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

May 30, 2018

$x = - \frac{6}{7}$, $y = 4 \frac{5}{7}$

#### Explanation:

We have
1) $3 y - 15 = x$
2) $y - 3 = - 2 x$

We can use substitution, i.e. insert $x = 3 y - 15$ from 1) into 2) and solve for y and afterwards x.

Here we have another way which is basically as simple, since we have
3) from 1): $2 x = 6 y - 30$ (we multiply each term with 2)
From 2): $- 2 x = y - 3$

Add 3) and 2 to get rid of the x terms:
4) $2 x - 2 x = 6 y - 30 + y - 3$
This gives $7 y - 33 = 0$ or $7 y = 33$
$y = \frac{33}{7}$
$x = 3 y - 15 = 3 \cdot \frac{33}{7} - 15 = \frac{99 - 15 \cdot 7}{7} = - \frac{6}{7}$

Check:
2)Left side: $y - 3 = \frac{33}{7} - 3 = 4 \frac{5}{7} - 3 = 1 \frac{5}{7}$
Right side: $- 2 x = - 2 \cdot \left(- \frac{6}{7}\right) = \frac{12}{7} = 1 \frac{5}{7}$