# How do you find the solution set for 2y=5x+10 and y^2-4y=5x+10?

Aug 11, 2015

$\left\{\begin{matrix}x = - 2 \\ y = 0\end{matrix}\right. \text{ }$ or $\text{ } \left\{\begin{matrix}x = \frac{2}{5} \\ y = 6\end{matrix}\right.$

#### Explanation:

Your system of equations looks like this

$\left\{\begin{matrix}2 y = 5 x + 10 \\ {y}^{2} - 4 y = 5 x + 10\end{matrix}\right.$

Right from the start, you can say that this system of equations can be written as one equation in $y$, since you have

$\left(5 x + 10\right) = 2 y \text{ }$ and $\text{ } 5 x + 10 = {y}^{2} - 4 y$

This is equivalent to

${\underbrace{{y}^{2} - 4 y}}_{\textcolor{b l u e}{= 5 x + 10}} = {\overbrace{2 y}}^{\textcolor{\mathmr{and} a n \ge}{= 5 x + 10}}$

Rearrange this equation into classic quadratic form

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

${y}^{2} - 6 y = 0$

You can factor this equation to get

$y \cdot \left(y - 6\right) = 0$

The two solutions will thus be $y = 0$ and $y = 6$.

Use these values of $y$ in one of the two original equations to get the values of $x$.

• when $y = o$, you have

$5 x + 10 = {0}^{2} - 4 \cdot 0$

$5 x + 10 = 0$

$5 x = - 10 \implies x = \frac{- 10}{5} = - 2$

• when $y = 6$, you have

$5 x + 10 = {6}^{2} - 4 \cdot 6$

$5 x + 10 = 12$

$5 x = 2 \implies x = \frac{2}{5}$

The two solution sets for this system of equations are

$\left\{\begin{matrix}x = - 2 \\ y = 0\end{matrix}\right. \text{ }$ or $\text{ } \left\{\begin{matrix}x = \frac{2}{5} \\ y = 6\end{matrix}\right.$