# How do you solve this system of linear equations: 3x-2y=8 and 2x-3y=-6?

Dec 6, 2015

$x = \frac{36}{5}$, $y = \frac{34}{5}$

#### Explanation:

Multiply the first equation by $2$ and the second one by $3$:

$6 x - 4 y = 16$
$6 x - 9 y = - 18$

Subtract the equations:

$5 y = 34 \setminus \implies y = \frac{34}{5}$

Now that $y$ is known, we can deduce $x$ from any of the two equations:

$6 x - 4 y = 16 \setminus \implies 6 x - 4 \cdot \frac{34}{5} = 16 \setminus \implies 6 x - \frac{136}{5} = 16$

So, $x = \frac{16 + \frac{136}{5}}{6} = \frac{36}{5}$