# How do you solve the system of equations -5x - 7y = - 12 and 7x + 6y = - 6?

Feb 16, 2017

See the entire solution process below:

#### Explanation:

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

$7 x + 6 y = - 6$

$- \textcolor{red}{7 x} + 7 x + 6 y = - \textcolor{red}{7 x} - 6$

$0 + 6 y = - \textcolor{red}{7 x} - 6$

$6 y = - \textcolor{red}{7 x} - 6$

$\frac{6 y}{\textcolor{red}{6}} = \frac{- \textcolor{red}{7 x} - 6}{\textcolor{red}{6}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{6}}} y}{\cancel{\textcolor{red}{6}}} = \frac{- 7 x}{6} - \frac{6}{\textcolor{red}{6}}$

$y = - \frac{7}{6} x - 1$

Step 2) Substitute $- \frac{7}{6} x - 1$ for $y$ in the first equation and solve for $x$:

$- 5 x - 7 y = - 12$ becomes:

$- 5 x - 7 \left(- \frac{7}{6} x - 1\right) = - 12$

$- 5 x + \left(7 \times \frac{7}{6} x\right) + \left(7 \times 1\right) = - 12$

$- 5 x + \frac{49}{6} x + 7 = - 12$

$\left(\frac{6}{6} \times - 5\right) x + \frac{49}{6} x + 7 - \textcolor{red}{7} = - 12 - \textcolor{red}{7}$

$- \frac{30}{6} x + \frac{49}{6} x + 0 = - 19$

$\frac{19}{6} x = - 19$

$\frac{\textcolor{red}{6}}{\textcolor{b l u e}{19}} \times \frac{19}{6} x = \frac{\textcolor{red}{6}}{\textcolor{b l u e}{19}} \times - 19$

$\frac{\cancel{\textcolor{red}{6}}}{\cancel{\textcolor{b l u e}{19}}} \times \frac{\textcolor{b l u e}{\cancel{\textcolor{b l a c k}{19}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{6}}}} x = \frac{\textcolor{red}{6}}{\cancel{\textcolor{b l u e}{19}}} \times - \textcolor{b l u e}{\cancel{\textcolor{b l a c k}{19}}}$

$x = - 6$

Step 3) Substitute $- 6$ for $x$ in the solution to the second equation at the end of Step 1 and calculate $y$:

$y = - \frac{7}{6} x - 1$ becomes:

$y = \left(- \frac{7}{6} \times - 6\right) - 1$

$y = \left(- \frac{7}{\textcolor{red}{\cancel{\textcolor{b l a c k}{6}}}} \times - \textcolor{red}{\cancel{\textcolor{b l a c k}{6}}}\right) - 1$

$y = 7 - 1$

$y = 6$

The solution is: $x = - 6$ and $y = 6$ or $\left(- 6 , 6\right)$