Write a system of equations to describe the situation below, solve using substitution, and fill in the blanks. $\newline$ Bill wants to take group fitness classes at a nearby gym, but needs to start by selecting a membership plan. With the first membership plan, Bill can pay $\$ 60$ per month, plus $\$ 1$ for each group class he attends. Alternately, he can get the second membership plan and pay $\$ 6$ per month plus $\$ 3$ per class. If Bill attends a certain number of classes in a month, the two membership plans end up costing the same total amount. How many classes per month is that? What is that total amount? $\newline$ If Bill attends $\square$ classes per month, each membership plan costs $\$ \square$ .

View step-by-step help

Home

Math Problems

Algebra 1

Solve a system of equations using substitution: word problems

Full solution

Q. Write a system of equations to describe the situation below, solve using substitution, and fill in the blanks.

\newline

Bill wants to take group fitness classes at a nearby gym, but needs to start by selecting a membership plan. With the first membership plan, Bill can pay

\$ 60

per month, plus

\$ 1

for each group class he attends. Alternately, he can get the second membership plan and pay

\$ 6

per month plus

\$ 3

per class. If Bill attends a certain number of classes in a month, the two membership plans end up costing the same total amount. How many classes per month is that? What is that total amount?

\newline

If Bill attends

\square

classes per month, each membership plan costs

\$ \square

Define Variables: Let's define the variables: $\newline$ Let $x$ be the number of classes Bill attends per month. $\newline$ Let $y$ be the total cost for each membership plan. $\newline$ For the first membership plan: $\newline$ The total cost is $\$60$ plus $\$1$ per class. $\newline$ So the equation for the first plan is: $y = 60 + 1x$
First Membership Plan: For the second membership plan: $\newline$ The total cost is $\$6$ plus $\$3$ per class. $\newline$ So the equation for the second plan is: $y = 6 + 3x$
Second Membership Plan: Now we need to find the number of classes ( extit{x}) where the cost of both plans is the same. This means we set the two equations equal to each other: $\newline$ $60 + 1x = 6 + 3x$
Set Equations Equal: To solve for $x$ , we'll subtract $1x$ from both sides: $60 = 6 + 2x$
Subtract $1x$ : Now, subtract $6$ from both sides to isolate the term with $x$ : $54 = 2x$
Isolate Term with x: Divide both sides by $2$ to solve for $x$ : $x = 27$
Solve for x: Now that we have the value for $x$ , we can substitute it back into either of the original equations to find the total cost ( $y$ ). We'll use the first equation: $\newline$ $y = 60 + 1x$ $\newline$ $y = 60 + 1(27)$
Substitute $x$ Value: Calculate the total cost: $y = 60 + 27$ $y = 87$