# How do you solve the system 5x-7y=-16 and 2x+8y=26?

May 3, 2018

1) 5x-7y=-16
2) 2x+8y=26

$2 x = 26 - 8 y | \cdot \frac{1}{2}$
$x = 13 - 4 y$

$- 7 y = - 16 - 5 x$
$7 y = 16 + 5 x$
$7 y = 16 + 5 \left(13 - 4 y\right)$
$7 y = 16 + 65 - 20 y$
$7 y + 20 y = 16 + 65$
$27 y = 81 | \cdot \frac{1}{27}$

$y = 3$

$x = 13 - 4 \left(3\right)$
$x = 1$

$y = 3$ and $x = 1$

#### Explanation:

You can solve this system by finding what one variable equals from one of the equations, then put this into the other equation.

I went to find $y$ here in the start. Because I saw that locking $x$ by itself would be fair enough. It gave a clean $x = 13 - 4 y$, instead of fractions or such.

I then put what $x$ equals to into the other $y$ equation. So that I can find the integer value of $y$ without having any $x$ variables. Which gave the result of $y = 3$.

From there, we can place the $y = 3$ into the other equation and find the $x$ value, $x = 13 - 4 \left(3\right)$ instead of $x = 13 - 4 y$. Which gave the result of $x = 1$.

From that, we now know that:
$y = 3$ and $x = 1$