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

Question

1168 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-15T00:18:50

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

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

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

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

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

$x = - 3 and 1$ $x = -3 and 1$

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

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

$y = - 7 and y = 9$ $y = -7 and y = 9$

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

Hopefully this helps!

How do you solve y=x2+6x+2y=x^2+6x+2 and y=−x2+2x+8y=-x^2+2x+8 using substitution?