# How do you solve (1/2)x - (1/3)y = -3 and (1/8)x + (1/6)y = 0?

Jan 18, 2016

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

#### Explanation:

$\left(\frac{1}{2}\right) x - \left(\frac{1}{3}\right) y = - 3$ and $\left(\frac{1}{8}\right) x + \left(\frac{1}{6}\right) y = 0$

The minimum common multiple between 2,3,6 and 8 is 24. So
let's convert all these fractions in something/24:

$\left(\frac{12}{24}\right) x - \left(\frac{8}{24}\right) y = - \left(\frac{72}{24}\right)$ and $\left(\frac{3}{24}\right) x + \left(\frac{4}{24}\right) y = 0$

Then multiply all the equalities by 24:

$12 x - 8 y = - 72$ and $3 x + 4 y = 0$

If $3 x + 4 y = 0$ the double $6 x + 8 y$ is also 0, so we can
add it to the other equation without change the result:

$6 x + 8 y = 0$ plus $12 x - 8 y = - 72$ gives $18 x = - 72$

So, $x = - \frac{72}{18} = - 4$

and applying the value $x = - 4$ to the expression:

$3 x + 4 y = 0$

gives

$3 \cdot \left(- 4\right) + 4 y = 0$
$4 y = 12$
$y = 3$