# How do you solve the following system?: -4x -3y =5, x -2y = -15

May 17, 2018

$x = - 5 \mathmr{and} y = 5$

#### Explanation:

$- 4 x - 3 y = 5 - - - e q n 1$

$x - 2 y = - 15 - - - e q n 2$

From the $e q n 1$ rearrange the equation..

$- 4 x - 3 y = 5$

Multiplying through by minus$\left(-\right)$;

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

$4 x + 3 y = - 5 - - - e q n 1$

Now solving simultaneously..

$4 x + 3 y = - 5 - - - e q n 1$

$x - 2 y = - 15 - - - e q n 2$

Using Elimination Method..

Multiplying $e q n 1$ by $1$ and $e q n 2$ by $4$ respectively..

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

$4 \left(x - 2 y = - 15\right)$

$4 x + 3 y = - 5 - - - e q n 3$

$4 x - 8 y = - 60 - - - e q n 4$

Subtracting $e q n 4$ from $e q n 3$

(4x - 4x) + (3y - (-8y) = -5 - (-60)

$0 + 3 y + 8 y = - 5 + 60$

$11 y = 55$

$y = \frac{55}{11}$

$y = 5$

Substituting the value of $y$ into $e q n 2$

$x - 2 y = - 15 - - - e q n 2$

$x - 2 \left(5\right) = - 15$

$x - 10 = - 15$

$x = - 15 + 10$

$x = - 5$

Therefore;

$x = - 5 \mathmr{and} y = 5$