1Cademy - Solving an Exercise Calorie Word Problem Using a System of Equations by Elimination

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Problem-Solving Strategy for Systems of Linear Equations

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Solving an Exercise Calorie Word Problem Using a System of Equations by Elimination

Apply the seven-step problem-solving strategy for systems of linear equations to a real-world exercise calorie problem, using elimination to solve the resulting system.

Problem: When Jenna spent 10 minutes on the elliptical trainer and then did circuit training for 20 minutes, her fitness app says she burned 278 calories. When she spent 20 minutes on the elliptical trainer and 30 minutes circuit training she burned 473 calories. How many calories does she burn for each minute on the elliptical trainer? How many calories does she burn for each minute of circuit training?

Read the problem.
Identify what to find: the number of calories burned each minute on the elliptical trainer and each minute of circuit training.
Name the unknowns: Let $e$ = number of calories burned per minute on the elliptical trainer. Let $c$ = number of calories burned per minute while circuit training.
Translate into a system of equations. The first workout gives $10e + 20c = 278$ . The second workout gives $20e + 30c = 473$ . The system is:

$\left\{\begin{array}{l} 10e + 20c = 278 \\ 20e + 30c = 473 \end{array}\right.$

Solve using elimination. Both equations are in standard form. To create opposite coefficients of $e$ , multiply the first equation by $-2$ :

$-2(10e + 20c) = -2(278) \implies -20e - 40c = -556$

Add this to the second equation:

$-20e - 40c + 20e + 30c = -556 + 473$

$-10c = -83$

Divide both sides by $-10$ : $c = 8.3$ .

Substitute $c = 8.3$ into the first equation:

$10e + 20(8.3) = 278$

$10e + 166 = 278$

$10e = 112$

$e = 11.2$

Check: First workout: $10(11.2) + 20(8.3) = 112 + 166 = 278$ ✓. Second workout: $20(11.2) + 30(8.3) = 224 + 249 = 473$ ✓.
Answer: Jenna burns 11.2 calories per minute on the elliptical trainer and 8.3 calories per minute during circuit training.

This example shares the same structural pattern as calorie-counting problems involving food orders — two different combinations of the same two activities translate into two equations in standard form, making elimination a convenient method. A key difference in this problem is that the solution consists of decimal values ( $8.3$ and $11.2$ ) rather than whole numbers, which reinforces the advantage of algebraic methods over graphing when solutions are not integers.

Updated 2026-04-21

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OpenStax Elementary Algebra

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