# How do you solve the following system: 4x+3y= -1 , 4x-5y-23=0 ?

Aug 2, 2018

$x = - \frac{5}{32} \mathmr{and} y = - \frac{1}{8}$

#### Explanation:

The $x$ terms are the same in both equations.

If you subtract the two equations the $x$ terms will be eliminated.

$\textcolor{w h i t e}{\times \times \times x} 4 x + 3 y = - 1 \text{ } \ldots A$
$\textcolor{w h i t e}{\times \times \times x} 4 x - 5 y = \text{ "0" } \ldots B$

$A - B : \textcolor{w h i t e}{\times \times \times} 8 y = - 1$
$\textcolor{w h i t e}{\times \times \times \times \times \times} y = - \frac{1}{8}$

$\textcolor{w h i t e}{\times x} 4 x + 3 \left(- \frac{1}{8}\right) = - 1$

$\textcolor{w h i t e}{\times \times \times \times} 4 x - \frac{3}{8} = - 1$
$\textcolor{w h i t e}{\times \times \times \times \times x} 4 x = - 1 + \frac{3}{8}$
$\textcolor{w h i t e}{\times \times \times \times \times x} 4 x = - \frac{5}{8}$
$\textcolor{w h i t e}{\times \times \times \times \times \times} x = - \frac{5}{32}$

Check in $B$

$4 \left(- \frac{5}{32}\right) - 5 \left(- \frac{1}{8}\right)$

$- \frac{5}{8} + \frac{5}{8}$

$= 0$

Aug 3, 2018

$x = 2$ and $y = - 3$

#### Explanation:

$\left(4 x + 3 y\right) - \left(4 x - 5 y\right) = - 1 - 23$

$8 y = - 24$, so $y = \frac{- 24}{8} = - 3$

Hence,

$4 x + 3 \cdot \left(- 3\right) = - 1$

$4 x - 9 = - 1$

$4 x = 8$, thus $x = \frac{8}{4} = 2$