The previous example lent itself to the elimination method because the coefficients of y were readily eliminated through addition. But what if this is not the case and the coefficients of x or y are neither the same nor opposites? Carefully study the next example. Example Solve the following system by the elimination strategy . x – 2y = 7 3x + 4y = 1 Neither addition nor subtraction of these two equations will eliminate one of the variables. But recall from earlier that we can multiply (or divide ) both sides of an equation by the same non-zero number without changing the solution. So if we multiply both sides of the first equation by 3 , then subtract , we can eliminate x. Or , multiply both sides of the first equation by 2 then add to eliminate y . Both solutions follow.
10C – C5-6 – LG – Systems of Linear Equations Page 21 of 42 Method 1 We will multiply the first equation by 3 then subtract to eliminate x . 3(x – 2y) = 3(7) first equation is multiplied by 3 on both sides 3x – 6y = 21 – (3x + 4y = 1) 0x – 10y = 20 -10y = 20 y = -2 Substitute y = - 2 into either original equation. Let‟s use the first equation. x – 2y = 7 x – 2(-2) = 7 x + 4 = 7 x = 3 The solution is (3, -2) Method 2 Let‟s multiply the first equation by 2 then add to eliminate y . 2(x – 2y) = 2(7) first equation is multiplied by 2 on both sides 2x – 4y = 14 3x + 4y = 1 5x + 0y = 15 5x = 15 x = 3 Substitute into the first equation to solve for y : x – 2y = 7 (3) – 2y = 7 -2y = 4 y = -2 The solution is (3, -2). As you can see, both methods result in the same solution. Check the solution to this system on your own. Line up like terms vertically then subtract. Watch your signs! Line up like terms vertically then add. When using the elimination method of solving linear systems, if the signs of the coefficients of the variable to be eliminated are the same, the equations are subtracted. If the signs of the coefficients of the variable to be eliminated are opposite, the equations are added.
10C – C5-6 – LG – Systems of Linear Equations Page 22 of 42 In the previous example, we were required to multiply only one equation by a number in order to eliminate one variable. This is not always the case, however. Carefully read the next example. Example Solve the following system using the elimination strategy . 2x + 3y = 18 5x – 4y = -1 If we choose to eliminate x , the coefficient of x will need to be 10 in each equation. Why 10, you ask? Well, 10 is the least common multiple for 2 and 5! So, we must multiply the first equation by 5 and the second equation by 2 then subtract to eliminate x. Method 1 2x + 3y = 18 5x – 4y = -1 5(2x + 3y) = 5(18) 2(5x – 4y) = 2(-1) 10x + 15y = 90 – (10x – 8y = -2) 0x + 23y = 92 23y = 92 y = 4 Substitute into the second equation to solve for x: 5x – 4y = -1 5x – 4(4) = -1 5x – 16 = -1 5x = 15 x = 3 The solution is (3, 4). To eliminate y , the coefficient of y will need to be 12 in the first equation and -12 in the second equation. So we must multiply the first equation by 4 on both sides and the second equation by 3 on both sides. We then add the equations to eliminate y.