# How do you solve the system of equations 12x - 6y = - 6 and - 6x + 2y = 8?

Apr 19, 2017

See the entire solution process below:

#### Explanation:

Step 1) Solve the first equation for $y$:

$12 x - 6 y = - 6$

$\frac{12 x - 6 y}{\textcolor{red}{- 6}} = \frac{- 6}{\textcolor{red}{- 6}}$

$\frac{12 x}{\textcolor{red}{- 6}} - \frac{6 y}{\textcolor{red}{- 6}} = 1$

$- 2 x + y = 1$

$\textcolor{red}{2 x} - 2 x + y = \textcolor{red}{2 x} + 1$

$0 + y = 2 x + 1$

$y = 2 x + 1$

Step 2) Substitute $2 x + 1$ for $y$ in the second equation and solve for $x$:

$- 6 x + 2 y = 8$ becomes:

$- 6 x + 2 \left(2 x + 1\right) = 8$

$- 6 x + \left(2 \cdot 2 x\right) + \left(2 \cdot 1\right) = 8$

$- 6 x + 4 x + 2 = 8$

$- 2 x + 2 = 8$

$- 2 x + 2 - \textcolor{red}{2} = 8 - \textcolor{red}{2}$

$- 2 x + 0 = 6$

$- 2 x = 6$

$\frac{- 2 x}{\textcolor{red}{- 2}} = \frac{6}{\textcolor{red}{- 2}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{- 2}}} x}{\cancel{\textcolor{red}{- 2}}} = - 3$

$x = - 3$

Step 3) Substitute $- 3$ for $x$ in the solution to the first equation at the end of Step 1 and calculate $y$:

$y = 2 x + 1$ becomes:

$y = \left(2 \cdot - 3\right) + 1$

$y = - 6 + 1$

$y = - 5$

The solution is: $x = - 3$ and $y = - 5$ or $\left(- 3 , - 5\right)$