# How do you solve the system of equations algebraically x-y-z=7, x+2y+3z=-12, 3x-2y+7z=30?

##### 1 Answer
Mar 29, 2017

Multiply the first equation by -3 first. Then go ahead, x=2/26; y=-226/26 and z=46/26

#### Explanation:

When you multiply the first equation by -3, you will get:
$- 3 x + 3 y + 3 z = - 21$
Then combine this equation with the second. Yielding,
$y = 9 - 10 z$.

When you see y (in the first (original equation)), write this and for the second original equation do the same.

You will get $x + 9 z = 16$ and $x - 17 z = - 30$. You can multiply either the first equation by -1 or the second by -1. You will get
$z = \frac{46}{26}$

Since $y = 9 - 10 z$, you can find y now. $y = 9 - \left(\frac{460}{26}\right)$ or $y = - \frac{226}{26}$.

Finally, you can easily solve using any original equation. $x = \frac{2}{26}$.