1. SOLVING SYSTEMS of EQUATIONS MATH REVIEW

2. Suppose… … you want to solve a set of two linear equations: y = 5z – 4 and y = -4z + 2. There are two methods.

3. Method 1: Addition/Subtraction (Elimination) (multiply 1 st equation by 4) (×4) {y = 5z – 4} (multiply 2 nd equation by 5) (×5) {y = -4z + 2} (add both equations) 4y + 5y = 20z + (-20z) – 16 + 10 (solve for y) 9y = -6 note how the terms cancel out y = -2/3 (replace solution into one of the equations:) y = -4z + 2  z = (2 – y)/4 = (2 – (-2/3))/4 = 2/3 this sign means “therefore” Solution: y = -2/3 and z = +2/3

4. Method 2: Substitution Replace first equation into second: 5z – 4 = -4z + 2 Solve it for z: 5z + 4z = 4 + 2 9z = 6 z = 2/3 Substitute to find y: y = -4(2/3) + 2 = -2/3. Solution: y = -2/3 and z = +2/3

5. Your Turn! Solve: y = 2x – 7 (5, 3) solutions (-8, -8) 3y = 14 – x and 2y + 8 = x y – 16 = 3x

6. THE END © Lilian Wehner 2012