1Cademy - Solving a Salary Comparison Word Problem Using a System of Equations by Substitution

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Problem-Solving Strategy for Systems of Linear Equations

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Solving a Salary Comparison Word Problem Using a System of Equations by Substitution

Apply the seven-step problem-solving strategy for systems of linear equations to a real-world compensation problem, using substitution to determine when two salary options yield equal pay.

Problem: Heather has been offered two salary options as a trainer at a gym. Option A pays $25,000 plus $15 per training session. Option B pays $10,000 plus $40 per training session. How many training sessions would make the salary options equal?

Read the problem.
Identify what to find: the number of training sessions that would make the two salary options equal.
Name the unknowns: Let $s$ = Heather's salary and $n$ = the number of training sessions.
Translate into a system of equations. Each salary option is modeled as a base amount plus a per-session rate:

$\left\{\begin{array}{l} s = 25{,}000 + 15n \\ s = 10{,}000 + 40n \end{array}\right.$

Solve using substitution. Both equations are already solved for $s$ , so substitute $25{,}000 + 15n$ for $s$ in the second equation:

$25{,}000 + 15n = 10{,}000 + 40n$

Subtract $15n$ from both sides: $25{,}000 = 10{,}000 + 25n$ . Subtract $10{,}000$ : $15{,}000 = 25n$ . Divide by $25$ : $n = 600$ .

Check: Are 600 training sessions reasonable? Substituting $n = 600$ into both equations: Option A gives $s = 25{,}000 + 15(600) = 25{,}000 + 9{,}000 = 34{,}000$ . Option B gives $s = 10{,}000 + 40(600) = 10{,}000 + 24{,}000 = 34{,}000$ ✓.
Answer: The salary options would be equal for 600 training sessions.

This example demonstrates how a real-world compensation comparison naturally produces a system where both equations are already solved for the same variable ( $s$ ). Because both expressions equal $s$ , the substitution method reduces to setting the two right-hand sides equal to each other — a special case where Step 1 of the substitution procedure is already complete for both equations. The large numerical values (thousands) in this problem would make graphing impractical, further illustrating the advantage of algebraic methods.

Updated 2026-04-21

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OpenStax Elementary Algebra

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