# Solve Systems of Equations with Multiple Unknowns

When you’re solving equations with three variables (or unknowns), you do the same as you did previously with the substitution method. The only difference is that it has one more step.

I recommend that you split your page into three columns, with one equation in each column. Read the instructions below carefully!

Rule

### Howtosolveequationswiththreevariables

1.
Select one of the equations and solve it for one of the variables.
2.
Take the first equation and input it into the other two equations. Arrange them nicely in your columns.
3.
Now you have two equations with two variables. You can solve them both with the substitution method.
4.
Using the two answers you now have, you input them into the equation you started with to find the final variable.

Example 1

Solve the set of equations: $\begin{array}{lll}\hfill 2x-3y+4z& =24\phantom{\rule{2em}{0ex}}& \hfill \text{(1)}\phantom{\rule{0.33em}{0ex}}\\ \hfill x-2y+5z& =25\phantom{\rule{2em}{0ex}}& \hfill \text{(2)}\phantom{\rule{0.33em}{0ex}}\\ \hfill 3x+5y+3z& =5\phantom{\rule{2em}{0ex}}& \hfill \text{(3)}\phantom{\rule{0.33em}{0ex}}\end{array}$

1.
Select (2) and solve for $x$:
$\begin{array}{cc}x-2y+5z=25& \\ x=25+2y-5z& \end{array}$

$\begin{array}{llll}\hfill x-2y+5z& =25\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x& =25+2y-5z\phantom{\rule{2em}{0ex}}& \hfill \text{(4)}\phantom{\rule{0.33em}{0ex}}\end{array}$

2.

(1):

$\begin{array}{cc}2x-3y+4z=24& \\ 2\left(25+2y-5z\right)-3y+4z=24& \\ 50+4y-10z-3y+4z=24& \\ y-6z=-26& \end{array}$

$\begin{array}{llll}\hfill 2x-3y+4z& =24\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 2\left(25+2y-5z\right)-3y+4z& =24\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 50+4y-10z-3y+4z& =24\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill y-6z& =-26\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

This is equal to
 $y=6z-26.$ (5)

(3): $\begin{array}{llll}\hfill 3x+5y+3z& =5\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 3\left(25+2y-5z\right)+5y+3z& =5\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 75+6y-15z+5y+3z& =5\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

This simplifies to

 $11y-12z=-70$ (6)
3.
Solve the set of equations with the two new equations. Substitute (5) into (6): $\begin{array}{llll}\hfill 11y-12z& =-70\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 11\left(6z-26\right)-12z& =-70\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 66z-286-12z& =-70\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 54z& =216\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill z& =4\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Substitute this back into (5) and you get:

 $y=6\cdot 4-26=-2.$
4.
Substitute the two answers into (4) and solve: $\begin{array}{llll}\hfill x& =25+2y-5z\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =25+2\cdot \left(-2\right)-5\cdot 4\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =25-4-20\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =1.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

 $\left(x,y,z\right)=\left(1,-2,4\right).$