# How do you solve the following system: 3x + 8y = -2, 2x – 4y = 8?

Mar 4, 2018

See a solution process below:

#### Explanation:

Step 1) Solve each equation to $8 y$

• Equation 1:

$3 x + 8 y = - 2$

$3 x - \textcolor{red}{3 x} + 8 y = - 2 - \textcolor{red}{3 x}$

$0 + 8 y = - 2 - 3 x$

$8 y = - 2 - 3 x$

• Equation 2:

$2 x - 4 y = 8$

$2 x - \textcolor{red}{2 x} - 4 y = 8 - \textcolor{red}{2 x}$

$0 - 4 y = 8 - 2 x$

$- 4 y = 8 - 2 x$

$\textcolor{red}{- 2} \times - 4 y = \textcolor{red}{- 2} \left(8 - 2 x\right)$

$8 y = \left(\textcolor{red}{- 2} \times 8\right) - \left(\textcolor{red}{- 2} \times 2 x\right)$

$8 y = - 16 - \left(- 4 x\right)$

$8 y = - 16 + 4 x$

Step 2) Because the left side of both equations are the same we can equate the right side of each equation and solve for $x$:

$- 2 - 3 x = - 16 + 4 x$

$- 2 + \textcolor{b l u e}{16} - 3 x + \textcolor{red}{3 x} = - 16 + \textcolor{b l u e}{16} + 4 x + \textcolor{red}{3 x}$

$14 - 0 = 0 + \left(4 + \textcolor{red}{3}\right) x$

$14 = 7 x$

$\frac{14}{\textcolor{red}{7}} = \frac{7 x}{\textcolor{red}{7}}$

$2 = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{7}}} x}{\cancel{\textcolor{red}{7}}}$

$2 = x$

$x = 2$

Step 3) Substitute $2$ for $x$ in the solution to either equation in Step 1 and solve for $y$:

$- 4 y = 8 - 2 x$ becomes:

$- 4 y = 8 - \left(2 \times 2\right)$

$- 4 y = 8 - 4$

$- 4 y = 4$

$\frac{- 4 y}{\textcolor{red}{- 4}} = \frac{4}{\textcolor{red}{- 4}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{- 4}}} y}{\cancel{\textcolor{red}{- 4}}} = - 1$

$y = - 1$

The Solution Is:

$x = 2$ and $y = - 1$

Or

$\left(2 , - 1\right)$