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Mom's Basement University

Systems with Three Variables

What You’ll Learn

Section titled “What You’ll Learn”

In this lesson you’ll learn how to solve systems of three linear equations with three variables using elimination and how to interpret the solutions.

The Concept

Section titled “The Concept”

A system of three linear equations in three variables looks like:

{a1x+b1y+c1z=d1a2x+b2y+c2z=d2a3x+b3y+c3z=d3\begin{cases} a_1 x + b_1 y + c_1 z = d_1 \\ a_2 x + b_2 y + c_2 z = d_2 \\ a_3 x + b_3 y + c_3 z = d_3 \end{cases}

Solutions can be:

  • One unique solution (x, y, z)
  • No solution (inconsistent system)
  • Infinitely many solutions (dependent system)

Method (Elimination):

  1. Eliminate one variable from two pairs of equations to get two equations with two variables.
  2. Solve the resulting 2×2 system.
  3. Substitute back to find the third variable.
  4. Check the solution in all three original equations.

The flowchart shows the process visually. You start with 3 equations in 3 variables (blue), eliminate one variable to get 2 equations in 2 variables (green), solve down to 1 equation in 1 variable (orange), then back-substitute (purple) to find all three values and arrive at the solution (gold). Always check your answer in all three original equations at the end.

Worked Example

Section titled “Worked Example”

Example 1: No solution (inconsistent)

{x+2y+z=82x−y+3z=15x−3y+2z=1\begin{cases} x + 2y + z = 8 \\ 2x - y + 3z = 15 \\ x - 3y + 2z = 1 \end{cases}

Step 1: Eliminate x from equations 1 and 2. Compute Eq2 − 2(Eq1):

(2x−y+3z)−2(x+2y+z)=15−16⇒−5y+z=−1(2x - y + 3z) - 2(x + 2y + z) = 15 - 16 \quad\Rightarrow\quad -5y + z = -1

Eliminate x from equations 1 and 3. Compute Eq3 − Eq1:

(x−3y+2z)−(x+2y+z)=1−8⇒−5y+z=−7(x - 3y + 2z) - (x + 2y + z) = 1 - 8 \quad\Rightarrow\quad -5y + z = -7

Step 2: We now have −5y + z = −1 and −5y + z = −7. The left sides are identical but the right sides are different. This is a contradiction. The system has no solution.

Example 2: Unique solution

{x+y+z=62x−y+z=3x+2y−z=2\begin{cases} x + y + z = 6 \\ 2x - y + z = 3 \\ x + 2y - z = 2 \end{cases}

Step 1: Eliminate x using Eq1 with the other two equations.

Eq2 − 2(Eq1):

(2x−y+z)−2(x+y+z)=3−12⇒−3y−z=−9(2x - y + z) - 2(x + y + z) = 3 - 12 \quad\Rightarrow\quad -3y - z = -9

Eq3 − Eq1:

(x+2y−z)−(x+y+z)=2−6⇒y−2z=−4(x + 2y - z) - (x + y + z) = 2 - 6 \quad\Rightarrow\quad y - 2z = -4

Step 2: Solve the 2×2 system.

From −3y − z = −9, solve for z:

z=−3y+9z = -3y + 9

Substitute into y − 2z = −4:

y−2(−3y+9)=−4⇒y+6y−18=−4⇒7y=14⇒y=2y - 2(-3y + 9) = -4 \quad\Rightarrow\quad y + 6y - 18 = -4 \quad\Rightarrow\quad 7y = 14 \quad\Rightarrow\quad y = 2

Step 3: Back-substitute.

z=−3(2)+9=3z = -3(2) + 9 = 3

From Eq1: x + 2 + 3 = 6, so x = 1.

Step 4: Check in all three original equations.

  • Eq1: 1 + 2 + 3 = 6 ✓
  • Eq2: 2(1) − 2 + 3 = 3 ✓
  • Eq3: 1 + 2(2) − 3 = 2 ✓

Solution: x = 1, y = 2, z = 3.

Real-World Application

Section titled “Real-World Application”

Systems with three variables model real situations such as:

  • Mixing three ingredients to achieve a target concentration or cost
  • Balancing a budget with three income/expense categories
  • Solving for currents in a three-loop electrical circuit
  • Finding production levels of three products to meet demand and profit goals

Example: a company produces three products. The system can represent total materials used, labor hours, and revenue targets.

Quiz

Section titled “Quiz”
A system of three linear equations can have:
The first step when solving a 3×3 system is usually to:
If after elimination you get 0 = 5, the system has:
After finding values for x, y, and z, you should:
Solve: $x + y = 5$, $y + z = 7$, $x + z = 6$. What is $x$?