# How do you solve the system of equations 9x + 8y = 0 and 3x - 8y = - 48?

Apr 19, 2017

$\left(- 4 , \frac{9}{2}\right)$

#### Explanation:

Since there is a $- 8 y$ and a $+ 8 y$, let's add the equations.

$9 x + 8 y = 0$
$3 x - 8 y = - 48$

$12 x = - 48$
$x = - 4$

Plug this in, and:
$9 \left(- 4\right) + 8 y = 0$

$8 y = 36$

$y = \frac{9}{2}$

Thus, our answer is $\left(- 4 , \frac{9}{2}\right)$.

Apr 19, 2017

$x = - 4 \text{ , } y = 4.5$

#### Explanation:

$9 x + 8 y = 0 \text{ , } \left(1\right)$
$3 x - 8 y = - 48 \text{ , } \left(2\right)$

$\text{let us sum the equation (1) and (2).}$

$9 x + \cancel{8 y} + 3 x - \cancel{8 y} = 0 - 48$

$12 x = - 48$

$\text{let us divide both sides of equation by 12.}$

$\frac{\cancel{12} x}{\cancel{12}} = \frac{- 48}{12}$

$x = - 4$

$\text{let us write x=-4 in the equation (1).}$

$9 \cdot \left(- 4\right) + 8 y = 0$

$- 36 + 8 y = 0$

$8 y = 36$

$\frac{\cancel{8} y}{\cancel{8}} = \frac{36}{8}$

$y = \frac{36}{8} = \frac{9}{2} = 4.5$