Method 2 When no equation in a linear system has a variable with coefficient 1, it is helpful if there are two like terms where one term is a multiple of the other term. 5x 3y = 18 – 4x 6y = 18 – In equation , the term can be written as 2(–3y): –6y 4x + 2(–3y) = 18 Solve equation for –3y. 5x 3y = 18 – = 18 5x –3y – Substitute for in equation . –3y 4x + 2(–3y) = 18 4x + 2(18 5x) = 18 Simplify, then solve for x. – 4x + 36 10x = 18 – = –6x –18 = 3 x Substitute = 3 into equation . x 5x 3y = 18 – 5(3) 3y = 18 – 15 3y = 18 – = 3 Solve for y. –3y = (Solution continues.) y –1
Method 2 When no equation in a linear system has a variable with coefficient 1, it is helpful if there are two like terms where one term is a multiple of the other term. 5x – 3y = 18 4x – 6y = 18 In equation , the term –6y can be written as 2(–3y): 4x + 2(–3y) = 18 Solve equation for –3y. 5x – 3y = 18 –3y = 18 – 5x Substitute for –3y in equation . 4x + 2(–3y) = 18 4x + 2(18 – 5x) = 18 Simplify, then solve for x. 4x + 36 – 10x = 18 –6x = –18 x = 3 Substitute x = 3 into equation . 5x – 3y = 18 5(3) – 3y = 18 15 – 3y = 18 –3y = 3 Solve for y. y = –1 (Solution continues.)