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Liner equation, The purpose of this section is to look at the solution of elementary simultaneous linear equations., Before we do that, let's just have a look at a relatively straightforward single equation. The, equation we are going to look at is, 2r – y = 3, This is a linear equation. It is a linear equation because there are no terms involving r2, y? or, xx y, or indeed any higher powers of r and y. The only terms we have got are terms in r, terms, in y and some numbers. So this is a linear equation., We can rearrange it so that we obtain y on its own on the left hand side. We can add y to each, side so that we get, 2x = 3 + y, Now let's take 3 away from each side., 2r – 3 = y, This gives us an expression for y: namely y = 2r – 3., Suppose we choose a value for r, say r 1, then y will be equal to:, y = 2 x 1 – 3 = -1, Suppose we choose a different value for r, say r = 2., y = 2 x 2 –3 = 1, Supppose we choose another value for r, say r = 0., y = 2 x 0 – 3 = -3, For every value of r we can generate a value of y., We can plot these as points on a graph. We can plot the first as the point (1, –1). We can plot, the second one as the point (2,1), and the third one as the point (0, -3) and so on. Plotting, the points on a graph, as shown in Figure 1, we see that these three points lie on a straight line., This is the line with equation y = 2x, calling the equation a linear equation., 3. It is a straight line and this is another reason for, 2-, y = 2r – 3, 1-, -1-, -2-, -3, Figure 1. Graph of y = 2x – 3.

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Suppose we take a second linear equation 3r + 2y = 8 and plot its graph on the same figure. A, quick way to achieve this is as follows., When r = 0, 2y = 8, so y = 4. Therefore the point (0,4) lies on the line., %3D, When y = 0, 3r = 8, so r = -= 2-. Therefore the point (2½,0) lies on the line., 3, %3D, %3D, Because this is a linear equation we know its graph is a straight line, so we can obtain this by, joining up the points. Both straightline graphs are shown in Figure 2., 3.r+2y = 8, %3D, y = 2r – 3, -1-, -2-, Figure 2. Graphs of y = 2r – 3 and 3r + 2y = 8, %3D, When we solve a pair of simultaneous equations what we are actually looking for is the intersec-, tion of two straight lines because it is this point that satisfies both equations at the same time., From Figure 2 we see that this occurs at the point where r = 2 and y = 1., Of course it could happen that we have two parallel lines; they would never meet, and hence, the simultaneous equations would not have a solution. We shall observe this behaviour in one, of the examples which follows.

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2. Solving simultaneous equations - method of substitution, How can we handle the two equations algebraically so that we do not have to draw graphs? We, are going to look at two methods of solution. In this Section we will look at the first method -, the method of substitution., Let us return to the two equations we met in Section 1., 2r - y = 3, (1), 3r + 2y, 8., (2), By rearranging Equation (1) we find, y = 2x – 3, (3), We can now substitute this expression for y into Equation (2)., 3r + 2(2г - 3), 8., 3x + 4x – 6 =, 8, 7r – 6 =, 7x =, 14, x = 2, Finally, using Equation (3), y = 2 x 2 – 3 = 1. So r = 2, y = 1 is the solution to the pair of, simultaneous equations., This solution should always be checked by substituting back into both original equations to, ensure that the left- and right- hand sides are equal for these values of r and y. So, with r = 2,, y = 1, the left-hand side of Equation (1) is 2(2) – 1 = 3, which is the same as the right-hand, side. With r = 2, y = 1, the left-hand side of Equation (2) is 3(2) +2(1) = 8, which is the same, as the right-hand side., %3D, Example, Let's have a look at another example using this particular method., The example we are going to use is, 7r + 2y =, 47, (1), 5x – 4y, 1, (2), %3D, Now we need to make a choice. WVe need to choose one of these two equations and re-arrange it, to obtain an expression for y, or if we wish, for r, The choice is entirely ours and we have to make the choice based upon what we feel will be, the simplest. Looking at a pair of equations like this, it is often difficult to know which is the, simplest.

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Let's choose Equation (2) and rearrange it to find an expression for r., 5x – 4y = 1, 5x, 1+ 4y, by adding 4y to each side, %3D, 1+ 4y, by dividing both sides by 5, We now use this expression for r and substitute it in Equation (1)., 1+ 4y, + 2y, 47, Now multiply throughout by 5. Why? Because we want to get rid of the fraction and the way, to do that is to multiply everything by 5., 7(1 + 4y) + 10y = 235, Now we need to multiply out the brackets, 7+ 28y + 10y, = 235, Gather the y's and subtract 7 from each side to get, 38y = 228, So, 228, y =, = 6, 38, So we have established that y = 6. Having done this we can substitute it back into the equation, that we first had for r., 1+ 4y, 1+ 24, and so, r = 5, So again, we have our pair of values - our solution to the pair of simultancous equations. In order, to check that our solution is correct these values should be substituted into both equations to, ensure they balance. So, with r = 5, y = 6, the left-hand side of Equation (1) is 7(5)+2(6) = 47,, which is the same as the right-hand side. With r 5, y = 6, the left-hand side of Equation (2), is 5(5) – 4(6) = 1, which is the same as the right-hand side., Exercises, 1. Solve the following pairs of simultaneous equations:, 2.r + 3, 3r, 1, 6r +, 4, a), b), 2.r + 4y, 5x + 2y, 5x, 3, 10, 1, %3D, 2x + y, 4r + 3y, f), 2x, 3y, 1, 1, %3D, d), 2r + 5y, e), 3r, 6., 35, + 5y, %3D, %3D, 51, I|||, I|||, I|||

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3. Solving simultaneous equations - method of elimination, We illustrate the second method by solving the simultaneous linear equations:, 7x + 2y =, 47, (1), 5x - 4y, 1, (2), We are going to multiply Equation (1) by 2 because this will make the magnitude of the coeffi-, cients of y the same in both equations. Equation (1) becomes, 14.r + 4y = 94, (3), If we now add Equation (2) and Equation (3) we will find that the terms involving y disappear:, 5x, +, 14x, 4y, 1, %3D, + 4y, 94, %3D, 19r, 95, and so, 95, 19, Now that we have a value for r we can substitute this into Equation (2) in order to find y., Substituting, 5x – 4y = 1, 5х5 — 4у, = 1, 25 = 4y + 1, 24 = 4y, y = 6, The solution is r = 5, y = 6., 4. Examples, Solve the simultaneous equations, 3т + 7у, 27, (1), 5r + 2y, 16, (2), We will multiply Equation (1) by 5 and Equation (2) by 3 because this will make the coefficients, of r in both equations the same., 15л + 35у, 135, (3), %3D, 15r + 6у, 48, (4), If we now subtract Equation (4) from Equation (3) we can eliminate the terms involving r.