# What is the solution set for y = x^2 - 6 and y = -2x - 3?

Aug 11, 2015

$\left\{\begin{matrix}x = - 3 \\ y = 3\end{matrix}\right. \text{ }$ or $\text{ } \left\{\begin{matrix}x = 1 \\ y = - 5\end{matrix}\right.$

#### Explanation:

Notice that you were given two equations that deal with the value of $y$

$y = {x}^{2} - 6 \text{ }$ and $\text{ } y = - 2 x - 3$

In order for these equations to be true, you need to have

${x}^{2} - 6 = - 2 x - 3$

Rearrange this equation into classic quadratic form

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

You can use the quadratic formula to determine the two solutions

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

${x}_{1 , 2} = \frac{- 2 \pm \sqrt{16}}{2} = \frac{- 2 \pm 4}{2} = \left\{\begin{matrix}{x}_{1} = \frac{- 2 - 4}{2} = - 3 \\ {x}_{2} = \frac{- 2 + 4}{2} = 1\end{matrix}\right.$

Now take these values of $x$ to one of the orignal equations and find the corresponding values of $y$.

• when $x = - 3$, you have

$y = {\left(- 3\right)}^{2} - 6 = 3$

• when $x = 1$, you have

$y = {1}^{2} - 6 = - 5$

So, the two possible solution sets are

$\left\{\begin{matrix}x = - 3 \\ y = 3\end{matrix}\right. \text{ }$ or $\text{ } \left\{\begin{matrix}x = 1 \\ y = - 5\end{matrix}\right.$