# How do you solve the system of equations 7x + 5y + 7z = 60, 15x + 2z = - 195, and 5z = 75?

Aug 6, 2018

$x = - 15$
$y = 12$
$z = 15$

#### Explanation:

First solve for $z$ using the third equation

$z = \frac{75}{5} = 15$

Now that we know $z$, we can plug it into the second equation and solve for $x$

$15 x + 2 \left(15\right) = - 195$

$15 x = - 225$

$x = - 15$

Now we can find $y$ using the values of $x$ and $z$ and plugging them into the first equation

$7 \left(- 15\right) + 5 y + 7 \left(15\right) = 60$

$5 y = 60$

$y = 12$

Aug 6, 2018

$x = - 15$
$y = 12$
$z = 15$

#### Explanation:

This question is far easier than it appears at first.

The key idea is that if a a linear equation has $1$ variable, there will be one solution, but as soon as there are $2$ variable, $2$ equations are required and likewise for $3$ variables, there must be $3$ equations.

We have all of these scenarios presented here.

Solve the $3 r d$ equation first as it only has $1$ variable,

$5 z = 75$
$\textcolor{b l u e}{z = 15} \text{ } \leftarrow$ now use this value for $z$ in the $2 n d$ equation:

$15 x + 2 \textcolor{b l u e}{z} = - 195$
$15 x + 2 \textcolor{b l u e}{\left(15\right)} = - 195$
$15 x + \textcolor{b l u e}{30} = - 195$
$15 x = - 225$
$\textcolor{g r e e n}{x = - 15} \text{ } \leftarrow$ use this value for $x$ in the first equation

$7 \textcolor{g r e e n}{x} + 5 y + 7 \textcolor{b l u e}{z} = 60$
$7 \textcolor{g r e e n}{\left(- 15\right)} + 5 y + 7 \textcolor{b l u e}{\left(15\right)} = 60$

$- 105 + 5 y + 105 = 60$
$5 y = 60$
$y = 12$