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

## Replaced a $+$ with $=$ for the first equation; hopefully, this is what was intended.

Nov 13, 2015

$x = - 2$, $y = - 1$.

#### Explanation:

From the second equation, we know that $y = 3 + x$.

So, the first equation becomes

$4 x + 5 \textcolor{g r e e n}{y} = - 3 \to 4 x + 5 \textcolor{g r e e n}{\left(3 + x\right)} = - 3$

Expand the left memeber: $4 x + 15 + 5 x = 9 x + 15$

So, $9 x + 15 = - 3 \setminus \iff 9 x = - 18 \iff x = - 2$

Knowing this, we obtain $y$, since $y = 3 + x = 3 - 2 = 1$

Nov 13, 2015

$\left(x , y\right) = \left(- 2 , 1\right)$

#### Explanation:

Given
[1]$\textcolor{w h i t e}{\text{XXX}} 4 x + 5 y = - 3$
[2]$\textcolor{w h i t e}{\text{XXX}} - x + y = 3$

Multiply [2] by $4$
[3]$\textcolor{w h i t e}{\text{XXX}} - 4 x + 4 y = 12$

[4]$\textcolor{w h i t e}{\text{XXX}} 9 y = 9$

Divide [4] by $9$
[5]$\textcolor{w h i t e}{\text{XXX}} y = 1$

Substitute $1$ for $y$ in [2]
[6]$\textcolor{w h i t e}{\text{XXX}} - x + \left(1\right) = 3$

Subtract $1$ from both sides of [6]
[7]$\textcolor{w h i t e}{\text{XXX}} - x = 2$

Multiply both sides of [7] by $\left(- 1\right)$
[8]$\textcolor{w h i t e}{\text{XXX}} x = - 2$