# Solving Three Variable Equations

When first introduced to systems of equations, you probably learned to solve a system of two-variable equations by graphing. But solving equations with three variables or more requires a new set of tricks, namely the techniques of elimination or substitution.

## An Example System of Equations

Consider this system of three, three-variable equations:

Equation #1:

\(2x + y + 3z = 10\)

Equation #2:

\(5x – y – 5z = 2\)

Equation #3:

\(x + 2y – z = 7\)

## Solving by Elimination

Look for places where adding any two equations together will make at least one of the variables cancel itself out.

### 1. Choose Two Equations and Combine

Choose any two of the equations and combine them to eliminate one of the variables. In this example, adding Equation #1 and Equation #2 will cancel out the *y* variable, leaving you with the following new equation:

New Equation #1:

\(7x – 2z = 12\)

### 2. Repeat Step 1 With Another Set of Equations

Repeat Step 1, this time combining a *different* set of two equations but eliminating the *same* variable. Consider Equation #2 and Equation #3:

Equation #2:

\(5x – y – 5z = 2\)

Equation #3:

\(x + 2y – z = 7\)

In this case the *y* variable doesn't immediately cancel itself out. So before you add the two equations together, multiply both sides of Equation #2 by 2. This gives you:

Equation #2 (modified):

\(10x – 2y – 10z = 4\)

Equation #3:

\(x + 2y – z = 7\)

Now the 2*y* terms will cancel each other out, giving you another new equation:

New Equation #2:

\(11x – 11z = 11\)

### 3. Eliminate Another Variable

Combine the two new equations you created, with the goal of eliminating yet another variable:

New Equation #1:

\(7x – 2z = 12\)

New Equation #2:

\(11x – 11z = 11\)

No variables cancel themselves out just yet, so you'll have to modify both equations. Multiply both sides of the first new equation by 11, and multiply both sides of the second new equation by −2. This gives you:

New Equation #1 (modified):

\(77x – 22z = 132\)

New Equation #2 (modified):

\(-22x + 22z = -22\)

Add both equations together and simplify, which gives you:

\(x = 2\)

### 4. Substitute the Value Back In

Now that you know the value of *x*, you can substitute it into the original equations. This gives you:

Substituted Equation #1:

\(y + 3z = 6\)

Substituted Equation #2:

\(-y – 5z = -8\)

Substituted Equation #3:

\(2y – z = 5\)

### 5. Combine Two Equations

Choose any two of the new equations and combine them to eliminate another one of the variables. In this case, adding Substituted Equation #1 and Substituted Equation #2 makes *y* cancel out nicely. After simplifying, you'll have:

\(z = 1\)

### 6. Substitute the Value In

Substitute the value from Step 5 into any one of the substituted equations, and then solve for the remaining variable, *y.* Consider Substituted Equation #3:

Substituted Equation #3:

\(2y – z = 5\)

Substituting in the value for *z* gives you 2*y* – 1 = 5, and solving for *y* brings you to:

\(y = 3\)

So the solution for this system of equations is *x* = 2, *y* = 3 and *z* = 1.

## Solving by Substitution

You can also solve the same system of equations using another technique called substitution. Here's the example again:

Equation #1:

\(2x + y + 3z = 10\)

Equation #2:

\(5x – y – 5z = 2\)

Equation #3:

\(x + 2y – z = 7\)

### 1. Choose a Variable and Equation

Pick any variable and solve any one equation for that variable. In this case, solving Equation #1 for *y* works out easily to:

\(y = 10 – 2x – 3z\)

### 2. Substitute That Into Another Equation

Substitute the new value for *y* into the other equations. In this case, choose Equation #2. This gives you:

Equation #2: 5*x* – (10 – 2*x* – 3*z*) *-* 5z = 2

Equation #3: *x* + 2(10 – 2*x* – *3z*) – *z* = 7

Make your life easier by simplifying both equations:

Equation #2:

\(7x – 2z = 12\)

Equation #3:

\(-3x – 7z = -13\)

### 3. Simplify and Solve for Another Variable

Choose one of the remaining two equations and solve for another variable. In this case, choose Equation #2 and *z*. This gives you:

\(z = \frac{7x – 12}{2}\)

### 4. Substitute This Value

Substitute the value from Step 3 into the final equation, which is #3. This gives you:

\(-3x – 7 × \frac{7x – 12}{2} = -13\)

Things get a little messy here but once you simplify, you'll be back to:

\(x = 2\)

### 5. Back-Substitute This Value

"Back-substitute" the value from Step 4 into the two-variable equation you created in Step 3:

\(z = \frac{7x – 12}{2}\)

This lets you solve for *z.* (In this case, *z* = 1).

Next, back-substitute both the *x* value and the *z* value into the first equation that you'd already solved for *y*. This gives you:

\(y = 10 – (2 × 2) – (3 × 1)\)

...and simplifying gives you the value *y* = 3.

## Always Check Your Work

Note that both methods of solving the system of equations brought you to the same solution: (*x* = 2, *y* = 3, *z* = 1). Check your work by substituting this value into each of the three equations.

### Cite This Article

#### MLA

Maloney, Lisa. "Solving Three Variable Equations" *sciencing.com*, https://www.sciencing.com/solving-three-variable-equations-13712183/. 5 December 2020.

#### APA

Maloney, Lisa. (2020, December 5). Solving Three Variable Equations. *sciencing.com*. Retrieved from https://www.sciencing.com/solving-three-variable-equations-13712183/

#### Chicago

Maloney, Lisa. Solving Three Variable Equations last modified March 24, 2022. https://www.sciencing.com/solving-three-variable-equations-13712183/