# How do you solve the following system:  1/2x - 2y = 4 , 5x+3y=-1 ?

May 7, 2017

$x = \frac{20}{23} \mathmr{and} y = - \frac{41}{23}$

#### Explanation:

The approach is either to use substitution or elimination.

For substitution, you need to have a single variable such as $x$

$\frac{1}{2} x - 2 y = 4 \text{ and } 5 x + 3 y = - 1$

Multiply the first equation by $2$

$2 \times \frac{1}{2} x - 2 \times 2 y = 2 \times 4$

$x - 4 y = 8$

$\textcolor{b l u e}{x = \left(4 y + 8\right)} \text{ }$Substitute for $x$ in the other equation:

$\textcolor{w h i t e}{m l . m} 5 \textcolor{b l u e}{x} + 3 y = - 1$
$\textcolor{w h i t e}{m m m} \downarrow$
$\textcolor{w h i t e}{m} 5 \textcolor{b l u e}{\left(4 y + 8\right)} + 3 y = - 1$

$20 y + 40 + 3 y = - 1$

$23 y = - 1 - 40$

$23 y = - 41$

$y = \frac{- 41}{23}$

Now find $x \text{ by using } x = 4 y + 8$

$x = 4 \left(\frac{- 41}{23}\right) + 8$

$x = \frac{20}{23}$

Substitute to check:

$5 x + 3 y$

$5 \left(\frac{20}{23}\right) + 3 \left(- \frac{41}{23}\right)$

$= \frac{100}{23} - \frac{123}{23}$

$= - 1$