# How do you solve the following linear system 4x - 3y = -5 , 4x +- 4y = -4 ?

For equations $4 x - 3 y = - 5$ and $4 x + 4 y = - 4$, solution $\left(- \frac{8}{7} , \frac{1}{7}\right)$
For equations $4 x - 3 y = - 5$ and $4 x - 4 y = - 4$, solution $\left(- 2 , - 1\right)$

#### Explanation:

We will use $\textcolor{b l u e}{\text{method of elimination of variables}}$

For equations $4 x - 3 y = - 5$ and $4 x + 4 y = - 4$

$4 x + 4 y = - 4$
$4 x - 3 y = - 5$

Perform subtraction to obtain

$7 y = 1$

then $y = \frac{1}{7}$
Using $4 x + 4 y = - 4$ and $y = \frac{1}{7}$

We have, $4 x + 4 \left(\frac{1}{7}\right) = - 4$

$x + \frac{1}{7} = - 1$

$x = - \frac{8}{7}$

For equations $4 x - 3 y = - 5$ and $4 x + 4 y = - 4$, solution $\left(- \frac{8}{7} , \frac{1}{7}\right)$

Again, we will use $\textcolor{b l u e}{\text{method of elimination of variables}}$

For equations $4 x - 3 y = - 5$ and $4 x - 4 y = - 4$

$4 x - 4 y = - 4$
$4 x - 3 y = - 5$

Perform subtraction to obtain

$y = - 1$

then $y = - 1$
Using $4 x - 4 y = - 4$ and $y = - 1$

We have, $4 x - 4 \left(- 1\right) = - 4$

$x - \left(- 1\right) = - 1$

$x + 1 = - 1$

$x = - 2$

For equations $4 x - 3 y = - 5$ and $4 x - 4 y = - 4$, solution $\left(- 2 , - 1\right)$
graph{(4x-3y+5)(4x+4y+4)(4x-4y+4)=0 [-3, 2, -5, 5]}

God bless....I hope the explanation is useful..