# How do you solve the following system?:  x + 2y = -2 , x – 3y = -12

Jun 20, 2018

See a solution process below:

#### Explanation:

Step 1) Solve each equation for $x$:

• Equation 1:

$x + 2 y = - 2$

$x + 2 y - \textcolor{red}{2 y} = - 2 - \textcolor{red}{2 y}$

$x + 0 = - 2 - 2 y$

$x = - 2 - 2 y$

• Equation 2:

$x - 3 y = - 12$

$x - 3 y + \textcolor{red}{3 y} = - 12 + \textcolor{red}{3 y}$

$x - 0 = - 12 + 3 y$

$x = - 12 + 3 y$

Step 2) Because the left side of both equations are the same we can equate the right sides and solve for $y$:

$- 2 - 2 y = - 12 + 3 y$

$- 2 + \textcolor{b l u e}{12} - 2 y + \textcolor{red}{2 y} = - 12 + \textcolor{b l u e}{12} + 3 y + \textcolor{red}{2 y}$

$10 - 0 = 0 + \left(3 + \textcolor{red}{2}\right) y$

$10 = 5 y$

$\frac{10}{\textcolor{red}{5}} = \frac{5 y}{\textcolor{red}{5}}$

$2 = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{5}}} y}{\cancel{\textcolor{red}{5}}}$

$2 = y$

$y = 2$

Step 3) Substitute $2$ for $y$ in the solutions to either of the equations in Step 1 and calculate $x$:

$x = - 12 + 3 y$ becomes:

$x = - 12 + \left(3 \times 2\right)$

$x = - 12 + 6$

$x = - 6$

The Solution Is:

$x = - 6$ and $y = 2$

Or

$\left(- 6 , 2\right)$