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All courses Math II · A-CED.1 2 of 73
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 per visit, and the total cost is $65. Let v represent the number of visits. Write and solve a one-variable equation to find 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 variable and write the equation

\[20+5v=65\]

Let \(v\) be the number of visits. The gym charges \(20\) once and \(5\) for each of \(v\) visits, so the equation is \(20+5v=65\).

02

Remove the fixed fee

\[20+5v-20=65-20;5v=45\]

Subtract \(20\) from both sides to keep the equation balanced and isolate the visit charges.

03

Divide to solve

\[\frac{5v}{5}=\frac{45}{5};v=9\]

Divide both sides by \(5\). This gives \(v=9\), so the payment covers \(9\) visits.

04

Check and format the response

\[20+5(9)=65;20,5,65;9\]

Substitution gives \(20+45=65\), so the solution checks. The typed packet records the fixed fee, rate, and total before the semicolon and the solution after it.

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Another way

  1. Reason directly with the costs: \(65-20=45\) remains for visits, and \(45\div5=9\). The same reasoning produces the model parameters \(20,5,65\) and solution \(9\).

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Common mistake

Do not compute \(65\div5\) before accounting for the fixed fee. The \(20\) is charged once, so subtract it before dividing the remaining cost by \(5\).