Question
Any help would be appreciated. Thank you.
What are the solutions of the system?
y = x2 + 3x – 4
y = 2x + 2
Options:
(–3, 6) and (2, –4)
(–3, –4) and (2, 6)
(–3, –4) and (–2, –2)
no solution
What are the solutions of the system?
y = x2 + 3x – 4
y = 2x + 2
Options:
(–3, 6) and (2, –4)
(–3, –4) and (2, 6)
(–3, –4) and (–2, –2)
no solution
Answers
Steve
Even if you can't solve the equations, you can always plug in the choices to see which works. Algebraically, we need
x^2 + 3x - 4 = 2x + 2
x^2 + x - 6 = 0
(x+3)(x-2) = 0
x = -3 or 2
y = 2x+2 = -4 or 6
so, (b)
x^2 + 3x - 4 = 2x + 2
x^2 + x - 6 = 0
(x+3)(x-2) = 0
x = -3 or 2
y = 2x+2 = -4 or 6
so, (b)