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

Mar 15, 2016

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

$- y = 11 \mathmr{and} y = - 11$

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

$- x = 8 \mathmr{and} x = - 8$

Therefore, $x = - 8$ and $y = - 11$

#### Explanation:

To solve this problem, you must first solve for one variable ($x$) and then the other ($y$). To solve for $y$, we eliminate the $x$ variable by multiplying the first equation by -4 on both sides:

$- 4 \left(x - y\right) = - 4 \cdot 3$ ----> $- 4 x + 4 y = - 12$

Then we add the two equations:

$- 4 x + 4 y = - 12$
+$4 x - 5 y = 23$

This gives us $\left(- 4 x + 4 x\right) + \left(4 y - 5 y\right) = \left(23 - 12\right)$ = $- y = 11$ or $y = - 11$

To solve for $x$ we then eliminate the $y$ variable by multiplying the first equation by -5 on both sides:

$- 5 \left(x - y\right) = - 5 \cdot 3$ ----> $- 5 x + 5 y = - 15$

Then we add the two equations:

$- 5 x + 5 y = - 15$
+ $4 x - 5 y = 23$

This gives us $\left(- 5 x + 4 x\right) + \left(5 y - 5 y\right) = 8$ = $- x = 8$ or $x = - 8$

You can check the answer by substituting -8 for $x$ and -11 for $y$, and you will find that both equations are satisfied.