How do you solve $y = 3 x^{2} - x - 2, y = - 2 x + 2$ $y=3x^2-x-2, y=-2x+2$ using substitution?

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Answer 1 · 2016-12-23T02:29:51

You can simply substitute the second equation into the first or vice versa.

$- 2 x + 2 = 3 x^{2} - x - 2$ $-2x + 2 = 3x^2 - x - 2$

$0 = 3 x^{2} + x - 4$ $0 = 3x^2 + x - 4$

$0 = 3 x^{2} - 3 x + 4 x - 4$ $0 = 3x^2 - 3x + 4x - 4$

$0 = 3 x (x - 1) + 4 (x - 1)$ $0 = 3x(x -1) + 4(x - 1)$

$0 = (3 x + 4) (x - 1)$ $0 = (3x + 4)(x - 1)$

$x = - \frac{4}{3} and 1$ $x = -4/3 and 1$

Case 1: $x = - \frac{4}{3}$ $x= -4/3$

$y = - 2 (- \frac{4}{3}) + 2$ $y = -2(-4/3) + 2$

$y = \frac{8}{3} + 2$ $y = 8/3 + 2$

$y = \frac{14}{3}$ $y = 14/3$

Case 2: $x = 1$ $x= 1$ :

$y = - 2 x + 2$ $y = -2x + 2$

$y = - 2 + 2$ $y = -2 + 2$

$y = 0$ $y = 0$

The solution set is therefore $(1, 0)$ $(1, 0)$ and $(- \frac{4}{3}, \frac{14}{3})$ $(-4/3, 14/3)$ .

Hopefully this helps!

How do you solve y=3x^2-x-2, y=-2x+2y=3x2−x−2,y=−2x+2y=3x^2-x-2, y=-2x+2 using substitution?