# How do you solve x + 6y = 28 and 2x - 3y = -19?

Jun 4, 2018

Double the second equation, and then add the two equations together linearly, solving for $x$, and then solve for $y$ in either. You will find that $x = - 2$ and $y = 5$

#### Explanation:

First, line up the equations with one above the other:

$x + 6 y = 28$
$2 x - 3 y = - 19$

next, multiply the ENTIRE bottom equation by 2:

$x + 6 y = 28$
$\textcolor{red}{4} x - \textcolor{red}{6} y = \textcolor{red}{- 38}$

$\left(x + 6 y = 28\right)$
ul(+(4x-6y=-38)
$\textcolor{red}{\left(4 + 1\right)} x + \textcolor{red}{\left(6 - 6\right)} y = \textcolor{red}{\left(28 - 38\right)}$

$\textcolor{red}{5} x + \textcolor{red}{0} y = \textcolor{red}{- 10}$

$5 x = - 10$

Now, solve for $x$:

$\frac{\cancel{5} x}{\textcolor{red}{\cancel{5}}} = \frac{- 10}{\textcolor{red}{5}}$

$\textcolor{b l u e}{x = - 2}$

Since we now have a solution for $x$, we can plug it back into either equation to solve for $y$. Doing it in both proves that our solution for $x$ is valid:

Equation 1:

$\textcolor{b l u e}{x} + 6 y = 28$

$\textcolor{b l u e}{- 2} + 6 y = 28$

$\textcolor{b l u e}{\cancel{- 2}} + 6 y \textcolor{red}{\cancel{+ 2}} = 28 \textcolor{red}{+ 2}$

$6 y = 30$

$\frac{\cancel{6} y}{\textcolor{red}{\cancel{6}}} = \frac{30}{\textcolor{red}{6}}$

$\textcolor{g r e e n}{y = 5}$

Equation 2:

$2 \textcolor{b l u e}{x} - 3 y = - 19$

$2 \textcolor{b l u e}{\left(- 2\right)} - 3 y = - 19$

$- 4 - 3 y = - 19$

$\cancel{- 4} - 3 y \textcolor{red}{\cancel{+ 4}} = - 19 \textcolor{red}{+ 4}$

$- 3 y = - 15$

$\frac{\cancel{- 3} y}{\textcolor{red}{\cancel{- 3}}} = \frac{- 15}{\textcolor{red}{- 3}}$

$\textcolor{g r e e n}{y = 5}$

Both equations support the solution, we're done!