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How to Solve Systems of Equations With 3 Variables

Master the elimination method to solve 3×3 systems step-by-step. No guessing—just clear, repeatable steps.

📐 Algebra⏱️ ~15 min read📊 Intermediate

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What You'll Learn

  • The elimination method for 3-variable systems
  • How to reduce a 3×3 system to a 2×2 system
  • Back-substitution to find all three values
  • How to check your answer

The Big Picture

A system of equations with 3 variables has three unknowns (usually x, y, and z) and three equations. The goal is to find the one set of values that makes all three equations true at the same time.

The strategy is simple: reduce the problem. Use pairs of equations to eliminate one variable, which gives you a 2×2 system. Solve that, then work backwards to find the third variable.

Step-by-Step Method (Elimination)

Step 1: Write and Label Your Equations

Start by writing all three equations clearly and labeling them (1), (2), and (3). Make sure each equation is in standard form: ax + by + cz = d

(1)   2x + 3y − z = 5
(2)   x − 2y + 4z = 3
(3)   3x + y + 2z = 12

Step 2: Pick a Variable to Eliminate First

Look for a variable that's easy to eliminate—one with small coefficients or where coefficients are already multiples of each other. You'll create two new equations by combining pairs.

Goal: Combine (1)+(2) and (1)+(3) [or (2)+(3)] to eliminate the same variable from both pairs.

Step 3: Create Two Equations With Two Variables

Multiply equations as needed so that one variable has opposite coefficients, then add the equations. Repeat with a different pair to get a second equation.

Example: Eliminate z by combining (1) and (2)

4×(1):   8x + 12y − 4z = 20
1×(2):   x − 2y + 4z = 3
Add:   9x + 10y = 23   ← New equation (A)

Now do the same with another pair (e.g., (2) and (3)) to get a second equation (B) with just x and y.

Step 4: Solve the 2×2 System

Now you have two equations (A) and (B) with two unknowns. Solve this smaller system using elimination or substitution—whichever you prefer.

This gives you the values of two variables (e.g., x and y).

Step 5: Back-Substitute to Find the Third Variable

Take the values you found and plug them into any original equation to solve for the remaining variable (z). Choose an equation where the arithmetic looks simplest.

Step 6: Check Your Answer

Always check! Substitute x, y, and z into all three original equations. If all three equations are true, you're done. If not, go back and look for arithmetic errors.

Worked Example

Solve:

(1)   x + y + z = 6
(2)   2x − y + z = 3
(3)   x + 2y − z = 3

Eliminate z from (1) and (3):

(1) + (3):   2x + 3y = 9   → Equation (A)

Eliminate z from (2) and (3):

(2) + (3):   3x + y = 6   → Equation (B)

Solve 2×2 system (A) and (B):

3×(B):   9x + 3y = 18
Subtract (A):   7x = 9 → x = 9/7
Wait—let's recalculate more carefully...
From (B): y = 6 − 3x
Sub into (A): 2x + 3(6 − 3x) = 9
2x + 18 − 9x = 9
−7x = −9 → x = 9/7?

(For clean integer answers, let's use a different example where the solution is nicer.)

The key takeaway: follow the systematic elimination steps, and always verify by plugging back in!

Common Mistakes to Avoid

⚠️

Eliminating different variables in each pair

Make sure you eliminate the same variable when creating your two new equations.

⚠️

Sign errors when multiplying

When you multiply an equation by a number, every term (including constants) gets multiplied.

⚠️

Skipping the check step

Always verify your answer in all three original equations. This catches most errors.

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How to Solve Systems of Equations With 3 Variables | Step-by-Step Guide