the graph paper, mark the points accurately and neatly using a sharp pencil. Also take at, least three points satisfying the two equations in order to obtain the correct straight line of, the equation. Since joining any two points gives a straight line and if one of the points is, computed incorrect will give a wrong line and taking third point will give a correct line., The point where the two straight lines will intersect will give the values of the two, variables, i.e., the solution of the two linear equations. State the solution point., 35., , Solve the following system of equations graphically:, Shade the region between the lines and the y-axis, 3x 4 y 7, (i), 5x 2 y 3, (ii), , 4x y 4, 3x 2 y 14, , Sol:, The given system of equations is, 3x 4 y 7, 5x 2 y 3
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, Now,, , 3x 4 y 7, , 3x 7 4 y, , 4 y 3x 7, 3x 7, , y, 4, When x 1, we have, 3 1 7, 1, 4, When x 3, we have, y, , y, , 3 3 7, , 4, 4, Thus, we have the following table:, x, 1, -3, y, , -1, , -4, , We have,, 5x 2 y 3, , 2 y 3 5x, 3 5x, , y, 2, When x 1, we have, , 3 5 1, 1, 2, When x 3, we have, y, , 3 5 3, 6, 2, Thus, we have the following table:, y, , x, , 1, , 3, , y, , -1, , -6, , Graph of the given system of equations:
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , Clearly, the two lines intersect at A 1, 1 Hence, x 1, y 1 is the solution of the given, system of equations., We also observe that the required shaded region is ABC, The given system of equations is, 4x y 4, 3x 2 y 14, Now,, 4x y 4, 4x 4 y, y 4x 4, , , , , When x 0, we have, y 4 0 4 4, When x 1, we have, , y 4 1 4 8, Thus, we have the following table:, x, , 0, , -1, , y, , -4, , -8, , We have,, 3x 2 y 14, , 2 y 14 3x, 14 3x, , y, 2
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, When x 0, we have, , 14 3 0, 7, 2, When x 0, we have, y, , 14 3 4, 1, 2, Thus, we have the following table:, y, , x, , 0, , 4, , y, , 7, , 1, , Graph of the given system of equations:, , Clearly, the two lines intersect at A 2, 4 . Hence, x 2, y 4 is the solution of the given, system of equations., We also observe ABC is the required shaded region., 36., , Represent the following pair of equations graphically and write the coordinates of points, where the lines intersects y-axis, x 3y 6, 2 x 3 y 12, Sol:, The given system of equations is, x 3y 6, 2 x 3 y 12, Now,
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , x 3y 6, , 3y 6 x, 6 x, , y, 3, When x 0, we have, , 60, 2, 3, When x 3, we have, y, , 63, 1, 3, Thus, we have the following table:, y, , x, , 0, , 3, , y, , 2, , 1, , We have,, 2 x 3 y 12, , , , , 2 x 12 3x, 3 y 2 x 12, 2 x 12, , y, 3, When x 0, we have, 2 0 12, 4, 3, When x 6, we have, y, , 2 6 12, 0, 3, Thus, we have the following table:, y, , x, , 0, , 6, , y, , -4, , 0, , Graph of the given system of equations:
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , We observe that the lines represented by the equations x 3 y 6 and 2 x 3 y 12 meet yaxis at B 0, 2 and C 0, 4 respectively., Hence, the required co-ordinates are 0, 2 and, 37., , 0, 4 ., , Given the linear equation 2x + 3y − 8 = 0, write another linear equation in two variables, such that the geometrical representation of the pair so formed is (i) intersecting lines (ii), Parallel lines (iii) coincident lines, Sol:, (i) For the two lines a1 x b1 x c1 0 and a2 x b2 x c2 0, to be intersecting, we must have, a1 b1, , a2 b2, , So, the other linear equation can be 5 x 6 y 16 0, As, , a1 2 b1 3 1 c1, 8 1, , , , , a2 5 b2 6 2 c2 16 2, , (ii) For the two lines a1 x b1 x c1 0 and a2 x b2 x c2 0, to be parallel we must have, a1 b1 c1, , a2 b2 c2, , So, the other linear equation can be 6 x 9 y 24 0,, As, , a1 2 1 b1 3 1 c1 8 1, , , , , a2 6 3 b2 9 3 c2 24 3
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, (iii) For the two lines a1 x b1 x c1 0 and a2 x b2 x c2 0, to be coincident, we must, have, , a1 b1 c1, , a2 b2 c2, , So, the other linear equation can be 6 x 9 y 24 0,, As, , a1 2 1 b1 3 1 c1, 8 1, , , , , , a2 8 4 b2 12 4 c2 32 4, , Concept insight: In orders to answer such type of problems, just remember the conditions, for two lines to be intersecting parallel, and coincident, This problem will have multiple answers as their can be marry equations satisfying the, required conditions., , Exercise 3.3, Solve the following systems of equations:, 1. 11x + 15y + 23 = 0, 7x – 2y – 20 = 0, Sol:, The given system of equation is, 11x 15 y 23 0, ... i , , 7 x 2 y 20 0, , ... ii , , From (ii), we get, 2 y 7 x 20, 7 x 20, , y, 2, 7 x 20, Substituting y , in (i) we get, 2, 7 x 20 , 11x 15 , 23 0, 2 , 105 x 300, , 11x , 23 0, 2, 22 x 105 x 300 46, , 0, 2, , 127 x 254 0, , 127 x 254, 254, , x, 2, 127
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , 7 x 20, we get, 2, 7 2 20, , y, 2, 14 20, , 2, 6, , 2, 3, Hence, the solution of the given system of equations is x 2, y 3., Putting x 2 in y , , 2., , 3x – 7y + 10 = 0, y – 2x – 3 = 0, Sol:, The given system of equation is, 3x 7 y 10 0, ... i , , y 2x 3 0, , ... ii , , From (ii), we get, y 2x 3, Substituting y 2 x 3 in (i) we get, , 3x 7 2 x 3 10 0, , , , 3x 14 x 21 10 0, 11x 11, 11, , x, 1, 11, Putting x 1 in y 2 x 3, we get, , , y 2 1 3, , , , 2 3, 1, y 1, , Hence, the solution of the given system of equations is x 1, y 1., 3., , 0.4x + 0.3y = 1.7, 0.7x + 0.2y = 0.8, Sol:
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, The given system of equation is, 0.4 x 0.3 y 1.7, ... i , , 0.7 x 0.2 y 0.8, , ... ii , , Multiplying both sides of (i) and (ii), by 10, we get, 4 x 3 y 17, ... iii , , 7x 2 y 8, , ... iv , , From (iv), we get, 7x 8 2 y, 8 2y, , 7x, 7, 8 2y, Substituting x , in (iii), we get, 7, 8 2y , 4, 3 y 17, 7 , 32 8 y, , 3 y 17, 7, , 32 29 y 17 7, , , , , 29 y 119 32, 29 y 87, 87, , y, 3, 29, 8 2y, Putting y 3 in x , , we get, 7, 8 23, x, 7, 86, , 7, 14, , 7, 2, Hence, the solution of the given system of equation is x 2, y 3., , 4., , x, y 0.8, 2, Sol:, x, y 0.8, 2
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, And, , 7, y, x, 2, , 10, , 7 2, 10, 2x y, x 2 y 1.6 and 7 10 x 5 y, , x 2 y 1.6 and, , Multiply first equation by 10, 10 x 20 y 16 and 10 x 5 y 7, Subtracting the two equations, 15 y 9, , 9 3, , 15 5, 6 2, 3, x 1.6 2 1.6 , 5 5, 5, 2 3, Solution is , , 5 5, y, , 5., , 7(y + 3) – 2 (x + 3) = 14, 4(y – 2) + 3 (x – 3) = 2, Sol:, The given system of equations id, 7 y 3 2 x 3 14, ... i , , 4 y 2 3 x 3 2, , ... ii , , From (i), we get, 7 x 21 2 x 4 14, , , , 7 y 14 4 21 2 x, 2x 3, , y, 7, From (ii), we get, 4 y 8 3x 9 2, , 4 y 3x 17 2 0, , , 4 y 3x 19 0, , Substituting y , , ... iii , , 2x 3, in (iii), we get, 7
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , 2x 3 , 4, 3x 19 0, 7 , 8 x 12, , 3x 19 0, 7, , 8 x 12 21x 133 0, , 29 x 145 0, , 29 x 145, 145, , x, 5, 29, 2x 3, Putting x 5 in y , , we get, 7, 25 7, y, 7, 10 3, , 7, 7, , 7, 1, , y 1, Hence, the solution of the given system of equations is x 5, y 1., 6., , 𝑥, 7, 𝑥, , 𝑦, , +3=5, 𝑦, , −9=6, 2, , Sol:, The given system of equation is, x y, 5, ... i , 7 3, x y, 6, ... ii , 2 9, From (i), we get, 3x 7 y, 5, 21, , 3x 7 y 105, , , , 3x 105 7 y, 105 7 y, , x, 3, From (ii), we get
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , 9x 2 y, 6, 18, 9 x 2 y 108, , , , ... iii , , 105 7 y, in (iii), we get, 3, 105 7 y , 9, 2 y 108, 3, , , 948 63 y, , 2 y 108, 3, , 945 63 y 6 y 108 3, , 945 69 y 324, , 945 324 69 y, , 69 y 621, 621, , y, 9, 69, 1105 7 y, Putting y 9 in x , , we get, 3, 105 7 9 105 63, x, , 3, 3, 42, , x, 14, 3, Hence, the solution of thee given system of equations is x 14, y 9., Substituting x , , 7., , 𝑥, , 6, , 𝑦, , + 4 = 11, , 3, 5𝑥, , 𝑦, , −3=7, , Sol:, The given system of equations is, x y, 11, ... i , 3 4, 5x y, 7, ... ii , 6 3, From (i), we get, 4x 3y, 11, 12, , 4 x 3 y 132, ... iii , From (ii), we get
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , , , 5x 2 y, 7, 6, 5 x 2 y 42, , ... iv , , Let us eliminate y from the given equations. The coefficients of y in the equations(iii) and, (iv) are 3 and 2 respectively. The L.C.M of 3 and 2 is 6. So, we make the coefficient of y, equal to 6 in the two equations., Multiplying (iii) by 2 and (iv) by 3, we get, 8 x 6 y 264, ... v , , 15 x 6 x 126, , ... vi , , Adding (v) and (vi), we get, 8 x 15 x 264 126, , , , 23x 138, 138, , x, 6, 23, Substituting x 6 in (iii), we get, 4 6 3 y 132, , 3 y 132 24, , 3 y 108, 108, , y, 36, 3, Hence, the solution of the given system of equations is x 6, y 36., 4u 3 y 8, 6u 4 y 5, , 8., Sol:, Taking, , 1, u, then given equations become, x, 4u 3 y 8, ... i , 6u 4 y 5, , ... ii , , From (i), we get, 4u 8 3 y, 8 3y, , u, 4, 8 3y, Substituting u , in (ii), we get, 4, From (ii), we get
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , 8 3y , 6, 4 y 5, 4 , 3 8 3 y , , 4 y 5, 2, 24 9 y, , 4 y 5, 2, 24 9 y 8 y, , 5, 2, , 24 17 y 10, , 17 y 10 24, , 17 y 34, 34, , y, 2, 17, 8 3y, Putting y 2, in u , , we get, 4, 8 3 2 8 6 2 1, u, , , 4, 4, 4 2, 1, Hence, x 2, u, So, the solution of the given system of equation is x 2, y 2., , 9., , x, , y, 4, 2, , x, 2y 5, 3, Sol:, The given system of equation is, y, ..(i), x 4, 2, x, 2y 5, ..(ii), 3, From (i), we get, 2x y, 4, 2, 2x y 8, , y 8 2x, From (ii), we get, x 6 y 15, , ..(iii)
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, Substituting y 8 2 x in (iii), we get, , x 6 8 2 x 15, , , , , x 48 12 x 15, 11x 15 48, 11x 33, 33, , x, 3, 11, Putting x 3, in y 8 2 x, we get, , , , y 8 23, 86, 2, y2, , Hence, solution of the given system of equation is x 3, y 2., , 10., , 3, 2, 3, 2x y , 2, x 2y , , Sol:, The given system of equation is, 3, ..(i), x 2y , 2, 3, ..(ii), 2x y , 2, Let us eliminate y from the given equations. The Coefficients of y in the given equations, are 2 and 1 respectively. The L.C.M of 2 and 1 is 2. So, we make the coefficient of y equal, to 2 in the two equations., Multiplying (i) by 1 and (ii) by 2, we get, 3, x 2y , ..(iii), 2, 4x 2 y 3, ..(iv), Subtracting (iii) from (iv), we get
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , 4x x 2 y 2 y 3 , , 3, 2, , 63, 2, 3, , 3x , 2, 3, , x, 23, 1, , x, 2, 1, Putting x , in equation (iv), we get, 2, 1, 4 2y 3, 2, , 2 2y 3, , , 3x , , , , 2y 3 2, 1, y, 2, , , , 1, 1, Hence, solution of the given system of equation is x , y ., 2, 2, , 11., , 2x 3y 0, 3x 8 y 0, Sol:, , 2x 3y 0, , ... i , , 3x 8 y 0, , ... ii , , From equation (i), we obtain:, 3y, ... iii , 2, Substituting this value in equation (ii), we obtain:, x, , , 3y , 3 , 8 y 0, 2, , , 3y, , 2 2y 0, 2, 3, , y, 2 2 0, 2, , , y0
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, Substituting the value of y in equation (iii), we obtain:, x0, x 0, y 0, y7, 2 10, 11, x 11, 2y , 10, 7, Sol:, The given systems of equation is, y7, 3x , 2 10, ... i , 11, x 11, 2y , 10, ... ii , 7, From (i), we get, 33x y 7 22, 10, 11, , 33x y 15 10 11, 3x , , 12., , , 33x 15 110 y, , y 33x 95, From (ii) we get, 14 y x 11, 109, 7, , 14 y x 11 10 7, , 14 y x 11 70, , 14 y x 70 11, , , 14 y x 59, , ... iii , , Substituting y 33x 95 in (iii), we get, , 14 33x 95 x 59, , , , , 462 x 1330 x 59, 463x 59 1330, 463x 1389, 1389, , x, 3, 463, Putting x 3, in y 33x 95, we get
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , , , , y 33 3 95, y 99 95, 4, y4, , Hence, solution of the given system of equation is x 3, y 4., , 3, 9, y, 7, 3 x 2, y 0, y, 2x , , 13., , Sol:, The given systems of equation is, 3, 2x 9, ... i , y, 7, 3x 2, y 0, ... ii , y, Taking, , 1, u, the given equations becomes, y, , 2 x 3u 9, , ... iii , , 3x 7u 2, , ... iv , , From (iii), we get, 2 x 9 3u, 9 3u, , x, 2, 9 3u, Substituting x , in (iv), we get, 2, 9 3u , 3, 7u 2, 2 , 27 9u 14u, , 2, 2, , 27 23u 2 2, , 23u 4 27, 23, , u, 1, 23, 1 1, 1, Hence, y , u 1
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , 9 3u, , we get, 2, 9 3 1 9 3 6, x, , 3, 2, 2, 2, , x3, Hence, solution of the given system of equation is x 3, y 1., Putting u 1 in x , , 14., , 0.5 x 0.7 y 0.74, 0.3x 0.5 y 0.5, , Sol:, The given systems of equations is, 0.5 x 0.7 y 0.74, i , , 0.3x 0.5 y 0.5, , ii , , Multiplying (i) and (ii) by 100, we get, 50 x 70 y 74, ... iii , , 30 x 50 y 50, , ... iv , , From (iii), we get, 50 x 74 70 y, 74 70 y, , x, 50, 74 70 y, Substituting x , in equation (iv), we get, 50, 74 70 y , 30 , 50 y 50, 50 , , , , 3 74 70 y , , 50 y 50, 5, 222 210 y, , 50 y 50, 5, , 222 210 y 250 y 250, , 40 y 250 222, , 40 y 28, 28 14 7, , y, , , 0.7, 40 20 10, 74 70 y, , we get, Putting y 0.7 in x , 50
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , 74 70 0 7, 50, 74 49, , 50, 25, , 50, 1, , 2, 0.5, Hence, solution of the given system of equation is x 0 5, y 0.7, x, , 15., , 1, 1, , 3, 7x 6 y, 1, 1, , 5, 2x 3y, Sol:, 1, 1, , 3, 7x 6 y, 1, 1, , 5, 2x 3y, Multiplying (2) by, , .... 1, .... 2 , , 1, , we get, 2, , 1, 1 5, , , .... 3, 4x 6 y 2, Solving (1) and (3), we get, 1, 1, , 3, 7x 6 y, 1, 1 5, , , 4x 6 y 2, 1, 1, 5, , 3, 7x 4x, 2, 47 65, , , 28x, 2, 11 11, , , 28 x 2, 11 2, 1, x, , 28 11 14, , (Adding the equations)
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , 1, , we get, 14, 1, 1, , 3, 1 6y, 7 , 14 , 1, 2, 3, 6y, 1, , 3 2 1, 6y, 1, y, 6, , When x , , (Using (1)), , Thus, the solution of given equation is x , , 16., , 1, 2𝑥, 1, , 1, , and y ., 14, 6, , 1, , + 3𝑦 = 2, 1, , + 2𝑦 =, 3𝑥, , 13, 6, , Sol:, Let, , 1, 1, u and v, the given equations become, x, y, , , , , u v, 2, 2 3, 3u 2v, 2, 6, 3u 2v 12, , .... i , , u v 13, , 3 2 6, 2u 3v 13, , , 6, 6, 6, v 3, 2, 1 1, 1 1, Hence, x and y , u 2, v 3, And,, , 1, 1, So, the solution of the given system o equation is x , y ., 2, 3
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , x y, 2, xy, x y, 6, xy, , 17., , Sol:, The given system of equation is, x y, 2, xy, x, y, , , 2, xy xy, 1 1, , 2, ..... i , y x, , x y, 6, xy, x, y, 6, xy xy, 1 y, 6, y x, , And,, , Taking, , ... ii , , 1, 1, v and u, the above equations become, y, x, , vu 2, , ..... iii , , v u 6, , ...... iv , , Adding equation (iii) and equation (iv), we get, v u v u 26, , 2v 8, 8, , v 4, 2, Putting v 4 in equation (iii), we get, 4u 2, , , , u 2 4 2, 1 1 1, 1 1, Hence, x , , and y , u 2 2, v 4, So, the solution of the given system of equation is x , , 1, 1, ,y, 2, 4
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, 18., , 15, 𝑢, , 2, , + 𝑣 = 17, , Sol:, , 1, 1, x and y, then, the given system of equations become, u, v, 15 x 2 y 17, ... i , , Let, , 36, ... ii , 5, From (i), we get, 2 y 17 15 x, x y , , 17 15 x, 2, 17 15x, Substituting y , in equation (ii), we get, 2, 17 15 x 36, x, , 2, 5, 2 x 17 15 x 36, , , 2, 5, 13x 17 36, , , 2, 5, , 5 13x 17 36 2, , , y, , , , , , 65 x 85 72, 65 x 72 85, 65 x 13, 13 1, , 65x , , 65 5, 1, Putting x in equation (ii), we get, 5, 1, 36, y, 5, 5, 36 1, , y, , 5 5, 36 1 35, , , 7, 5, 5, 1, 1 1, Hence, u 5 and v ., x, y 7, , 1, So, the solution off the given system of equation is u 5, v ., 7
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, 19., , 3, , 1, , 2, , 3, , − 𝑦 = −9, 𝑥, +𝑦 =5, 𝑥, Sol:, , 1, 1, u and, v, Then, the given system of equations becomes, x, y, , Let, , 3u v 9, , ...... i , , 2u 3v 5, , ...... ii , , Multiplying equation (i) by 3 an equation (ii) by 1, we get, 9u 3v 27, ......... iii , , 2u 3v 5, , .......... iv , , Adding equation (i) and equation (ii), we get, 9u 2u 3v 3v 27 5, , 11u 22, 22, , u, 2, 11, Putting u 2 in equation (iv), we get, 2 2 3v 5, , , , , 4 3v 5, 3v 5 4, 9, , v 3, 3, 1 1 1, 1 1, Hence, x , and y ., u 2 2, v 3, So, the solution of the given system of equation is x , , 20., , 2, , 1, 1, ,y ., 2, 3, , 5, , +𝑦 =1, 𝑥, 60, 𝑥, , +, , 40, 𝑦, , = 19, 𝑥 ≠ 0, 𝑦 ≠ 0, , Sol:, Taking, , 1, 1, u and v, the given becomes, x, y, , 2u 5v 1, , ...... i , , 60u 40u 19, , ...... ii , , Let us eliminate ‘u’ from equation (i) and (ii), multiplying equation (i) by 60 and equation, (ii) by 2, we get
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , 120u 300v 60, , ...... iii , , 120u 80v 38, , ...... iv , , Subtracting (iv) from (iii), we get, 300v 80v 60 38, , , , 220v 22, 22, 1, , v, , 220 10, 1, Putting v in equation (i), we get, 10, 1, 2u 5 1, 10, 1, , 2u 1, 2, 1, , 2u 1 , 2, 2 1 1, , 2u , , 2, 2, 1, , 2u , 2, 1, , u, 4, 1, 1, Hence, x 4 and y 10, u, v, So, the solution of the given system of equation is x 4, y 10., 21., , 1, 5𝑥, 1, 3𝑥, , 1, , + 6𝑦 = 12, 3, , − 7𝑦 = 8, 𝑥 ≠ 0, 𝑦 ≠ 0, , Sol:, Taking, , 1, 1, u and v, the given equations become\, x, y, , , , u v, 12, 5 6, 6u 5v, 12, 30, 6u 5v 360, , , And,, , u 3v, 8, 3 7, , ..... i
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , , , 7u 9v, 8, 21, 7u 9v 168, , ...... ii , , Let us eliminate ‘v’ from equation (i) and (ii), Multiplying equation (i) by 9 and equation, (ii) by 5, we get, 54u 45v 3240, ...... iii , , 35u 45v 840, , ...... iv , , Adding equation (i) adding equation (ii), we get, 54u 35u 3240 840, , , , 89u 4080, 4080, , u, 89, 4080, Putting u , in equation (i), we get, 89, 4080, 6, 5v 360, 89, 24480, , 5v 360, 89, 24480, , 5v 360 , 89, 32040 24480, , 5v , 89, 7560, , 5v , 89, 7560, , v, 5 89, 1512, , v, 89, 1, 89, 1, 89, Hence, x , and y , u 4080, v 1512, So, the solution of the given system of equation is x , , 89, 89, ,y, ., 4080, 1512
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , 2 3 9, , x y xy, 4 9 21, , where x 0, y 0, x y xy, , 22., , Sol:, The system of given equation is, 2 3 9, , x y xy, 4 9 21, , where x 0, y 0, x y xy, , ..... i , ..... ii , , Multiplying equation (i) adding equation (ii) by xy , we get, , 2 y 3x 9, , .... iii , , 4 y 9 x 21, , .... iv , , From (iii), we get, 3x 9 2 y, 9 2y, , x, 3, 9 2y, Substituting x , in equation (iv), we get, 3, 9 2y , 4x 9 , 21, 3 , , , , , , , 4 y 3 9 2 y 21, , 4 y 27 6 y 21, 2 y 21 27, 2 y 6, 6, , y, 3, 2, 9 2y, Putting y 3 in x , , we get, 3, 9 23, x, 3, 96, , 3, 3, , 3, 1, Hence, solution of the system of equation is x 1, y 3
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , 23., , 6, 𝑥+𝑦, 1, , 7, , = 𝑥−𝑦 + 3, 1, , 2(𝑥+𝑦), , = 3(𝑥−𝑦), where x + y ≠ 0 and x – y ≠ 0, , Sol:, Let, , 1, 1, u and, v. Then, the given system of equation becomes, x y, x y, 6u 7v 3, , , , And,, , , , 6u 7v 3, , ....... i , , u v, , 2 3, 3u 2v, 3u 2v 0, , ...... ii , , Multiplying equation ii by 2, and equation (i) by 1, we get, , 6u 7v 3, , ....... iii , , 6u 4v 0, , ........ iv , , Subtracting equation (iv) from equation (iii), we get, 7 4v 3, , , 3v 3, , v 1, Putting v 1 in equation (ii), we get, 3u 2 1 0, , , , , , 3u 2 0, 3u 2, 2, u, 3, , Now,, , u, , 2, 3, , 1, 2, , x2 3, 3, , x y , 2, And, v 1, 1, , 1, x y, , , .... v
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , , , x y 1, , ...... vi , , Adding equation (v) and equation (vi), we get, 3, 2x , 1, 2, 3 2, , 2x , 2, 5, , 2x , 2, 5, , x, 4, 5, Putting x , in equation (vi), we get, 4, 5, y 1, 4, 5, , 1 y, 4, 5 4, , y, 4, 1, , y, 4, 1, , y, 4, Hence, solution of the system of equation is x , , xy, 6, , x y 5, xy, 6, yx, , 24., , Sol:, The given system of equation is, xy, 6, , x y 5, , , , 5 xy 6 x y , , , , 5 xy 6 x 6 y, , And,, , xy, 6, yx, , .... i , , 5, 1, ,y ., 4, 4
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , , , xy 6 y x , , , , xy 6 y 6 x, , ..... ii , , Adding equation (i) and equation (ii), we get, 6 xy 6 y 6 y, , 6 xy 12 y, 12 y, , x, 2, 6y, Putting x 2 in equation (i), we get, 5 2 y 6 2 6 y, , , , , , 10 y 12 6 y, 10 y 6 y 12, 4 y 12, 12, , y, 3, 4, Hence, solution of the given system of equation is x 2, y 3., 25., , 22, 𝑥+𝑦, 55, 𝑥+𝑦, , 15, , + 𝑥−𝑦 = 5, 45, , + 𝑥−𝑦 = 14, , Sol:, Let, , 1, 1, u and, v. Then, the given system of equation becomes, x y, x y, 22u 15v 5, , ... i , , 55u 45v 14, , ... ii , , Multiplying equation (i) by 3, and equation (ii) by 1, we get, 66u 45v 15, ..... iii , , 55u 45v 14, , ...... iv , , Subtracting equation (iv) from equation (iii), we get, 66u 55u 15 4, , 11u 1, 1, , u, 11, 1, Putting u in equation (i), we get, 11
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , 1, 15v 5, 11, , 2 15v 5, , 15v 5 2, , 15v 3, 3 1, , v , 15 5, 1, Now, u , x y, 22 , , , , 1, 1, , x y 11, , , , x y 11, , And, , , , , .... v , , 1, x y, 1, 1, , x y 5, v, , x y 5, , ..... vi , , Adding equation (v) and equation (vi), we get, 2 x 11 5, , 2 x 16, 16, , x, 8, 2, Putting x 8 in equation (v), we get, 8 y 11, , y 11 8 3, Hence, solution of the given system of equation is x 8, y 3., 26., , 5, , 2, , 15, , 7, , − 𝑥−𝑦 = −1, 𝑥+𝑦, + 𝑥−𝑦 = 10, 𝑥+𝑦, Sol:, Let, , 1, 1, u and, v . Then, the given system off equations becomes, x y, x y, 5u 2v 1, , ...... i , , 15u 7v 10, , ..... ii , , Multiplying equation (i) by 7, and equation (ii) by 2, we get
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , 35u 14v 7 ..... iii , 30u 14v 20, , ...... iv , , Adding equation (iii) and equation (iv), we get, , 35u 30u 7 20, , , , , 65u 13, 13 1, u, , 65 5, , 1, Putting u in equation (i), we get, 5, , , , , , , 1, 5 2 v 1, 5, 1 2v 1, 2v 1 1, 2v 2, 2, v, 1, 2, , 1, x y, 1, 1, , x y 5, , Now, u , , , , , x y 5, , and, v , , , , ...... v , , 1, 1, x y, , x y 1, , ....... vi , , Adding equation (v) and equation (vi), we get, 2x 5 1, , 2x 6, 6, , x 3, 2, Putting x 3 in equation (v), we get, 3 y 5, , y 53 2, Hence, solution of the given system of equation is x 3, y 2.
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, 27., , 3, , 2, , 9, , 4, , + 𝑥−𝑦 = 2, 𝑥+𝑦, − 𝑥−𝑦 = 1, 𝑥+𝑦, Sol:, Let, , 1, 1, u and, v. Then, the given system of equation becomes, x y, x y, 3u 2v 2, , ... i , , 9u 4v 1, , ... ii , , Multiplying equation (i) by 3, and equation (ii) by 1, we get, 6u 4v 4, ..... iii , , 9u 4v 1, , ...... iv , , Adding equation (iii) and equation (iv), we get, 6u 9u 4 1, , , , 15u 5, 5 1, , u , 15 3, 1, Putting u in equation (i), we get, 3, 1, 3 2v 2, 3, , 1 2v 2, , 2v 2 1, 1, , v, 2, 1, u, x y, Now,, , , 1, 1, , x y 3, , , , x y 3, , v, And,, , ...... v , , 1, x y, , , , 1, 1, , x y 2, , , , x y 2, , ...... vi , , Adding equation (v) and equation (vi), we get
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , 2x 3 2, 5, , x, 2, 5, x, 2 in equation (v), we get, Putting, 5, y3, 2, 5, , y 3, 2, 65 1, , y, , 2, 2, , 5, 1, Hence, solution of the given system of equation is x , y ., 2, 2, 28., , 1, , 5, , 2(𝑥+2𝑦), , + 3(3𝑥−2𝑦) =, , −3, , 5, , 2, , 4(𝑥+2𝑦), , 3, , 61, , − 5(3𝑥−2𝑦) = 60, , Sol:, , 1, 1, u and, v., x 2y, 3x 2 y, Then, the given system of equation becomes, u 5v 3, , 2 3, 2, 3u 10v 3, , , 6, 2, 3 6, , 3u 10v , 2, , 3u 10v 9, ..... i , Let, , And,, , 5u 3v 61, , 4 5 60, , 25u 12v 61, , 20, 60, 61, , 25u 12v , ....... ii , 3, Multiplying equation (i) by 12, and equation (ii) by 10, we get, 36u 120v 108, ...... iii , , , 610, ...... iv , 3, Adding equation (iii) and equation (iv), we get, 250u 120v
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, 610, 108, 3, 610 324, , 286u , 3, 286, , 286u , 3, 1, , u, 3, 1, Putting u in equation (i), we get, 3, 1, 3 10v 9, 3, , 1 10v 9, 36u 250u , , , , 10v 9 1, 10, , v, 1, 10, 1, Now, u , x 2y, 1, 1, , , x y 3, , , x 2y 3, , .......... v , , 1, 3x 2 y, 1, 1, 3x 2 y, , And, v , , , , , 3x 2 y 1, , Putting x , , , , , , ....... vi , , 1, in equation (v), we get, 2, , 1, 2y 3, 2, 1, 2y 3, 2, 6 1, 2y , 2, 5, y, 4, , 1, 5, Hence, solution of the given system of equations is x , y ., 2, 4
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, 29., , 5, , 2, , 1, , 10, , 2, , 5, , − 𝑦−1 = 2, 𝑥+1, + 𝑦−1 = 2 , 𝑤ℎ𝑒𝑟𝑒 𝑥 ≠ −1 𝑎𝑛𝑑 𝑦 ≠ 1, 𝑥+1, Sol:, , 1, 1, u and, v., x 1, y 1, Then, the given system of equations becomes, 1, , 5u 2v , ........ i , 2, 5, , 10u 2 y , ........ ii , 2, Adding equation (i) equation (ii), we get, 1 5, 5u 10u , 2 2, 1 5, , 15u , 2, 6, , 15u 3, 2, 3 1, , u , 15 5, 1, Putting u in equation (i), we get, 5, 1, 1, 5 2v , 5, 2, 1, , 1 2v , 2, 1, , 2v 1, 2, 1 2, , 2v , 2, 1, , 2v , 2, 1 1, , v, , 4 4, 1, Now, u , x 1, Let
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , 1, 1, , x 1 5, , x 1 5, , x 5 1 4, 1, And, v , y 1, 1, 1, , , y 1 4, , y 1 4, , y 4 1 5, , , Hence, solution of the give system of equation is x 4, y 5., , 30., , x y 5 xy, 3x 2 y 13xy, , Sol:, The give system of equation is, x y 5 xy, ...... i , , 3x 2 y 13xy, , ..... ii , , Multiplying equation (i) by 2 and equation (ii) by , we get, , 2 x 2 y 10 xy, , .... iii , , 3x 2 y 13xy, , .... iv , , Subtracting equation (iii) from equation (iv), we get, 3x 2 x 13xy 10 xy, , x 3xy, x, , y, 3x, 1, , y, 3, 1, Putting y in equation (i), we get, 3
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , x y 5 x , , 1, 3, , 1 5x, , 3 3, 1 5x, , x, 3 3, 1 5 x 3x, , 3, 3, 1 2x, 2x 1, 1, x, 2, x, , , , , , , , 1, 1, Hence, solution of the given system of equations is x , y ., 2, 3, , x y 2 xy, , 31., , x y, 6 x 0, y 0, xy, Sol:, The system of the given equation is, x y 2 xy, ....... i , And,, , x y, 6, xy, x y 6 xy, , ....... ii , , Adding equation (i) and equation (ii), we get, 2 x 2 xy 6 xy, , 2 x 8 xy, 2x, y, 8x, 1, , y, 4, 1, Putting y in equation (i), we get, 4,
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, 33., , 2, , 3, , 5, , 1, , + 3𝑥−2𝑦 =, 3𝑥+2𝑦, 3𝑥+2𝑦, , 17, 5, , + 3𝑥−2𝑦 = 2, , Sol:, Let, , 1, 1, u and, v. Then, the given system of equation becomes, 3x 2 y, 3x 2 y, 17, 2u 3v , ....... i , 5, 5u v 2, ....... ii , , Multiplying equation (ii) by 3, we get, 17, 15u 2u 6 , 5, 30 17, , 13u , 5, 13, , 13u , 5, 13, 1, , u, , 5 13 5, 1, Putting u in equation (ii), we get, 5, 1, 5 v 2, 5, , 1 v 2, , , , , v 2 1, v 1, 1, Now, u , 3x 2 y, 1, 1, , , 3x 2 y 5, , , , 3x 2 y 5, , ..... iv , , 1, 3x 2 y, 1, 1, 3x 2 y, , And, v , , , , , 3x 2 y 1, , ....... v , , Adding equation (iv) and (v), we get
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , 6x 1 5, , 6x 6, , x 1, Putting x 1 in equation (v), we get, 3 1 2 y 5, , 2y 5 3, , 2y 2, 2, , y 1, 2, Hence, solution of the given system of equation is x 1, y 1., 34., , 4, 𝑥, 3, 𝑥, , + 3𝑦 = 14, − 4𝑦 = 23, , Sol:, 4, 3 y 14, x, 3, 4 y 23, x, 1, Let p, x, The given equations reduce to:, 4 p 3 y 14, , , , 4 p 3 y 14 0, , ..... 1, , 3 p 4 y 23, , , 3 p 4 y 23 0, , ..... 2 , , Using cross-multiplication method, we obtain, p, y, 1, , , 69 56 42 92 16 9, , p, y 1, , , 125 50 25, p, 1 y 1, , , 125 25 50 25, p 5, y 2, p, x, , 1, 5, , 1, 5, x
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, 99 x 101y 499, 101x 99 y 501, , 35., , Sol:, The given system of equation is, 99 x 101y 499, , ..... i , , 101x 99 y 501, , .... ii , , Adding equation (i) and equation (ii), we get, 99 x 101x 101y 99 y 499 501, , 200 x 200 y 1000, , , 200 x y 1000, , , , 1000, 5, 200, x y 5, , , , x y , , ..... iii , , Subtracting equation (i) by equation (ii), we get, 101x 99 x 99 y 101y 501 499, , , , 2x 2 y 2, , , , 2 x y 2, , , , x y , , , , 2, 2, x y 1, , ....... iv , , Adding equation (iii) and equation (iv), we get, 2x 5 1, 6, , x 3, 2, Putting x 3 in equation (iii), we get, 3 y 5, , y 53 2, Hence, solution of the given system of equation is x 3, y 2., , 36., , 23x 29 y 98, 29 x 23 y 110, , Sol:, The given system of equation is, 23x 29 y 98, , ..... i , , 29 x 23 y 110, , ..... ii
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, Adding equation (i) and equation (ii), we get, 23x 29 x 29 y 23 y 98 110, , 52 x 52 y 208, , , , 52 x y 208, , , , x y , , , , 208, 4, 52, x y4, , ........ iii , , Subtracting equation (i) by equation (ii), we get, 29 x 23x 23 y 29 y 110 98, , 6 x 6 y 12, , , , 6 x y 12, , , , x y , , , , 12, 2, 6, x y 2, , ...... iv , , Adding equation (iii) and equation (iv), we get, 2x 2 4 6, 6, , x 3, 2, Putting x 3 in equation (iv), we get, 3 y 2, , y 2 3 1, Hence, solution of the given system of equation is x 3, y 1., , 37., , x yz 4, x 2 y 2z 9, 2 x y 3z 1, Sol:, We have,, x yz 4, , ...... i , , x 2 y 2z 9, , ...... ii , , 2 x y 3z 1, , ..... iii , , From equation (i), we get, z 4 x y, , z x y 4, Subtracting the value of z in equation (ii), we get
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , x 2 y 2 x y 4 9, , , , x 2 y 2x 2 y 8 8, 3x 4 y 9 8, , , , 3x 4 y 17, , ..... iv , , Subtracting the value of z in equation (iii), we get, 2x y 3 x y 4 1, , , , , 2 x y 3x 3 y 12 1, x 4 y 1 12, , , , x 4 y 11, , ...... v , , Adding equations (iv) and (v), we get, 3 x x 4 y 4 y 17 11, , 2x 6, 6, , x 3, 2, Putting x 3 in equation (iv), we get, 3 3 4 y 17, , , , , , 9 4 y 17, 4 y 17 9, 4 y 8, 8, , y, 2, 4, Putting x 3 and y 2 in z x y 4, we get, z 3 2 4, , z 5 4, , z 1, Hence, solution of the giving system of equation is x 3, y 2, z 1., , 38., , x yz 4, x yz 2, 2 x y 3z 0, Sol:, We have,, x yz 4, , ...... i , , x y z 2, , ...... ii , , 2 x y 3z 0, , ...... iii , , From equation (i), we get
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , , z 4 x y, z x y 4, , Substituting z x y 4 in equation (ii), we get, , x y x y 4 2, , , , , , x yx y4 2, 2y 4 2, 2 y 2 4 2, 2 y 2, 2, , y, 1, 2, Substituting the value of z in equation (iii), we get, 2x y 3 x y 4 0, , , , 2 x y 3x 3 y 12 0, 5 x 2 y 12 0, , , , 5 x 2 y 12, , ....... iv , , Putting y 1 in equation (iv), we get, , 5 x 2 1 12, , , , 5 x 2 12, 5 x 12 2 10, 10, , x 2, 5, Putting x 2 and y 1 in z x y 4, we get, z 2 1 4, 2 1 4, 3 4, 1, Hence, solution of the giving system of equation is x 2, y 1, z 1., 39., , 44, , 30, , 55, , 40, , + 𝑥−𝑦 = 4, 𝑥+𝑦, + 𝑥−𝑦 = 13, 𝑥+𝑦, Sol:, , 1, 1, u and, v., x y, x y, Then, the system of the given equations becomes, Let
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , 44u 30v 10, , ..... i , , 55u 40v 13, , ..... ii , , Multiplying equation (i) by 4 and equation (ii) by 3, we get, 176u 120v 40, ....... iii , , 165u 120v 39, , ....... iv , , Subtracting equation (iv) by equation (iii), we get, 176 165u 40 39, , 11u 1, 1, , u, 11, 1, Putting u in equation (i), we get, 11, 1, 44 30v 10, 11, 4 30v 10, , 30v 10 4, , 30v 6, 6 1, , v, , 30 5, 1, Now, u , x y, , , , 1, 1, , x y 11, , , , x y 11, , ........ v , , Adding equation (v) and (vi), we get, 2 x 11 5, , 2 x 16, 16, , x, 8, 2, Putting x 8 in equation (v),, we get, 8 y 11, , y 11 8 3, Hence, solution of the given system of equations is x 8, y 3.
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, 40., , 4, 15 y 21, x, 3, 4y 5, x, Sol:, The given system of equation is, 4, 15 y 21, ...... i , x, 3, 4y 5, ...... ii , x, Multiplying equation (i) by 3 and equation (ii) by 4, we get, , 12, 15 y 21, x, 12, 16 y 20, x, , ...... iii , ...... iv , , Subtracting equation (iii) from equation (iv), we get, , , , 12 12, 16 y 15 y 20 21, x x, y 1, , Putting y 1 in equation (i), we get, , , , , , , , , , , , 4, 5 1 7, x, 4, 5 7, x, 4, 75, x, 4, 12, x, 4 12 x, 4, x, 12, 4, x, 12, 1, x, 3, , 1, Hence, solution of the given system of equation is x , y 1., 3
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , 41., , 1 1, 2 3 13, x y, 1 1, 5 4 2, x y, Sol:, Let us write the given pair of equation as, 1 1, 2 3 13, 1, x y, 1 1, 5 4 2, 2, x y, These equation are not in the form ax by c 0. However, if we substitute, , 1, 1, p and q in equations (1) and (2), we get, x, y, 2 p 3q 13, 5 p 4q 2, So, we have expressed the equations as a pair of linear equations. Now, you can use any, method to solve these equations, and get p 2, q 3, You know that p , , 1, 1, and q ., x, y, , Substitute the values of p and q to get, , 1, 1, 1, 1, 2, i.e., x and 3 i.e., y ., x, 2, y, 3, 42., , 5, 𝑥−1, , 1, , + 𝑦−2 = 2, , Sol:, x = 4, y = 5, Detailed answer not given in website, 43., , 10, 2, , 4, x y x y, 15, 5, , 2, x y x y, Sol:, 10, 2, , 4, x y x y
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , 15, 5, , 2, x y x y, 1, 1, Let, p and, q, x y, x y, The given equations reduce to:, 10 p 2q 4, , , 10 p 2q 4 0, , ..... 1, , 15 p 5q 2, , , 15 p 5q 2 0, , ..... 2 , , Using cross-multiplication method, we obtain:, , p, q, 1, , , 4 20 60 20 50 30, p, q, 1, , , 16 80 80, p, 1, q, 1, , and, , 16 80, 80 80, 1, p and q 1, 5, 1, 1, 1, p, and q , 1, x y 5, x y, x y 5, , ...... 3, , x y 1, , ...... 4 , , Adding equation (3) and (4), we obtain:, , 2x 6, x3, Substituting the value of x in equation (3), we obtain:, , y2, x 3, y 2, , 44., , 1, 3𝑥+𝑦, 1, , 1, , 3, , + 3𝑥−𝑦 = 4, , 2(3𝑥+𝑦), , 1, , 1, , − 2(3𝑥−𝑦) = − 8, , Sol:, Let us put, , 1, 1, p and, q. Then the given equations, x 1, y2
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , 1, 1 , 5, 2, , x 1 y 2, , ...... 1, , 1 1 , 6, ...... 2 , 1, 3, x 1 y 2 , ........ 3, Can be written as: 5 p q 2, , 6 p 3q 1, , ...... 4 , , Equations (3) and (4) from a pair of linear equations in the general form. Now, you can use, 1, 1, any method to solve these equations. We get p and q ., 3, 3, 1, Substituting, for p, we have, x 1, 1, 1, ,, x 1 3, i.e., x 1 3, i.e., x 4., Similarly, substituting, , 1, for q, we get, y2, , 1, 1, , y2 3, i.e., x 1 3, i.e., x 4, Similarly, substituting, , 1, for q, we get, y2, , 1, 1, , y2 3, i.e., 3 y 2, i.e., y 5, Hence, x 4, y 5 is the required solution of the given pair of equations., , 45., , 2, 3, , 2, x, y, 4, 9, , 1, x, y, Sol:, 2, 3, , 2, x, y, , 4, 9, , 1, x, y
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, Let, , 1, 1, q, p and, y, x, , The given equations reduce to:, 2 p 3q 2, ..... 1, , 4 p 9q 1, , ..... 2 , , Multiplying equation (1) by (3), we obtain:, , 6 p 9q 6, , ..... 3, , Adding equation (2) and (3), we obtain:, 10 p 5, 1, p, 2, Putting the value of p in equation (1), we obtain:, 1, 2 3q 2, 2, 3q 1, , q, , 1, 3, , p, , 1, 1, , x 2, , x 2, x4, 1, 1, q, , y 3, , y 3, y9, x 4, y 9, , 7x 2 y, 5, xy, 8x 7 y, 15, xy, , 46., , Sol:
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , , , , , 7x 2 y, 5, xy, 7 2, 5, ... 1, y x, 8x 7 y, 15, xy, 8 7, 15, ..... 2 , y x, , 1, 1, p and q, x, y, The given equations reduce to:, 2 p 7q 5, Let, , , , 2 p 7q 5 0, , .... 3, , 7 p 8q 15, , , 7 p 8q 15 0, , .... 4 , , Using cross multiplication method, we obtain:, p, q, 1, , , 105 40 35 30 16 49, , p, 1, q, 1, , ,, , 65 65 65 65, p 1, q 1, 1, 1, 1, q 1, x, y, x 1, y 1, p, , 47., , 152 x 378 y 74, 378 x 152 y 604, , Sol:, 152 x 378 y 74, , .... 1, , 378 x 152 y 604, , .... 2 , , Adding the equations (1) and (2), we obtain:, 226 x 226 y 678, , x y 3, , ..... 3, , Subtracting the equation (2) from equation (1), we obtain, 530 x 530 y 530
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , x y 1, , ..... 4 , , Adding equations (3) and (4), we obtain:, 2x 4, , x2, Substituting the value of x in equation (3), we obtain:, y 1, , Exercise 3.4, Solve each of the following systems of equations by the method of cross-multiplication:, 1., , x 2 y 1 0, 2 x 3 y 12 0, , Sol:, The given system of equation is, x 2 y 1 0, 2 x 3 y 12 0, Here,, a1 1, b1 2, c1 1, , a2 2, b2 3 and c2 12, By cross-multiplication, we get, x, y, 1, , , , 2 12 1 3 1 12 1 2 1 3 2 2, , x, y, 1, , , 24 3 12 2 3 4, x, y, 1, , , , 21 14 7, Now,, x, 1, , 21 7, 21, , x, 3, 7, And,,
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, y, 1, , 14 7, y 1, , , 14 7, 14, , y, 2, 7, Hence, the solution of the given system of equations is x 3, y 2., , 2., , 3x 2 y 25 0, 2 x y 10 0, , Sol:, The given system of equation is, 3x 2 y 25 0, 2 x y 10 0, Here,, a1 3, b1 2, c1 25, a2 2, b2 1 and c2 10, , By cross-multiplication, we have, x, y, 1, , , , 2 10 25 1 3 10 25 2 3 1 2 2, x, y, 1, , , , 20 25 30 50 3 4, x, y, 1, , , , 5 20 1, x, 1, Now,, , 5 1, 5, x, 5, 1, And,, y, 1, , 20 1, y, , 1, 20, , y 20, Hence, x 5, y 20 is the solution of the given system of equations., , 3., , 2 x y 35 0, 3x 4 y 65 0
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, Sol:, The given system of equations may be written as, 2 x y 35 0, 3x 4 y 65 0, Here,, a1 2, b1 1, c1 35, a2 3, b2 4 and c2 65, , By cross multiplication, we have, x, y, 1, , , , 1 65 35 4 2 65 35 3 2 4 1 3, , x, y, 1, , , 65 140 130 105 8 3, x, y 1, , , , 75 25 5, x, y 1, , , , 75 25 5, Now,, y 1, , 25 5, 25, , y, 5, 5, Hence, x 15, y 5 is the solution of the given system of equations., , , 4., , 2x y 6 0, x y20, , Sol:, The given system of equations may be written as, 2x y 6 0, x y20, Here,, a1 2, b1 1, c1 6, , a2 1, b2 1 and c2 2, By cross multiplication, we get
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , , , x, y, 1, , , 1 2 6 1 2 2 6 1 2 1 1 1, , x, y, 1, , , 2 6 4 6 2 1, x y 1, , , , 4 2 1, x y, , , 1, 4 2, Now,, x, 1, 4, , x 4 1 4, , , And,, , , , y, 1, 2, y 1 2, , , , , y 2, y2, , Hence, x 4, y 2 is the solution of the given system of the equations., , 5., , x y, 2, xy, x y, 6, xy, Sol:, The given system of equations is, x y, 2, xy, x, y, , , 2, xy xy, 1 1, , 2, y x, 1 1, , 2, ...... i , x y, And,
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , , , , , x y, 6, xy, x, y, , 6, xy xy, 1 1, 6, y x, 1 1, 6, x y, , Taking u , , ...... ii , , 1, 1, and v , we get, x, y, , uv 2uv2 0, , ..... iii , , And, u v 6 u v 6 0, , ..... iv , , Here,, a1 1, b1 1, c1 2, , a2 1, b2 1 and c2 6, By cross multiplication, u, v, 1, , , , 1 6 2 1 1 6 2 1 1 1 11, u, v, 1, , , 6 2 6 2 1 1, u v 1, , , 4 8 2, u 1, Now, , 4 2, 4, u , 2, 2, v 1, And,, , 8 2, 8, v , 4, 2, v 4, , , v4, 1 1, 1 1, , and y , u 2, v 4, 1, 1, Hence, x , y is the solution of the given system of equations., 2, 4, Now, x
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, 6., , ax by a b, bx ay a b, Sol:, The given system of equations is, ax by a b, .... i , , .... ii , , bx ay a b, Here,, a1 a, b1 b, c1 b a, , a2 b, b2 a and c2 a b, By cross multiplication, we get, x, y, 1, , , , a b b b a a a b a b a b a a b b, x, y, 1, 2, 2 2, 2, 2, ab b ab a, a ab b ab a b, x, y, 1, 2, 2 2 2 2, 2, b a, a b, a b, Now,, x, 1, 2 2, 2, 2, b a, a b, 2, 2, b a, , x 2 2, a b, , , 2, , , , , , b2 a 2, , a, , a, , a, , And,, , , , , , 2, , b, , 2, , b2, , 2, , b2, , 2, , , , , , , , , x 1, y, 1, 2 2, 2, 2, a b, a b, 2, a b2, y , a 2 b2, , , , , , , , y 1, y 1, , Hence, x 1, y 1 is the solution of the given system of the equations.
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, 7., , x ay b 0, ax by c 0, , Sol:, The given system of equations may be written as, x ay b 0, ax by c 0, Here,, a1 1, b1 a, c1 b, , a2 a, b2 b and c2 c, By cross multiplication, we get, x, y, 1, , , , a c b b 1 c b a 1 b a a, , x, y, 1, , , 2, ac b, c ab b a 2, Now,, x, 1, , 2, ac b, b a 2, ac b 2, , x, b a 2, b 2 ac, , x, a2 b, , , , , , , , , , , b 2 ac, a2 b, , And, , , , , y, 1, , c ab b a 2, ab c, y , a2 b, , , , y, , , , ab c, a2 b, , ac b2, ab c, ,y 2, Hence, x 2, is the solution of the given system of the equations., a b, a b, , 8., , ax by a 2, bx ay b 2, Sol:
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, The system of the given equations may be written as, , ax by a 2 0, bx ay b 2 0, Here,, a1 a, b1 b, c1 a 2, , a2 b, b2 a and c2 b 2, By cross multiplication, we get, x, y, 1, , , , 2, 2, 2, 2, b b a a a b a b a a b b, , , , , , , , , , x, y, 1, , 2 2, 3, 2, 2, b a, ab a b a b, Now,, x, 1, 2 2, 3, 3, b a, a b, 3, 3, a b, , x 2 2, a b, a b a 2 ab b 2 , , a b a b , , , 3, , , , a 2 ab b 2, ab, , And,, , , , , y, 1, 2 2, 2, 2, ab a b a b, a 2b ab 2, y 2 2, a b, 2, ab a 2b, y 2 2, a b, ab b a , , a b a b , , , , , ab a b , , a b a b , ab, ab, , Hence, x , , a 2 ab b2, ab, ,y, is the solution of the given system of the equations., ab, a b
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, 9., , 𝑥, 𝑎, , 𝑦, , +𝑏 =2, , 𝑎𝑥 − 𝑏𝑦 = 𝑎2 − 𝑏 2, Sol:, The system of the given equations may be written as, 1, 1, x y 2 0, a, b, ax by b 2 a 2 0, Here,, 1, 1, a1 , b1 , c1 2, a, b, a2 a, b2 b and c2 b 2 a 2, By cross multiplication, we get, x, y, 1, , , , 1, 1, b 1 a 1, b 2 a 2 2 b , b 2 a 2 2 a, , b, a, a, b, x, y, 1, 2, 2, , 2, 2, b a, b a, b a, , 2b, 2b, a b, b, b, x, y, 1, 2, 2, 2, 2, 2, 2, 2, b a 2b, b a 2b, b a 2, b, a, ab, x, y, 1, , 2, 2, 2, 2, 2, a b, b a, b a 2, b, a, ab, Now,, x, 1, 2, 2, 2, a b, b a 2, b, ab, 2, 2, a b, ab, , x, 2, b, b a 2, And,, y, 1, 2, 2, 2, b a, b a 2, a, ab, 2, 2, b a, ab, , y , 2 2, a, b a
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , , b, y , , , , y b, , 2, , , , , , a2 b, , b a2, 2, , , , Hence, x a, y b is the solution of the given system of the equations., 10., , 𝑥, , 𝑦, , +𝑏 =𝑎+𝑏, 𝑎, Sol:, The given system of equation may be written as, 1, 1, x y a b 0, a, b, 1, 1, x 2 y 2 0, 2, a, b, Here,, 1, 1, a1 , b2 , c1 a b , a, b, 1, 1, a2 2 , b2 2 , and c2 2, a, b, By cross multiplication, we get, x, y, 1, , , , 1, 1, 1, 1, 1 1 1 1, 2 2 x a b , 2 2 x a b , , b, b, a, a, a b2 a 2 b, x, y, 1, , , , 2 a 1, 2 1 b, 1, 1, 2, 2 2 2, b b b, a a a, ab a b, x, y, 1, , , , a 1, 1 b, 1, 1, , 2, 2, 2, 2, b b, a a, ab a b, x, y, 1, , , , a b a b a b, b2, a2, a 2b 2, a b, 1, a b, 1, x 2 , a 2 and y 2 , b2, a b, a b, b, a, 2 2, ab, a 2b 2, Hence, x a 2 , y b 2 is the solution of the given system of the equtaions., , 11., , x y, , a b, ax by a 2 b 2
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, Sol:, x y, , a b, ax by a 2 b 2, Here a1 , , 1, 1, , b1 , c1 0, a, b, , , , a2 a, b2 b, c2 a 2 b 2, , , , By cross multiplication, we get, x, y, , 1, 1, a 2 b2 b 0 , a 2 b2, b, a, , , , , , , , a 0, , , , 1, 1, 1, b a , a, b , , x, y, 1, 2 2 , 2, b a, a b, a b, , a b, b, a, a 2 b2 a 2 b2, b, x, 2 b 2 a, b a, b a, , a b, ab, 2, 2, 2, a b, a b2, a, y, 2 b 2 b, b a, b a, , a b, ab, Solution is (a, b), 2, , 12., , 5, 𝑥+𝑦, 15, 𝑥+𝑦, , 2, , − 𝑥−𝑦 = −1, 7, , + 𝑥−𝑦 = 10, 𝑤ℎ𝑒𝑟𝑒 𝑥 ≠ 0 𝑎𝑛𝑑 𝑦 ≠ 0, , Sol:, , 1, 1, u and, v. Then, the given system of equations becomes, x y, x y, 5u 2v 1, 15u 7v 10, Here, a1 5, b1 2, c1 1, a2 15, b2 7 and c2 10, Let, , By cross multiplication, we get, u, u, 1, , , , 2 10 1 7 5 10 115 5 7 2 15
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , u, v, 1, , , 20 7 50 15 35 30, u, v, 1, , , 13 65 65, u, v, 1, , , 13 65 65, Now,, u, 1, , 13 65, 13 1, , u, , 65 5, And,, v, 1, , 65 65, 65, , v, 1, 65, Now,, 1, u, x y, 1, 1, , , ...... i , x y 5, , , And,, v, , 1, x y, , , , 1, 1, x y, , , , x y 1, , ...... ii , , Adding equation (i) and (ii), we get, 2x 5 1, , 2x 6, 6, , x 3, 2, 13., , 2 3, 13, x y, 5 4, 2, where x 0 and y 0, x y, Sol:
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, The given system of equation is, 2 3, 13, x y, , 5 4, 2, where x 0 and y 0, x y, 1, 1, Let u and v, Then, the given system of equations becomes, x, y, 2u 3v 13, 5u 4v 2, Here,, a1 2, b1 3, c1 13, a2 5, b2 4 and c2 2, By cross multiplication, we have, u, v, 1, , , , 3 2 13 4 2 2 13 5 2 4 3 5, , u, v, 1, , , 6 52 4 65 8 15, u, v, 1, , , , 46 69 23, Now,, u, 1, , 46 23, 46, , u, 2, 23, And, v, 1, , 69 23, 69, , v, 3, 23, Now,, 1 1, x , u 2, And,, 1 1, y , v 3, 1, 1, Hence, x , y is the solution of the given system of equations., 2, 3,
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, Hence, x , , 18., , 5b 2a, a 10b, is the solution of the given system of equations., ,y, 10ab, 10ab, , ab , ab , , , xa b , ya b, , a b , ab , , , , x y 2a 2, , Sol:, The given system of equation is, ab , ab , , , xa b , ya b, , a b , ab , , , , ...…(i), , x y 2a 2, , …...(ii), , From equation (i), we get, ab , ab , , , xa b , ya b, 0, a b , ab , , , , , , a b 2 ab , a b 2 ab , y, 0, x, , , , , a b, ab, , , , , , , , a 2 b 2 2ab ab , a 2 b 2 2ab ab , x, y, 0, a b, ab, , , , , , a 2 b 2 ab , a 2 b 2 ab , , x, y, 0, a b, ab, , , , , From equation (ii), we get, x y 2a 2 0, , .....(iii), , Here,, , a 2 b 2 ab , a 2 b 2 ab, a1 , , b1 , , c1 0, a b, ab, , , a2 1, b2 1 and c2 2a 2, , By cross multiplication, we get, x, y, 1, , , , 2, 2, a 2 b 2 ab , a 2 b 2 ab , 2, a 2 b 2 ab a b ab, 2a 2 , , 0, , 1, , 2, a, , , 0, , 1, , , , , , , , , ab, a b, a b, a b, , , , , , , , , , , , , , , , , x, y, 1, , 2 2, 2, 2, 2, a b ab a 2 b 2 ab, 2 a b ab , 2 a b ab , , 2a , 2a , , a, , b, ab, a, , b, a, , b, , , , , 2, , , , , ,
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , x, y, 1, , , a 2 b 2 ab , a 2 b 2 ab a b a 2 b 2 ab a b a 2 b 2 ab, 2a 2 , 2a 2 , , , ab, a b, , , , , a b a b , x, y, 1, , , 3 3, 2, 2, 2, 2, a b a 3 b3, 2 a b ab , 2 a b ab , 2a , 2a , , a b a b , ab, a b, , , , , , , , , , , , , , , , x, y, 1, , , 2, 2, 2, 2a 3, a b ab , 2 a b ab , 2a 2 , 2, a, , , , ab, a b, , , , a b a b , Now,, x, 1, , 2, 2, 2a 3, 2 a b ab , 2a , , ab, , a b a b , , , , , 2, , x, , , , , , , 2a 2 a 2 b2 ab, ab, , a b a b , , a b a 2 b2 ab , , 2a 3, , a, , a b, a, 3, , 3, , , , , , a3 b3 a b a 2 b2 ab , , , , And,, , y, 1, , 2, 2a 3, a b ab , 2a 2 , , a b, , a b a b , 2, , , , y, , , , , , 2a 2 a 2 b 2 ab, , a b a b , , a b, a b a 2 b2 ab, , , , , , 2a 3, , a, a b, a, 3, , 3, , , , , , a 3 b3 a b a 2 b 2 ab , , , , a 3 b3, a 2 b2, ,y, is the solution of the given system of equaatiions., a, a, The given system of equation id, ab , ab , , , …(i), xa b , ya b, , a b , ab , , , x y 2a 2, …(ii), Hence, x , ,
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, From equation (i), w get, ab , ab , , , xa b , ya b, 0, a b , ab , , , , , , a b 2 ab , a b 2 ab , y, 0, x, , , , , a b, ab, , , , , , , , a 2 b 2 2ab ab , a 2 b 2 2ab ab , x, y, 0, a b, ab, , , , , , a 2 b 2 ab , a 2 b 2 ab , , x, y, 0, a b, ab, , , , , From equation (ii), we get, x y 2a 2 0, , …..(iii), , …..(iv), , Here,, , a 2 b 2 ab , a 2 b 2 ab, a1 , , b1 , , c1 0, a b, ab, , , a2 1, b2 1 and c2 2a 2, , By cross multiplication we get, x, y, 1, , , , 2, 2, 2, 2, a 2 b2 ab , a b ab a 2 b2 ab , 2 a b ab , 2, , 2a , 0 1, 0 1 2a , a b, a b, a b , ab, , , , , , x, y, 1, , , 2 2, 2, 2, 2, 2, a b ab a 2 b 2 ab, a b ab , 2 a b ab , , 2a 2 , 2, a, , , , a b, ab, ab, a b, , , , , , , , , , , , , , , , , , x, y, 1, , , 2, 2, 2, 2, 2, 2, a b ab a 2 b 2 ab, 2 a b ab , 2 a b ab , , 2a , 2a , , a, , b, ab, a, , b, a, , b, , , , , , x, y, 1, , 3 3, 2, 2, 2, a b a 3 b3, a b ab , 2 a b ab , 2a 2 , 2, a, , , , a b a b , ab, a b, , , , , x, y, 1, , , , 2, 2, 2, 2, 2a 3, a b ab , 2 a b ab , 2a 2 , 2, a, , , , ab, a b, , , , a b a b , , , Now,, , 2
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , x, 1, , 2, 2a 3, 2 a b ab , 2a , a b a b, , , ab, , , 2, , x, , , , , 2a 2 a 2 b2 ab, ab, , 2a 3, , a b a 2 b2 ab , a, , a b, a, And,, , , a b a b , , 3, , 3, , , , , , , , , , a 2 b2 a b a 2 b2 ab , , , , y, 1, , 2, 2a 3, a b ab , 2a 2 , , a b, , a b a b , 2, , y, , , , , , , 2a 2 a 2 b2 ab, a b, , a b a b , , a b a 2 b2 ab , , 2a 3, , a, , a b, a, 3, , 3, , a3 b3 a b a 2 b2 ab , , , , a 2 b2, a 3 b3, ,y, Hence, x , is the solution of the given system of equation., a, a, bx cy a b, 19., , 1 , 1 , 2a, 1, 1, ax , , , cy , , a b a b , ba ba ab, Sol:, The given system of equation is, bx cy a b, 1 , 1 , 2a, 1, 1, ax , , , cy , , a b a b , ba ba ab, From equation (ii), we get, bx cy a b 0, , ....(i ), .....(ii ), .... iii , , From equation (ii), we get, , a b a b , b a b a 2a, ax , 0, , cy , a b a b , b a b a a b
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, y, 1, , 2, 4ab, 4abc, 4ab 2 b, , y, , 4abc c, a, b, Hence, x , y is the solution of the given system of the equations., b, c, , a b x a b y 2a 2 2b 2, a b x y 4ab, , 20., , Sol:, The given system of equation is, a b x a b y 2a 2 2b 2, , ……(i), , a b x y 4ab, , ……(ii), , From equation (i), we get, , a b x a b y 2a 2 2b2 0, a b x a b y 2 a 2 b2 0, , , , ……(iii), , From equation (ii), we get, a b x a b y 4ab 0, Here,, , ……(iv), , a1 a b, b1 a b, c1 2 a 2 b 2 , , a2 a b, b2 a b and c2 4ab, By cross multiplication, we get, x, y, 1, , , , 2, 2, 2, 2, 4ab a b 2 a b a b 4ab a b 2 a b a b a b a b a b a b , , , x, y, 1, , , 2, 2, 2 a b 2ab a b 4ab a b 2 a b a b a b a b a b a b , x, , , , 2 a b a b 2ab , , , , , , x, , 2 a b a b 2ab , , , , , , x, , 2 a b a b 2ab , , , , 2, , 2, , Now,, , 2, , 2, , 2, , 2, , y, 1, , 2 a b 2ab a b a b a b a b a b , y, , 2 a b 2ab a 2 b 2 2ab , , y, 1, , 2, 2, 2 a b a b 2b a b , , , , 1, , a b 2b
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , x, , 2 a b a 2 b2 2ab , , x, , , , 1, 2b a b , , 2 a b a 2 b2 2ab , 2b a b , , a 2 b2 2ab, b, 2, a b2 2ab, x, b, 2ab a 2 b2, , b, Now,, y, 1, , 2, 2, 2 a b a b 2ab a b , x, , 2 a b a 2 b2 , , , , y , , , , a b a 2 b2 , y, b a b, , 2b a b , , a b a 2 b2 , 2ab a 2 b2, ,y, Hence, x , is the solution of the given system of, b, b a b, equations., , y, 1, 4 4, 2, 2 2, a d b c, a b, 2 2, 2 2, a d b c, y , a 4 b4, a 2 d 2 b2c 2, y, a 4 b4, 2, , , , , 21., , a 2 x b2 y c 2, b2 x a 2 y d 2, Sol:, The given system of equations may be written as, , a2 x b2 y c2 0, b2 x a 2 y d 2 0, Here,
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , a1 a 2 , b1 b 2 , c1 c 2, a2 b 2 , b2 a 2 and c2 d 2, By cross multiplication, we have, x, y, 1, 2 2, 2 2 2 2 4 4, 2 2, b d a c, a d b c, a b, Now,, x, 1, 4 4, 2 2, 2 2, b d a c, a b, 2 2, 2 2, a c b d, , x, a 4 b4, And,, y, 1, 4 4, 2 2, 2 2, a d b c, a b, 2 2, 2 2, a d b c, , y , a 4 b4, a 2 d 2 b2c 2, , y, a 4 b4, Hence, x , , 22., , 57, 𝑥+𝑦, 38, 𝑥+𝑦, , a 2c 2 b 2 d 2 a 2 d 2 b 2c 2, is the solution of the given system of the equations., ,y, a 4 b4, a 4 b4, , 6, , + 𝑥−𝑦 = 5, 21, , + 𝑥−𝑦 = 9, , Sol:, , 1, 1, u and, v. Then, the given system of equations become, x y, x y, 57u 6v 5 57u 6v 5 0, 38u 21v 9 38u 21v 9 0, Here,, a1 57, b1 6, c1 5, Let, , a2 38, b2 21, and c2 9, By cross multiplication, we have, u, v, 1, , , , 54 105 513 190 1193 228, u, v, 1, , , , 51 323 969, u, v, 1, , , , 51 323 969
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , a 2u b 2 v 0, a 2bu b2 av a b 0, Here,, , a1 a 2 , b1 b2 , c1 0, a2 a 2b, b2 b2 a, and c2 a b , By cross multiplication, we have, u, v, 1, 2, 2, 3 2, 2, 2, b a b 0 b a a a b 0 a b a b a 2b 3, , , , u, v, 1, 2, 2 2, b a b a a b a b a b, 2, , Now,, u, 1, 2 2, b a b a b a b, 2, , , , b2 a b , u 2 2, a b a b, , , , u, , 1, a2, , And,, , v, 1, 2 2, a a b a b a b, 2, , , , a2 a b , v 2 2, a b a b, , , , v, , 1, b2, , Now,, 1, x a2, u, And,, 1, y b2, v, Hence, x a 2 , y b 2 is the solution of the given system of equations., 26., , mx – my = m2 + n2, x + y = 2m, Sol:, The given system of equations may be written as
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , mx ny m 2 n 2 0, x y 2m 0, , Here,, , a1 m, b1 n, c1 m2 n 2 , , a2 1, b2 1, and c2 2m, By cross multiplication, we have, x, y, 1, , , 2, 2, 2, 2, 2, 2mn m n 2m m n m n, x, y, 1, , , 2, 2, 2, 2, 2mn m n, m n, mn, x, y, 1, , , 2, 2, 2, m n m n m n, , , , , Now,, , x, , m n, , 2, , , , 1, mn, , m n , x, 2, , , , mn, x mn, , , And,, , y, 1, , 2, 2, m n, mn, 2, m n 2, y , mn, 2, m n2, y, mn, m n m n , y, mn, y mn, , , , , , , Hence, x m n, y m n is the solution of the given system of equation., 27., , 𝑎𝑥, 𝑏, , −, , 𝑏𝑦, 𝑎, , =𝑎+𝑏, , 𝑎𝑥 − 𝑏𝑦 = 2𝑎𝑏, Sol:, The given system of equation may be written as
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, a, b, x y a b 0, b, a, ax by 2ab 0, , Here,, , , , , , a, b, a1 , b1 , c1 a b , b, a, x, y, 1, , , 2, 2, 2, 2, 2b ab b, 2a a ab a b, x, y, 1, 2, , 2, b ab a ab a b, x, y, 1, , , b b a a a b b a, , Now,, x, 1, , b b a b a, , , , x, , b b a , b, ba, , And,, , y, 1, , a b a b a, , , , , a b a , ba, y a, y a, y , , Hence, x b, y a is the solution of the given system of equations., , 28., , b, a, x y a 2 b2 0, a, b, x y 2ab 0, Sol:, The given system of equation may be written as, b, a, x y a 2 b2 0, a, b, x y 2ab 0, Here,, b, a, a1 , b1 , c1 a 2 b 2 , a, b, a2 1, b2 1, and c2 2ab
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, By cross multiplication, we have, x, y, 1, , , a, a, b a, 2ab a 2 b 2 2ab a 2 b 2, , b, b, a b, x, y, 1, , , 2, 2, 2, 2, 2, 2, 2, b a2, 2a a b, 2b a b, ab, x, y, 1, , , , b2 a 2 b 2 a 2 b2 a 2, ab, Now,, x, 1, 2, 2, 2, b a2, b a, ab, ab, , x b2 a 2 2, b a2, , x ab, And,, y, 1, 2, 2, 2, b a2, b a, ab, ab, , y b 2 a 2 2, b a2, ab, , y b2 a 2 2, b a2, , y ab, , y ab, Hence, x ab, y ab is the solution of the given system of equations., , Exercise 3.5, In each of the following systems of equations determine whether the system has a unique, solution, no solution or infinitely many solutions. In case there is a unique solution, find it:, (1 −4), x 3y 3 0, 1., 3x 9 y 2 0, Sol:, The given system of equations may be written as
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , 4 k, , 2 2, i.e., k 4, Therefore, for all values of k, except 4, the given pair of equations will have a unique, solution., i.e.,, , 7., , 4x 5 y k, 2 x 3 y 12, , Sol:, The given system of equation is, 4x 5 y k 0, 2 x 3 y 12 0, The system of equation is of the form, a1 x b1 y c1 0, a2 x b2 y c2 0, , Where, a1 4, b1 5, c1 k, And, a2 2, b2 3, c2 12, For a unique solution, we must have, a1 b1, , a2 b2, , 4 5, , 2 3, , k is any real number., So, the given system of equations will have a unique solution for all real values of k., , , , 8., , x 2y 3, 5 x ky 7 0, , Sol:, The given system of equation is, x 2y 3 0, 5 x ky 7 0, The system of equation is of the form, a1 x b1 y c1 0, a2 x b2 y c2 0, , Where, a1 1, b1 2, c1 3, And, a2 5, b2 k , c2 7, For a unique solution, we must have
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , a1 b1, , a2 b2, 1 2, , 5 k, , k 10, So, the given system of equations will have a unique solution for all real values of k other, than 10., , , , Find the value of k for which each of the following systems of equations have definitely many, solution: (9-19), 2x 3y 5 0, 9., 6 x ky 15 0, Sol:, The given system of equation is, 2x 3y 5 0, 6 x ky 15 0, The system of equation is of the form, a1 x b1 y c1 0, a2 x b2 y c2 0, , Where, a1 2, b1 3, c1 5, And, a2 6, b2 k , c2 15, For a unique solution, we must have, a1 b1 c1, , a2 b2 c2, , 2 3, , 6 k, , 2k 18, 18, , k 9, 2, Hence, the given system of equations will have infinitely many solutions, if k 9., , , 10., , 4x 5 y 3, kx 15 y 9, , Sol:, The given system of equation is, 4x 5 y 3 0, kx 15 y 9 0, The system of equation is of the form
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , , , , , k 6, , 4 9, k 2, , 4 3, 2 4, k, 3, 8, k, 3, , 8, Hence, the given system of equations will have infinitely many solutions, if k ., 3, 12., , 8x 5 y 9, kx 10 y 18, , Sol:, The given system of equation is, 8x 5 y 9 0, kx 10 y 18 0, The system of equation is of the form, a1 x b1 y c1 0, a2 x b2 y c2 0, , Where, a1 8, b1 5, c1 9, And, a2 k , b2 10, c2 18, For a unique solution, we must have, a1 b1 c1, , a2 b2 c2, , , , 8 5, 9, , k 10 18, , Now,, , 8 5, , k 10, , 8 10 5 k, 8 10, , k, 5, , k 8 2 16, Hence, the given system of equations will have infinitely many solutions, if k 16.
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , a1 b1 c1, , a2 b2 c2, , , , 2, 3, k, , , k 1 k 1 3k, 2, 3, 3, k, , and, , k 1 k 1, k 1 3k, 2 k 2 3 k 1 and 3 3 k 2, , , , , , , 2k 4 3k 3 and 9 k 2, 4 3 3k 2k and 9 2 k, 7 k and 7 k, k 7 and k 7, , , , , , k 7 satisfies both the conditions, Hence, the given system of equations will have infinitely many solutions, if k 7., , Find the value of k for which the following system of equations has no solution: (20 – 25), kx 5 y 2, 20., 6x 2 y 7, Sol:, Given, kx 5 y 2, 6x 2 y 7, Condition for system of equations having no solution, a1 b1 c1, , a2 b2 c2, , k 5 2, , , 6 2 7, 2k 30, k 15, , , 21., , x 2y 0, 2 x ky 5 0, , Sol:, The given system of equation may be written as, x 2y 0, 2 x ky 5 0, The system of equation is of the form
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, a1 b1, 2 k, 4, , i.e., , k , a2 b2, k, 2, 3, , Hence, the given system of equations has no solutions, k , , 24., , 4, ., 3, , 2 x ky 11 0, 5x 7 y 5 0, , Sol:, The given system of equation is, 2 x ky 11 0, 5x 7 y 5 0, The system of equation is of the form, a1 x b1 y c1 0, a2 x b2 y c2 0, , Where, a1 2, b1 k , c1 11, And, a2 5, b2 7, c2 5, For a unique solution, we must have, a1 b1 c1, , a2 b2 c2, , , , , 2, , 5, 2, , 5, , k, 11, , 7 5, k, k, 11, and, , 7, 7 5, , Now,, , , , 2 k, , 5 7, 2 7 5k, , , , 5k 14, 14, , k, 5, 14, k 11, , Clearly, for, we have, 5, 7 5, Hence, the given system of equation will have no solution, if k , , 25., , kx 3 y 3, 12 x ky 6, , 14, 5
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, Sol:, kx 3 y 3, 12 x ky 6, For no solution, , , , a1 b1 c1, , a2 b2 c2, , k 2 3, , 12 k 6, k 3, , 12 k, k 2 36, k 6 i.e.,, , k 6, 6, , Also,, , 3 3, , k 6, 3 6, k, 3, k 6, k 6 satisfies both the condition, Hence, k 6, 26., , For what value of 𝛼, the following system of equations will be inconsistent?, 4 x 6 y 11 0, 2 x ky 7 0, Sol:, The given system of equation may be written as, 4 x 6 y 11 0, 2 x ky 7 0, The system of equation is of the form, a1 x b1 y c1 0, a2 x b2 y c2 0, , Where, a1 4, b1 6, c1 11, And, a2 2, b2 k , c2 7, For a unique solution, we must have, a1 b1 c1, , a2 b2 c2, Now,
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , 3, , , , , , , , , , And,, , 3, , , , , , 3 , , 3, , , , 3 3, 33 , 6, 6, , , 12, , , , 3, , , , , , , 36, 6, , , , 6, , 2, , , , 6, , Hence, the given system of equation will have no solution, if 6., 28., , Find the value of k for which the system, kx + 2y = 5, 3x + y = 1, has (i) a unique solution, and (ii) no solution., Sol:, The given system of equation may be written as, kx 2 y 5 0, 3x y 1 0, It is of the form, a1 x b1 y c1 0, a2 x b2 y c2 0, , Where, a1 k , b1 2, c1 5, And, a2 3, b2 1, c2 1, (i) The given system will have a unique solution, if, a1 b1, , a2 b2, , k 2, , 3 1, , k 6, So, the given system of equations will have a unique solution, if k 6, (ii) The given system will have no solution, if,
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, a1 2 b1, k, , , a2 3 b2 5, , And,, , c1 1 1, , , c2 7 7, , Clearly,, , a1 c1, , a2 c2, , So, whatever be the value of k, we cannot have, a1 b1 c1, , a2 b2 c2, Hence, there is no value of k, for which the given system of equations has infinitely many, solutions., 31., , For what value of k, the following system of equations will represent the coincident lines?, x 2y 7 0, 2 x ky 14 0, Sol:, The given system of equations may be written as, x 2y 7 0, 2 x ky 14 0, The given system of equations is of the form, a1 x b1 y c1 0, a2 x b2 y c2 0, , Where, a1 1, b1 2, c1 7, And a2 2, b2 k , c2 14, The given equations will represent coincident lines if they have infinitely many solutions,, The condition for which is, a1 b1 c1, 1 2 7, k4, a2 b2 c2, 2 k 14, Hence, the given system of equations will represent coincident lines, if k 4., 32., , Obtain the condition for the following system of linear equations to have a unique solution, ax by c, lx my n, Sol:, The given system of equations may be written as, ax by c 0, lx my n 0
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , a1 b1 c1, , a2 b2 c2, , , , , 2a 1, 3, 5, , , 3, b 1 2, 5, 3, 5, 2 2a 1 , and, , 2, b 1 2, 2 2a 1 5 3 and 3 2 5 b 1, , , , , 4a 2 15 and 6 5b 5, 4a 15 2 and 6 5 5b, 17, 11, , a, and, b, 4, 5, 17, 11, , a, and b , 4, 5, Hence, the given system of equations will have infinitely many solutions,, 17, 11, If a , and b ., 4, 5, 34., , Find the values of a and b for which the following system of linear equations has infinite, number of solutions:, 2x 3y 7, , a b x a b 3 y 4a b, Sol:, The given system of equations may be written as, 2x 3y 7 0, , a b x a b 3 y 4a b 0, It is of the form, a1 x b1 y c1 0, a2 x b2 y c2 0, , Where, a1 2, b1 3, c1 7, And a2 a b, b2 a b 3 , c2 4a b , The given system of equations will have infinite number of solutions, if
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , a1 b1 c1, , a2 b2 c2, , , 2, 3, 7, , , a b a b 3 4a b , , , , 2, 3, 3, 7, , and, , a b a b 3, a b 3 4a b, , , , 2 a b 3 3 a b and 3 4a b 7 a b 3, , , , , 6 3a 2a 3b 2b and 12a 7a 3b 7b 21, 6 a b and 5a 4b 21, , Now,, , a b 6, , a 6 b, Substituting the value of ‘a’ in 5a 4b 2, we get, 5 b 6 4b 21, , , , , 5b 30 4b 21, 9b 21 30, 9b 9, 9, , b, 1, 9, Putting b 1 in a b 6, we get, , a 1 6 1 6 5, Hence, the given system of equations will have infinitely many solutions,, If a 5 and b 1., 35., , Find the values of p and q for which the following system of linear equations has infinite, number of solutions:, 2x 3y 9, , p q x 2 p q y 3 p q 1, Sol:, The given system of equations may be written as, 2x 3y 9 0, , p q x 2 p q y 3 p q 1 0, It is of the form, a1 x b1 y c1 0, a2 x b2 y c2 0, , Where, a1 2, b1 3, c1 9
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, And a2 p q, b2 2 p q, c2 3 p q 1, The given system of equations will have infinite number of solutions, if, a1 b1 c1, , a2 b2 c2, , , , , , 2, 2, 9, , , p q 2 p q 3 p q 1, 2, 3, 3, , , p q 2 p q p q 1, 2, 3, 3, 3, , and, , p q 2p q, 2 p q p q 1, , , , 2 2 p q 3 p q and p q 1 2 p q, , , , , 4 p 2q 3 p 3q and 2 p p q q 1, p 5q 0 and p 2q 1, , , , , p 5q p 2q 1, [On adding], 3q 1, 1, , q, 3, 1, Putting q in p 5q, we get, 3, 1, p 5 0, 3, 5, , p, 3, Hence, the given system of equations will have infinitely many solutions,, 5, 1, If p and q , 3., 3, 36., , Find the values of a and b for which the following system of equations has infinitely many, solutions:, 2x 3 y 7, , a b x a b y 3a b 2, Sol:, 2x 3y 7 0, , a b x a b y 3a b 2 0, Here, a1 2, b1 3, c1 7, , a2 a b , b2 a b , c2 3a b 2
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , a1 b1 c1, , a2 b2 c2, , , , , , , , 2a 1 3, 5, , , 3, b 2 3, 2a 1 3, 5, , , 3, b2 3, 2a 1 5, 3, 5, and, , 3, 3, b2 3, 3 2a 1, 5 and 9 5 b 2 , 3, 2a 1 5 and 9 5b b 2 , , , , 2a 5 1 and 9 10 5b, 6, , a and 1 5b, 2, 1, , a 3 and b, 5, 1, , a 3 and b , 5, Hence, the given system of equations will have infinitely many solutions,, 1, If a 3 and b , 5, (ii), 2 x 2a 5 y 5, , 2b 1 x 9 y 15, Sol:, The given system of equations is, 2 x 2a 5 y 5 0, , 2b 1 x 9 y 15 0, It is of the form, a1 x b1 y c1 0, a2 x b2 y c2 0, , Where, a1 x b1 2a 5 , c1 5, And, a2 2b 1 , b2 9, c2 15, The given system of equations will be have infinite number of solutions, if
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , a1 b1 c1, , a2 b2 c2, , , , a 1, 3, 2, , , 6, 1 2b 6, a 1, 3, 1, , , 6, 1 2b 3, a 1 1, 3, 1, and, , b, 3, 1 2b 3, 3 a 1 6 and 3 3 1 2b, , , , a 1 2 and 9 1 2b, , , , , , , , , a 2 1 and 2b 1 9, a 3 and 2b 8, 8, , a 3 and b , 4, 2, Hence, the given system of equations will have infinitely many solutions,, If a 3 and b 4., (iv), 3x 4 y 12, , a b x 2 a b y 5a 1, Sol:, The given system of equations is, 3x 4 y 12 0, , a b x 2 a b y 5a 1 0, It is of the form, a1 x b1 y c1 0, a2 x b2 y c2 0, , Where, a1 3, b1 4, c1 12, And, a2 a b, b2 2 a b , c2 5a 1, The given system of equations will be have infinite number of solutions, if
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , , , 2, 3, 3, 7, , and, , a 1 a 2, a 2 3a, 2 a 2 3 a 1 and 3 3a 7 a 2 , , , , , 2a 4a 3 and 9a 7a 14, 2a 3a 3 and 9a 7a 14, , , , a 7 and 2a 14, , , , a 7 and a , , , , 14, 7, 2, , , a7, Hence, the given system of equations will have infinitely many solutions,, If a 7., , Exercise 3.6, 1., , 5 pens and 6 pencils together cost Rs 9 and 3 pens and 2 pencils cost Rs 5. Find the cost of, 1 pen and 1 pencil., Sol:, Let the cost of a pen be Rs x and that of a pencil be Rs y. Then,, 5x 6 y 9, ....... i , , and, , 3x 2 y 5, , ...... ii , , Multiplying equation (i) by 2 and equation (ii) by 6, we get, 10 x 12 y 18, ...... iii , , 18 x 12 y 30, , ...... iv , , Subtracting equation (iii) by equation (iv), we get, 18 x 10 x 12 y 12 y 30 18, , 8 x 12, 12 3, , x 1.5, 8 2, Substituting x 1.5 in equation (i), we get, 5 1.5 6 y 9, , , , , , 7.5 6 y 9, 6 y 9 7.5, 6 y 1.5, 1.5 1, , y, 0.25, 6 4, Hence, cost of one pen Rs 1.50 and cost of one pencil Rs 0.25
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, 2., , 7 audio cassettes and 3 video cassettes cost Rs 1110, while 5 audio cassettes and 4 video, cassettes cost Rs 1350. Find the cost of an audio cassette and a video cassette., Sol:, Let the cost of a audio cassette be Rs x and that of a video cassette be Rs y. Then,, , and, , 7 x 3 y 1110, , .... i , , 5 x 4 y 1350, , .... ii , , Multiplying equation (i) by 4 and equation (ii) by 3, we get, 28 x 12 y 4440, ...... iii , , 15 x 12 y 4050, , ...... iv , , Subtracting equation (iv) from equation (iii), we get, 28 x 15 x 12 y 12 y 4440 4050, , , , 13x 390, 390, , x, 30, 13, Substituting equation (iv) from equation (iii), we get, 28x 15 x 12 y 12 y 4440 4050, , , , 13x 390, 390, , x, 30, 13, Substituting x 30 in equation (i), w get, 7 30 3 y 1110, , 210 3 y 1110, , 3 y 1110 210, , 3 y 900, 900, , y, 300, 3, Hence, cost of one audio cassette Rs 30 and cost of one video cassette Rs 300, 3., , Reena has pens and pencils which together are 40 in number. If she has 5 more pencils and, 5 less pens, then nuimber of pencils would become 4 times the number of pens. Find the, original number of pens and pencils., Sol:, Let the number of pens be x and that of pencil be y. then,, x y 40, .......(i ), and, , y 5 4 x 5
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , , , , y 5 4 x 20, 5 20 4 x y, , , , 4 x y 25, , ...... ii , , Adding equation (i) and equation (ii), we get, x 4 x 40 25, , 5 x 65, 65, , x, 13, 5, Putting x 13 in equation (i), we get, 13 y 40, , , , y 40 13 27, , Hence, Reena has 13 pens 27 pencils., 4., , 4 tables and 3 chairs, together, cost Rs 2,250 and 3 tables and 4 chairs cost Rs 1950. Find, the cost of 2 chairs and 1 table., Sol:, Let the cost of a table be Rs x and that of a chairs be Rs y. Then,, 4 x 3 y 2, 250, , ..... i , , and , 3x 4 y 1950, .....(ii), Multiplying equation (i) by 4 and equation (ii) by 3, we get, 16 x 12 y 9000, ......(iii ), 9 x 12 y 5850, , ....... iv , , Subtracting equation (iv) by equation (iii), we get, 16 x 9 x 9000 5850, , 7 x 3150, 3150, , x, 450, 7, Putting x 450 in equation (i), we get, 4 450 3 y 2, 250, , , , , , , , 1800 3 y 2250, 3 y 2250 1800, 3 y 450, 450, y, 150, 3, 2 y 2 150 300, , Cost of 2 chairs Rs 300 and cost of 1 table Rs 450, The cost of 2 chairs and 1 table 300 450 Rs 750
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, 5., , 3 bags and 4 pens together cost Rs 257 whereas 4 bags and 3 pens together cost R 324., Find the total cost of 1 bag and 10 pens., Sol:, Let the cost of a bag be Rs x and that of a pen be Rs y . Then,, , and ,, , 3x 4 y 257, , ..... i , , 4 x 3 y 324, , ..... ii , , Multiplying equation (i) by 3 and equation (ii) by 4, we get, 9 x 12 y 770, .... iii , , 16 x 12 y 1296, , ...... iv , , Subtracting equation (iii) by equation (iv), we get, 16 x 9 x 1296 771, , 7 x 525, 525, , x, 75, 7, Cost of a bag Rs 75, Putting x 75 in equation (i), we get, 3 75 4 y 257, , , , , , 225 4 y 257, 4 y 257 225, 4 y 32, 32, , y, 8, 4, Cost of a pen Rs 8, Cost of 10 pens 8 10 Rs 80, Hence, the total cost of 1 bag and 10 pens 75 80 Rs 155., 6., , 5 books and 7 pens together cost Rs 79 whereas 7 books and 5 pens together cost Rs 77., Find the total cost of 1 book and 2 pens., Sol:, Let the cost of a book be Rs x and that of a pen be Rs y. Then,, 5 x 7 y 79, ..... i , , and ,, , 7 x 5 y 77, , .... ii , , Multiplying equation (i) by 5 and equation (ii) by 7, we get, 25 35 y 395, ...... iii , , 49 x 35 y 539, , ....... iv
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, Subtracting equation (iii) by equation (iv), we get, 49 x 25 x 539 395, , 24 x 144, 144, , x, 6, 24, Cost of a book Rs 6, Putting x 6 in equation (i), we get, 5 6 7 y 79, , , , , , 30 7 y 79, 7 y 79 30, 7 y 49, 79, , y, 7, 7, Cost of a pen Rs 7, Cost of 2 pens 2 7 Rs 14, Hence, the total cost of 1 book and 2 pens 6 14 Rs 20, 7., , A and B each have a certain number of mangoes. A says to B, “if you give 30 of your, mangoes, I will have twice as many as left with you.” B replies, “if you give me 10, I will, have thrice as many as left with you.” How many mangoes does each have?, Sol:, Suppose A has x mangoes and B has y mangoes, According to the given conditions, we have, x 30 2 y 30 , , , , , x 30 2 y 60, x 2 y 60 30, , , , x 2 y 90, , .... i , , And, y 10 3 x 10 , , , , , y 10 3x 30, 10 30 3x y, , , , 3x y 40, , .... ii , , Multiplying equation (i) by 3 and equation (ii) by 1, we get, 3x 6 y 270, ..... iii , , 3x y 40, , ...... iv , , Subtracting equation (iv) by equation (iii), we get
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , 6 y y 270 40, , , , 6 y y 310, 5 y 310, 310, , y, 62, 5, Putting x 62 in equation (i), we get, x 2 62 90, , x 124 90, , x 90 124, , x 34, Hence, A has 34 mangoes and B has 62 mangoes, 8., , On selling a T.V. at 5%gain and a fridge at 10% gain, a shopkeeper gains Rs 2000. But if, he sells the T.V. at 10% gain and the fridge at 5% loss. He gains Rs 1500 on the, transaction. Find the actual prices of T.V. and fridge., Sol:, Let the price of a T.V. be Rs x and that of a fridge be Rs y. Then, we have, 5 x 10 y, , 2000, 100 100, , 5 x 10 y 200000, , , , 5 x 2 y 200000, , , , x 2 y 400000, , ... i , , 10 x 5 y, , 1500, 100 100, 10 x 5 y 150000, , And,, , 5 2 x y 150000, 2 x y 30000, Multiplying equation (ii) by 2, we get, 4 x 2 y 6000, ..... iii , Adding equation (i) and equation (iii), we get, x 4 x 40000 60000, , , 5 x 100000, , x 20000, Putting x 20000 in equation (i), we get
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , 20000 2 y 40000, , 2 y 40000 20000, 20000, , y, 10000, 2, Hence, the actual price of T.V Rs 20, 000 and, the actual price of fridge Rs 10, 000, 9., , The coach of a cricket team buys 7 bats and 6 balls for Rs 3800. Later, he buys 3 bats and 5, balls for Rs 1750. Find the cost of each bat and each ball., Sol:, Let the cost of bat and a ball be x and y respectively, According to the given information, …….(1), 7 x 6 y 3800, 3x 5 y 1750, , ……..(2), , From (1), we obtain, 3800 7 x, ……..(3), y, 6, Substituting this value in equation (2), we obtain, 3800 7 x , 3x 5 , 1750, 6, , , 9500 35 x, 3x , , 1750, 3, 6, 35x, 9500, 3x , 1750 , 6, 3, 35x 5250 9500, 3x , , 6, 3, 17 x 4250, , , 6, 3, 17 x 8500, , x 500, , 4, , Substituting this in equation (3), we obtain, 3800 7 500, y, 6, 300, , 50, 6, Hence, the cost of a bat is Rs 500 and that of a ball is Rs 50., Concept Insight: Cost of bats and balls needs to be found so the cost of a ball and bat will, be taken as the variables. Applying the conditions of total cost of bats and balls algebraic
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, equations will be obtained. The pair of equations can then be solved by suitable, substitution., 10., , One says, “Give me a hundred, friend! I shall then become twice as rich as you.” The other, replies, “If you give me ten, I shall be six times as rich as you.” Tell me what is the amount, of their respective capital?, Sol:, Let the money with the first person and second person be Rs x and Rs y respectively., According to the question, x 100 2 y 100 , x 100 2 y 200, x 2 y 300, , ……(1), , 6 x 10 y 10 , 6 x 30 y 10, 6 x y 70, , …….(2), , Multiplying equation (2) by 2, we obtain, …….(3), 12 x 2 y 140, Subtracting equation (1) from equation (3), we obtain:, 11x 140 300, , 11x 440, x 40, Putting the value of x in equation (1), we obtain, 40 2 y 300, 40 300 2 y, 2 y 340, y 170, Thus, the two friends had Rs 40 and Rs 170 with them., 11., , A lending library has a fixed charge for the first three days and an additional charge for, each day thereafter. Saritha paid Rs 27 for a book kept for seven days, while Susy paid Rs, 21 for the book she kept for five days. Find the fixed charge and the charge for each extra, day., Sol:, Let the fixed charge for first three days and each day charge thereafter be Rs x and Rs y, respectively., According to the question,
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Class X, Chapter 3 – Pair of Linear Equations in Two Variables, Maths, ______________________________________________________________________________, , x 4 y 27, , ... 1, , x 2 y 21, , ... 2 , , Subtracting equation (2) from equation (1), we obtain:, 2y 6, , y3, Subtracting the value of y in equation (1), we obtain, x 12 27, x 15, Hence, the fixed charge is Rs 15 and the charge per day is Rs 3.
