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All courses Algebra I · A-CED.1 48 of 76
Create and solve one-variable equations and inequalities, including absolute-value, linear, quadratic, simple rational, and exponential cases.

Write and solve a linear equation for a fixed fee plus a rate

Problem
A gym charges a fixed \(\$20\) fee plus \(\$5~\text{per}~\text{visit}\) for a total cost of \(\$65\). Let \(v\) be the number of visits. Write and solve a linear equation for \(v\).
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Hint

Let \(v\) represent the number of visits. The repeated charge is \(5v\), so the fixed-fee model is \(20+5v=65\).

In a fixed-fee-plus-rate relationship, the fixed fee is added once and the rate is multiplied by the variable. Subtract \(20\) before dividing by \(5\).

Solution walkthrough

01

Define the cost quantities

\[\begin{aligned} v=\text{number}~\text{of}~\text{visits} \\ \text{fixed}~\text{fee}=20~\text{dollars} \\ \text{visit}~\text{cost}=5v~\text{dollars} \end{aligned}\]

The $20 fee is paid once. Each of v visits costs $5, so the variable charge is 5v dollars.

02

Write the total-cost equation

\[\begin{aligned} \text{fixed}~\text{fee}+\text{rate}*\text{visits}=\text{total} \\ 20+5v=65 \end{aligned}\]

A fixed-fee model adds the one-time fee to the repeated rate. Substituting 20 dollars, 5 dollars per visit, and the 65-dollar total gives 20+5v=65.

03

Isolate the visit count

\[\begin{aligned} 20+5v=65 \\ 5v=45 \\ v=9 \end{aligned}\]

Subtract 20 from both sides to remove the fixed fee, then divide both sides by 5 dollars per visit. The units leave 9 visits.

04

Interpret and check

\[20+5(9)=20+45=65\]

Nine visits reproduce the stated total, so the equation is 20+5v=65 and its solution is v=9 visits.

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Another way

  1. Reason with money directly: $65-$20=$45 remains for visits, and $45 divided by $5 per visit is 9 visits.

!

Common mistake

Do not divide the full $65 by $5. First subtract the one-time $20 fee; only the remaining $45 pays for visits.

Solution walkthrough video