Unit 4 · Lesson 4 of 4

Systems with three variables

About 25 minutes · 5 steps · one idea: cancel the same letter twice to get back to two variables

You'll be able to: solve a system of three equations in $x$, $y$, and $z$ by elimination.

1 Warm-up
2 The idea
3 Examples
4 Practice
5 Wrap-up

Step 1 of 5Warm-up

Two quick questions from earlier lessons. Both come back today. Answer from memory first.

Solve by elimination: $3x + 2y = 8$ and $x - 2y = 0$.

The $y$ terms are opposites. Add: $4x = 8$, so $x = 2$.

Then $2 - 2y = 0$, so $y = 1$.

Answer: $(2, 1)$.

Evaluate $2a - 3b + c$ when $a = 1$, $b = -2$, and $c = 4$.

Use parentheses: $2(1) - 3(-2) + 4$.

$= 2 + 6 + 4 = 12$.

Answer: $12$.

Step 2 of 5The idea

Some systems have three letters and three equations. A solution is now an ordered triple $(x, y, z)$ that makes all three equations true.

You already know how to solve two equations with two letters. So the plan is to cancel one letter twice. That leaves two equations with two letters, which you can solve by elimination like last time.

Three-variable systems, step by step:

Number the equations (1), (2), (3). Pick one letter to cancel, often the one with the easiest coefficients.
Use (1) and (2) to cancel that letter. Call the result (A).
Use a different pair, such as (1) and (3) or (2) and (3), to cancel the same letter. Call it (B).
Solve (A) and (B) together: two letters, two equations.
Put both values into any original equation to find the third letter. Check all three equations.

Shortcut: if an equation is already missing a letter, it counts as one of your two-letter equations.

Why this shows up on the placement test: three-variable systems are the top end of the MATH 1010 systems topic, and they are mostly careful bookkeeping with the elimination you already know.

Step 3 of 5Worked examples

Example 1: Is $(2, 1, -1)$ a solution of $x + y + z = 2$, $x - y + 2z = -1$, and $3x + y - z = 8$?

(1): $2 + 1 + (-1) = 2$. True.
(2): $2 - 1 + 2(-1) = -1$. True.
(3): $3(2) + 1 - (-1) = 6 + 1 + 1 = 8$. True.
All three are true, so yes. One false equation would mean no.

Example 2: Solve the system

\[\begin{aligned} x + y + z &= 6 &\quad (1)\\ 2x - y + z &= 3 &\quad (2)\\ x + 2y - z &= 2 &\quad (3) \end{aligned}\]

Cancel $z$. It has coefficients $+1$ or $-1$ everywhere, so no multiplying is needed.
Add (1) and (3): $2x + 3y = 8$. Call this (A).
Add (2) and (3): $3x + y = 5$. Call this (B). Same letter, $z$, canceled again.
Solve (A) and (B). From (B), $y = 5 - 3x$. Then $2x + 3(5 - 3x) = 8$, so $2x + 15 - 9x = 8$, $-7x = -7$, and $x = 1$.
$y = 5 - 3(1) = 2$. From (1): $1 + 2 + z = 6$, so $z = 3$.
Check (2): $2 - 2 + 3 = 3$. Check (3): $1 + 4 - 3 = 2$. The solution is $(1, 2, 3)$.

Step 4 of 5Practice

Work on paper, labeling your equations. Write triples like $(1, 2, 3)$, in the order $(x, y, z)$. If you miss twice, open the worked solution.

Is $(1, -2, 3)$ a solution of $x + y + z = 2$, $2x - y + z = 7$, and $x - y + 2z = 8$?

(1): $1 - 2 + 3 = 2$. True.

(2): $2 + 2 + 3 = 7$. True.

(3): $1 + 2 + 6 = 9$, not 8. False.

One equation fails, so the answer is no.

Solve: $x + y + z = 4$, $\;y + 2z = 5$, $\;z = 3$.

$z = 3$ is given. Put it into $y + 2z = 5$: $y + 6 = 5$, so $y = -1$.

Put both into the first: $x - 1 + 3 = 4$, so $x = 2$.

Answer: $(2, -1, 3)$.

Solve: $x + y + z = 6$, $\;x - y + z = 4$, $\;x + y - z = 2$.

Subtract the second from the first: $2y = 2$, so $y = 1$.

Subtract the third from the first: $2z = 4$, so $z = 2$.

Then $x + 1 + 2 = 6$, so $x = 3$.

Answer: $(3, 1, 2)$.

Solve: $x + y + z = 2$, $\;2x - y + 3z = 0$, $\;x + 2y - z = 5$.

Cancel $z$. Add (1) and (3): $2x + 3y = 7$ (A).

Multiply (3) by 3 to get $3x + 6y - 3z = 15$, and add it to (2): $5x + 5y = 15$, so $x + y = 3$ (B).

From (B), $x = 3 - y$. Then $2(3 - y) + 3y = 7$, so $6 + y = 7$ and $y = 1$, $x = 2$.

From (1): $2 + 1 + z = 2$, so $z = -1$. Answer: $(2, 1, -1)$.

Solve: $x + y + z = 5$, $\;2x + y - z = -4$, $\;x - y + 2z = 5$.

Cancel $y$. Add (1) and (3): $2x + 3z = 10$ (A). Add (2) and (3): $3x + z = 1$ (B).

From (B), $z = 1 - 3x$. Then $2x + 3(1 - 3x) = 10$, so $2x + 3 - 9x = 10$, $-7x = 7$, and $x = -1$.

$z = 1 - 3(-1) = 4$. From (1): $-1 + y + 4 = 5$, so $y = 2$.

Answer: $(-1, 2, 4)$.

Three numbers add up to 15. The second number is 3 more than the first. The third number is twice the first. Find the three numbers, written as (first, second, third).

Let the numbers be $a$, $b$, $c$: $a + b + c = 15$, $b = a + 3$, $c = 2a$.

Substitute: $a + (a + 3) + 2a = 15$, so $4a + 3 = 15$, $4a = 12$, and $a = 3$.

Then $b = 6$ and $c = 6$. Check: $3 + 6 + 6 = 15$.

Answer: $(3, 6, 6)$.

A museum sold 100 tickets in one day: child tickets for $5, adult tickets for $8, and senior tickets for $6. Sales were $640, and there were twice as many adult tickets as senior tickets. How many adult tickets were sold?

Let $c$, $a$, $s$ be child, adult, senior. Then $c + a + s = 100$, $5c + 8a + 6s = 640$, and $a = 2s$.

Replace $a$ with $2s$: $c + 3s = 100$ and $5c + 16s + 6s = 640$, so $5c + 22s = 640$.

From $c = 100 - 3s$: $5(100 - 3s) + 22s = 640$, so $500 + 7s = 640$, $7s = 140$, and $s = 20$.

Then $a = 40$ (and $c = 40$). Check: $200 + 320 + 120 = 640$. Answer: $40$ adult tickets.

Step 5 of 5Wrap-up

You can now check an ordered triple, and solve a three-variable system by canceling the same letter twice, solving the two-variable system that's left, and filling in the third letter.

Read or watch next: OpenStax Intermediate Algebra 2e, 4.4 Solve Systems of Equations with Three Variables (free textbook, with more practice at the end of the section).

Also good: OpenStax College Algebra 2e, 7.2 Systems of Linear Equations: Three Variables (the same idea from the College Algebra book). Quick rules: Lines and systems cheat sheet.

Something unclear? Ask your teacher bot a follow-up question, for example: "How do I pick which letter to cancel in a three-variable system?" It knows this course and can explain things another way.