# How do you find the solution of the system of equations y= x^2-4 and y= x-2?

May 17, 2015

You can isolate one variable in one equation and substitute its value in the other one. Let's do it.

Let's get the value of $y$ in the first equation and substitute it in the second one and, then, find $x$:

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

$\frac{1 \pm \sqrt{{1}^{2} - 4 \left(1\right) \left(- 2\right)}}{2} \cdot 1$
$\frac{1 \pm 3}{2}$
${x}_{1} = 2$ and ${x}_{2} = - 1$

Now we have these values for $x$, let's find the values of $y$ - using the first equation:

${y}_{1} = {\left(2\right)}^{2} - 4 = 0$
${y}_{2} = {\left(- 1\right)}^{2} - 4 = - 3$

$y = 0$ when $x = 2$
$y = - 3$ when $x = - 1$