# How do you solve 2y = 4 – 6x  and 3y + 9x = 6?

Mar 29, 2017

y = 2

#### Explanation:

$3 y + 9 x = 6$

Lets solve for y first

3y + 9x + 3y = 6 + 3y

9x = 6 + 3y

9x = 6 + 3y

x = 6 + 3y/9

using the value of x solve the other equation

$2 y = 4 - 6 \cdot \left(\frac{6 + 3 y}{9}\right)$

Solve for normal division that is $\frac{6 + 3}{9}$ but differently

$\frac{3 y}{9} + \frac{6}{9}$ = $\frac{1}{3} y + \frac{2}{3}$

Plug it into the equation

$2 y = 4 - 6 \cdot \frac{1}{3} y + \frac{2}{3}$

$2 y = 4 - \frac{6}{18} y + \frac{12}{18}$

Reduce the fractions to lower terms

$2 y = 4 - \frac{1}{3} y + \frac{2}{3}$

$2 y = \frac{1}{3} y + \frac{2}{3}$

Simplify

$4 - \frac{1}{3} y + \frac{2}{3} = 4 + \left(- \frac{1}{3} y\right) + \frac{2}{3}$

$2 y = 4 + \left(- \frac{1}{3} y\right) + \frac{2}{3}$

$2 y = - \frac{1}{3} y + \left(4 + \frac{2}{3}\right)$

$2 y = - \frac{1}{3} y + \frac{14}{3}$

Add $\frac{1}{3} y$ to both sides

$2 y + \frac{1}{3} y = - \frac{1}{3} y + \frac{14}{3} + \frac{1}{3} y$

$- \frac{1}{3} y + \frac{1}{3} y = 0$

$2 y + \frac{1}{3} y = \frac{14}{3}$

$\frac{7}{3} y = \frac{14}{3}$

$y = \frac{\frac{14}{3}}{\frac{7}{3}} = 2$