How do you solve the following system: y = 3x – 12, 3x + 8y = -2?

Mar 1, 2017

See the entire solution process below:

Explanation:

Step 1) Because the first equation is already solved for $y$, substitute $3 x - 12$ for $y$ in the second equation and solve for $x$: xx
) 3x + 8y = -2# becomes:

$3 x + 8 \left(3 x - 12\right) = - 2$

$3 x + 24 x - 96 = - 2$

$27 x - 96 = - 2$

$27 x - 96 + \textcolor{red}{96} = - 2 + \textcolor{red}{96}$

$27 x - 0 = 94$

$27 x = 94$

$\frac{27 x}{\textcolor{red}{27}} = \frac{94}{\textcolor{red}{27}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{27}}} x}{\cancel{\textcolor{red}{27}}} = \frac{94}{27}$

$x = \frac{94}{27}$

Step 2) Substitute $\frac{94}{27}$ for $x$ in the first equation and calculate $y$:

$y = 3 x - 12$ becomes:

$y = 3 \times \frac{94}{27} - 12$

$y = \frac{94}{9} - 12$

$y = \frac{94}{9} - \left(\frac{9}{9} \times 12\right)$

$y = \frac{94}{9} - \frac{108}{9}$

$y = - \frac{14}{9}$

The solution is $x = \frac{94}{27}$ and $y = - \frac{14}{9}$ or $\left(\frac{94}{27} , - \frac{14}{9}\right)$