How do solve the following linear system?: 2x + 4 = -4y, 7x + 6y + 11 = 0 ?

Jun 17, 2017

$\left(- \frac{5}{4} , - \frac{3}{8}\right)$

Explanation:

First, solve one of the equations for $x$. It doesn't matter which one.

$2 x + 4 = - 4 y$

$2 x = - 4 - 4 y$

$x = - 2 - 2 y$

Next, plug this equation for $x$ into the other equation

$7 \textcolor{red}{x} + 6 y + 11 = 0$

$7 \left(\textcolor{red}{- 2 - 2 y}\right) + 6 y + 11 = 0$

$- 14 - 14 y + 6 y + 11 = 0$

$- 14 - 8 y + 11 = 0$

$- 3 - 8 y = 0$

$- 8 y = 3$

$y = - \frac{3}{8}$

Finally, plug this solution for $y$ into the equation for $x$

$x = - 2 - 2 y$

$x = - 2 - 2 \left(- \frac{3}{8}\right)$

$x = - 2 + \frac{6}{8}$

$x = - \frac{16}{8} + \frac{6}{8} = - \frac{10}{8} = - \frac{5}{4}$