# How do you solve the following system?:  8x - 2y = - 4 , 2x - 3y = 5

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#### Explanation

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#### Explanation:

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1
May 21, 2018

See a solution process below:

#### Explanation:

Step 1) Solve each equation for $8 x$:

• Equation 1:

$8 x - 2 y = - 4$

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

$8 x - 0 = - 4 + 2 y$

$8 x = - 4 + 2 y$

• Equation 2:

$2 x - 3 y = 5$

$\textcolor{red}{4} \left(2 x - 3 y\right) = \textcolor{red}{4} \times 5$

$\left(\textcolor{red}{4} \times 2 x\right) - \left(\textcolor{red}{4} \times 3 y\right) = 20$

$8 x - 12 y = 20$

$8 x - 12 y + \textcolor{red}{12 y} = 20 + \textcolor{red}{12 y}$

$8 x - 0 = 20 + 12 y$

$8 x = 20 + 12 y$

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

$- 4 + 2 y = 20 + 12 y$

$- 4 - \textcolor{b l u e}{20} + 2 y - \textcolor{red}{2 y} = 20 - \textcolor{b l u e}{20} + 12 y - \textcolor{red}{2 y}$

$- 24 + 0 = 0 + \left(12 - \textcolor{red}{2}\right) y$

$- 24 = 10 y$

$- \frac{24}{\textcolor{red}{10}} = \frac{10 y}{\textcolor{red}{10}}$

$- \frac{12}{5} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{10}}} y}{\cancel{\textcolor{red}{10}}}$

$- \frac{12}{5} = y$

$y = - \frac{12}{5}$

Step 3) Substitute $- \frac{12}{5}$ for $y$ in either of the equations in Step 1 and calculate $x$:

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

$8 x = - 4 + \left(2 \times - \frac{12}{5}\right)$

$8 x = - 4 + \left(- \frac{24}{5}\right)$

$8 x = - 4 - \frac{24}{5}$

$8 x = \left(\frac{5}{5} \times - 4\right) - \frac{24}{5}$

$8 x = - \frac{20}{5} - \frac{24}{5}$

$8 x = - \frac{44}{5}$

$8 x \times \frac{1}{8} = - \frac{44}{5} \times \frac{1}{8}$

$\frac{8}{8} x = - \frac{44}{40}$

$x = - \frac{11}{10}$

The Solution Is:

$x = - \frac{11}{10}$ and $y = - \frac{12}{5}$

Or

$\left(- \frac{11}{10} , - \frac{12}{5}\right)$

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