# How do you solve the following linear system:  3x + y = 1 , x + 2y = -4 ?

Dec 14, 2015

Original Equations:
$3 x + y = 1$
$x + 2 y = - 4$

First, you isolate a variable. I chose to set the second equation equal to $y$.

$y = - 3 x + 1$

Next, plug the above equation for $y$ into the equation you haven't used yet.
$x + 2 \left(- 3 x + 1\right) = - 4$

Multiply it out and combine like terms.
$x - 6 x + 2 = - 4$
$- 5 x + 2 = - 4$

Now, solve for $x$
$- 5 x + 22 = - 4$
$- 5 x = - 6$
$x = \frac{6}{5}$

Plug in $x$ to solve for $y .$ I'm going to plug it into the equation I used first:
$y = - 3 x + 1$
$y = - 3 \left(\frac{6}{5}\right) + 1$
$y = - \frac{18}{5} + 1$
$y = - \frac{13}{5}$

The answer is $\left(\frac{6}{5} , - \frac{13}{5}\right)$