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 2y 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 2y – 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: 5x – (10 – 2x – 3z) - 5z = 2
Equation #3: x + 2(10 – 2x – 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/