How do you solve $y=x^2+6x+2$ and $y=-x^2+2x+8$ using substitution?

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Answer 1 · 2016-07-15T00:18:50

$-x^2 + 2x + 8 = x^2 + 6x + 2$

$0 = x^2 + x^2 + 6x - 2x + 2 - 8$

$0 = 2x^2 + 4x - 6$

$0 = 2(x^2 + 2x - 3)$

$0 = 2(x + 3)(x - 1)$

$x = -3 and 1$

$y = (-3)^2 + 6(-3) + 2" and "y = (1)^2 + 6(1) + 2$

$y = 9 - 18 + 2" and "y = 9$

$y = -7 and y = 9$

Hence, the solution set is ${-3, -7} and$ {1, 9}#.

Hopefully this helps!

How do you solve y=x^2+6x+2y=x^2+6x+2 and y=-x^2+2x+8y=-x^2+2x+8 using substitution?