# How do you solve the following system: 2x-y/2=4, 5x+2y=12 ?

Feb 3, 2016

$x = \frac{28}{13} , y = \frac{8}{13}$

#### Explanation:

This is done by the process of elimination but it is not the only method.

Begin by multiplying the first equation by 4 to get the coefficients in front of the $y$ equal. Multiplying the first equation by 4 gives:

$8 x - 2 y = 16$

Now we can add both equations together

$\left(5 x + 2 y\right) + \left(8 x - 2 y\right) = \left(16\right) + \left(12\right)$

So we get:

$13 x = 28 \to x = \frac{28}{13}$

Now replace $x$ with this value in either of the equations and solve for $y$. So using the first equation: $2 x - \frac{y}{2} = 4$.

$\to 2 \left(\frac{28}{13}\right) - \frac{y}{2} = 4$
$\frac{56}{13} - \frac{y}{2} = 4$

Rearrange and you should arrive at:

$y = 2 \left(\frac{56}{13} - 4\right) = 2 \frac{56 - 52}{13} = \frac{8}{13}$
$\therefore x = \frac{28}{13} , y = \frac{8}{13}$