How do you solve y= x^2-4 and y= x-2?

Aug 10, 2015

The answer is $x = - 1$ ; $x = 2$

Explanation:

Since

$\left\{\begin{matrix}y = {x}^{2} - 4 \\ y = x - 2\end{matrix}\right.$

we shall have

${x}^{2} - 4 = x - 2$

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

${x}^{2} - x - 2 = 0$

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

$x \left(x - 2\right) + 1 \left(x - 2\right) = 0$

$\left(x + 1\right) \left(x - 2\right) = 0$

This means that you have

$x + 1 = 0 \implies x = - 1$

or

$x - 2 = 0 \implies x = 2$

If you substitute the two values of $x$ in the original equations, they satisfy them simultaneously.