# How do solve the following linear system?: -4x + 4 = -4y, 7x + 6y + 11 = 0 ?

Jul 15, 2016

The solution is $\left(- \frac{5}{13} , \frac{8}{13}\right)$

#### Explanation:

This problem makes things fairly easy for you by providing a clear definition of $y$ in the first equation:

$- 4 x + 4 = - 4 y$
$x - 1 = y$

We can plug this into the other equation to solve for $x$ and then use it to find $y$:
$7 x + 6 \left(x - 1\right) + 11 = 0$
$7 x + 6 x - 6 + 11 = 0$
$13 x + 5 = 0$
$x = - \frac{5}{13}$

Then:
$- \frac{5}{13} + 1 = y$
$y = \frac{8}{13}$

So the solution is $\left(- \frac{5}{13} , \frac{8}{13}\right)$

A little additional info on the problem in concluding. In case it is not clear, the solution takes the form of a point, $\left(x , y\right)$. This is because solving a system of equations is the same as finding the point(s) at which those lines (or curves) intersect.