Simultaneous linear equations - Mathematics …

1 Simultaneous linearequationsmc- Simultaneous -2009-1 The purpose of this section is to look at the solution of Simultaneous linear equations . We willsee that solving a pair of Simultaneous equations is equivalent to finding the location of the pointof intersection of two straight order to master the techniques explained here it is vital that you undertake plenty of practiceexercises so that all this becomes second nature. To help youto achieve this, the unit includesa number of such reading this text, and/or viewing the video tutorial on this topic, you should be able to: solve pairs of Simultaneous linear equations recognise that this is equivalent to finding the point of intersection of two straight Simultaneous equations - method of Simultaneous equations - method of mathcentre 20091.

2 IntroductionThe purpose of this section is to look at the solution of elementary Simultaneous linear we do that, let s just have a look at a relatively straightforward single equation. Theequation we are going to look at is2x y= 3 This is a linear equation. It is a linear equation because there are no terms involvingx2, y2orx y, or indeed any higher powers ofxandy. The only terms we have got are terms inx, termsinyand some numbers. So this is alinear can rearrange it so that we obtainyon its own on the left hand side. We can addyto eachside so that we get2x= 3 +yNow let s take 3 away from each 3 =yThis gives us an expression fory: namelyy= 2x we choose a value forx, sayx= 1, thenywill be equal to:y= 2 1 3 = 1 Suppose we choose a different value forx, sayx= 2 2 3 = 1 Supppose we choose another value forx, sayx= 2 0 3 = 3 For every value ofxwe can generate a value can plot these as points on a graph.

3 We can plot the first as the point(1, 1). We can plotthe second one as the point(2,1), and the third one as the point(0, 3)and so on. Plottingthe points on a graph, as shown in Figure 1, we see that these three points lie on a straight is the line with equationy= 2x 3. It is a straight line and this is another reason for callingthe equation alinear 2x 3 Figure 1. Graph ofy= 2x mathcentre 2009 Suppose we take a second linear equation3x+ 2y= 8and plot its graph on the same figure. Aquick way to achieve this is as 0,2y= 8,soy= 4. Therefore the point(0,4)lies on the 0,3x= 8,sox=83= 223. Therefore the point(223,0)lies on the this is a linear equation we know its graph is a straight line, so we can obtain this byjoining up the points.

4 Both straightline graphs are shown inFigure 2x 33x+ 2y= 8 Figure 2. Graphs ofy= 2x 3and3x+ 2y= 8 When we solve a pair of Simultaneous equations what we are actually looking for is the intersectionof two straight lines because it is this point that satisfies both equations at the same time. FromFigure 2 we see that this occurs at the point wherex= 2andy= course it could happen that we have two parallel lines; they would never meet, and hence thesimultaneous equations would not have a solution. We shall observe this behaviour in one of theexamples which PointWhen solving a pair of Simultaneous linear equations we are,in fact, finding a common point -the point of intersection of the two mathcentre 20092.

5 Solving Simultaneous equations - method of substitutionHow can we handle the two equations algebraically so that we do not have to draw graphs? Weare going to look at two methods of solution. In this Section we will look at the first method -the method of us return to the two equations we met in Section y= 3(1)3x+ 2y= 8(2)By rearranging Equation (1) we findy= 2x 3(3)We can now substitute this expression foryinto Equation (2).3x+ 2(2x 3) = 83x+ 4x 6 = 87x 6 = 87x= 14x= 2 Finally, using Equation (3),y= 2 2 3 = 1. Sox= 2,y= 1is the solution to the pair ofsimultaneous solution should always be checked by substituting backinto both original equations to ensurethat the left- and right- hand sides are equal for these values ofxandy.

6 So, withx= 2,y= 1,the left-hand side of Equation (1) is2(2) 1 = 3, which is the same as the right-hand side. Withx= 2,y= 1, the left-hand side of Equation (2) is3(2) + 2(1) = 8, which is the same as theright-hand s have a look at another example using this particular example we are going to use is7x+ 2y= 47(1)5x 4y= 1(2)Now we need to make a choice. We need to choose one of these two equations and re-arrangeit to obtain an expression fory, or if we wish, forxThe choice is entirely ours and we have to make the choice based upon what we feel will bethe simplest. Looking at a pair of equations like this, it is often difficult to know which is mathcentre 2009 Let s choose Equation (2) and rearrange it to find an expression 4y= 15x= 1 + 4yby adding4yto each sidex=1 + 4y5by dividing both sides by 5We now use this expression forxand substitute it in Equation (1).

7 7(1 + 4y5)+ 2y= 47 Now multiply throughout by 5. Why? Because we want to get rid of the fraction and the wayto do that is to multiply everything by (1 + 4y) + 10y= 235 Now we need to multiply out the brackets7 + 28y+ 10y= 235 Gather they s and subtract 7 from each side to get38y= 228 Soy=22838= 6So we have established thaty= 6. Having done this we can substitute it back into the equationthat we first had + 4y5=1 + 245and sox= 5So again, we have our pair of values - our solution to the pair of Simultaneous equations . In orderto check that our solution is correct these values should be substituted into both equations toensure they balance.

8 So, withx= 5,y= 6, the left-hand side of Equation (1) is7(5)+2(6) = 47,which is the same as the right-hand side. Withx= 5,y= 6, the left-hand side of Equation (2)is5(5) 4(6) = 1, which is the same as the right-hand Solve the following pairs of Simultaneous equations :a)y= 2x+ 3y= 5x 3b)y= 3x 12x+ 4y= 10c)6x+y= 45x+ 2y= 1d)x 3y= 12x+ 5y= 35e)2x+13y= 13x+ 5y= 6f)4x+ 3y= 52x 34y= mathcentre 20093. Solving Simultaneous equations - method of eliminationWe illustrate the second method by solving the simultaneouslinear 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 ofythe same in both equations .

9 Equation (1) becomes14x+ 4y= 94(3)If we now add Equation (2) and Equation (3) we will find that theterms involvingydisappear:5x 4y= 1+14x+ 4y=9419x= 95and sox=9519= 5 Now that we have a value forxwe can substitute this into Equation (2) in order to 4y= 15 5 4y= 125 = 4y+ 124 = 4yy= 6 The solution isx= 5,y= ExamplesSolve the Simultaneous equations3x+ 7y= 27(1)5x+ 2y= 16(2)We will multiply Equation (1) by 5 and Equation (2) by 3 because this will make the coefficientsofxin both equations the + 35y= 135(3)15x+ 6y= 48(4)If we now subtract Equation (4) from Equation (3) we can eliminate the terms mathcentre 200915x+ 35y= 135 15x+ 6y=4829y= 87from whichy=8729= 3If we substitute this result in Equation (1)

10 We can + 7y= 273x+ 21 = 273x= 6x= 2As before, the solution should be checked by substitution into the original equations . So, withx= 2,y= 3, the left-hand side of Equation (1) is3(2) + 7(3) = 27, which is the same as theright-hand side. Withx= 2,y= 3, the left-hand side of Equation (2) is5(2) + 2(3) = 16, whichis the same as the right-hand the examples that we have looked at so far have all had whole number coefficients; let s havea look at a couple that don t look like the ones we have just the Simultaneous equationsx= 3yx3 y= 34 First of all let us rearrange the first equation so thatxandyterms are on the left.