How do you solve the system of equations 3x - \frac { 4} { 3} y = - 8 and - 4y = x + 16?

Aug 23, 2017

See a solution process below:

Explanation:

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

$- 4 y = x + 16$

$- 4 y - \textcolor{red}{16} = x + 16 - \textcolor{red}{16}$

$- 4 y - 16 = x + 0$

$- 4 y - 16 = x$

$x = - 4 y - 16$

Step 2) Substitute $\left(- 4 y - 16\right)$ for $x$ in the first equation and solve for $y$:

$3 x - \frac{4}{3} y = - 8$ becomes:

$3 \left(- 4 y - 16\right) - \frac{4}{3} y = - 8$

$\left(3 \cdot - 4 y\right) - \left(3 \cdot 16\right) - \frac{4}{3} y = - 8$

$- 12 y - 48 - \frac{4}{3} y = - 8$

$- 12 y - \frac{4}{3} y - 48 = - 8$

$\left(\frac{3}{3} \cdot - 12\right) y - \frac{4}{3} y - 48 = - 8$

$- \frac{36}{3} y - \frac{4}{3} y - 48 = - 8$

$- \frac{40}{3} y - 48 = - 8$

$- \frac{40}{3} y - 48 + \textcolor{red}{48} = - 8 + \textcolor{red}{48}$

$- \frac{40}{3} y - 0 = 40$

$- \frac{40}{3} y = 40$

$- \frac{\textcolor{red}{3}}{\textcolor{b l u e}{40}} \cdot - \frac{40}{3} y = - \frac{\textcolor{red}{3}}{\textcolor{b l u e}{40}} \cdot 40$

$- \frac{\cancel{\textcolor{red}{3}}}{\cancel{\textcolor{b l u e}{40}}} \cdot - \frac{\textcolor{b l u e}{\cancel{\textcolor{b l a c k}{40}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}}} y = - \frac{\textcolor{red}{3}}{\cancel{\textcolor{b l u e}{40}}} \cdot \textcolor{b l u e}{\cancel{\textcolor{b l a c k}{40}}}$

$- - y = - 3$

$y = - 3$

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

$x = - 4 y - 16$ becomes:

$x = \left(- 4 \cdot - 3\right) - 16$

$x = 12 - 16$

$x = - 4$

The Solution Is: $x = - 4$ and $y = - 3$ or $\left(- 4 , - 3\right)$