# How do you solve the following system: 2x-y/2=4, 5x+2y=20 ?

Mar 23, 2018

Use linear algebra to solve for x, and then plug the solution back in to solve for y. You will find that $x = \frac{36}{13}$ and $y = \frac{40}{13}$.

#### Explanation:

Let's align the equations, first and foremost:

$2 x - \frac{y}{2} = 4$
$5 x + 2 y = 20$

To eliminate y, we'll multiply the entire top equation by 4, which will make that $- \frac{y}{2}$ into $- 2 y$. We'll then add the two equations together and get a system where it's only x and constants:

$\left(2 x - \frac{y}{2} = 4\right) \times 4 \Rightarrow 8 x - 2 y = 16$

$\left(8 x - 2 y = 16\right)$
$\underline{+ \left(5 x + 2 y = 20\right)}$
$13 x = 36$

color(red)(x=36/13

Now, let's plug it back into one of the equations to solve for y:

$5 x + 2 y = 20 \Rightarrow 5 \left(\frac{36}{13}\right) + 2 y = 20$

$\left(\frac{5}{2}\right) \left(\frac{36}{13}\right) + y = 10 \Rightarrow \frac{5 \cdot 18}{13} + y = 10$

$y = 10 - \frac{90}{13} \Rightarrow y = \frac{130}{13} - \frac{90}{13} = \frac{130 - 90}{13}$

$\textcolor{b l u e}{y = \frac{40}{13}}$