# How do you solve the system 1/3 x - 1/12 y = -7 and 1/3 x + 1/6 y = 11?

Mar 16, 2018

x = -3 and y = 72

#### Explanation:

So we are given with $2$ equations

equation 1 is $\frac{1}{3} x - \frac{1}{12} y$ = -7

equation 2 is $\frac{1}{3} x + \frac{1}{6} y$ = 11

equation 1 can be written as $4 x - y = - 84$
(multiplied the whole equation by 12)

equation 2 can be written as $2 x + y = 66$
(multiplied whole equation by 6)

now we add both the equations

==> $4 x - y + 2 x + y = - 84 + 66$
==> $6 x = - 18$

==> $x = - \frac{18}{6}$

==> $x = - 3$

now we substitute $x = - 3$ in any one of the equations
we will get $y = 72$

Mar 19, 2018

$x = - 3 \mathmr{and} y = 72$

#### Explanation:

Both equations have the same $x$- coeffiecients.

Rewrite both equations:

$\frac{1}{3} x = \frac{1}{12} y - 7 \text{ "and" } \frac{1}{3} x = - \frac{1}{6} y + 11$

$\frac{1}{3} x = \frac{1}{3} x$ , so equate the right sides of the equations.

$\frac{1}{12} y - 7 = - \frac{1}{6} y + 11 \text{ } \leftarrow$ multiply by $12$

$y - 84 = - 2 y + 132$

$y + 2 y = 132 + 84$

$3 y = 216$

$y = 72$

Substitute to find $x$

$\frac{1}{3} x = \frac{1}{12} \left(72\right) - 7$

$\frac{1}{3} x = 6 - 7$

$\frac{1}{3} x = - 1$

$x = - 3$