What are the solutions of the system?
y = x^2 + 4x - 5
y = 2x - 2

A. (–3, –8) and (1, 0)

B. (–3, –8) and (–1, –4)

C. (–3, 0) and (1, –8)

D. no solution

A, B, C, or D? Please explain (:

Question

What are the solutions of the system?
y = x^2 + 4x - 5
y = 2x - 2

A. (–3, –8) and (1, 0)	

B. (–3, –8) and (–1, –4)	

C. (–3, 0) and (1, –8)	

D. no solution	

A, B, C, or D? Please explain (:

anonymous · Answer

substitute one into the other
2x-2 = x^2 +4x -5 
Then you work out as a regular quadratic
x^2+2x-3
(x+3)(x-1)
You substitute these x values into the original equation to get your y values