# How do solve the following linear system?:  x-4y=-24 , x+y=-1 ?

Jan 17, 2016

The lines cross at $\left(x , y\right) \to \left(- \frac{28}{5} , + \frac{23}{5}\right) \to \left(- 5 \frac{3}{5} , + 4 \frac{3}{5}\right)$

#### Explanation:

Given:
$\textcolor{w h i t e}{. .} \textcolor{b r o w n}{x - 4 y = - 24}$......................(1)
$\textcolor{w h i t e}{. .} \textcolor{b r o w n}{x + y = - 1}$..........................(2)
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Consider equation (2)

Take the $y$ over to the other side of the = and change its sign

$x = - 1 - y$............................................(3)

Using equation (3) substitute for $x$ in equation (1) giving

$x - 4 y = - 24 \text{ becomes } \to - 1 - y - 4 y = - 24$

Multiply everything by (-1) to change the signs

$1 + y + 4 y = 24$

$1 + 5 y = 24$

Move the left hand side 1 to the right and change its sign

$5 y = 24 - 1 \text{ = } 23$

Divide both sides by 5 giving

$\textcolor{g r e e n}{y = \frac{23}{5}}$
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Substitute $y = \frac{23}{5}$ in equation (2) giving:

$x + y = - 1 \text{ becomes } \to x + \frac{23}{5} = - 1$

Move the $\frac{23}{5}$ to the other side of = and change its sign

$x = - 1 - \frac{23}{5} \text{ = } - \frac{28}{5}$

$\textcolor{g r e e n}{x = - \frac{28}{5}}$
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So the lines cross at $\left(x , y\right) \to \left(- \frac{28}{5} , \frac{23}{5}\right) \to \left(- 5 \frac{3}{5} , + 4 \frac{3}{5}\right)$