1) An aerobics class is being offered at the ABC body gym. Nonmembers pay $4 per class while member pay a monthly fee of $10 plus $2 per class.
a. Create a linear system to represent the two options.
b. At how many classes will the cost be the same? What is that cost?
c. When does being a member cost effective?
a. y=4x
y=2x+10
b. six classes
It costs $24
c. Being a member cost is more effective when you go to more than six classes a month.
a. To create a linear system, we need to define two variables: x, the number of classes, and y, the cost. We can then represent the two options as equations.
For nonmembers:
y = 4x
For members:
y = 10 + 2x
b. To find the number of classes when the cost is the same, we can set the two equations equal to each other and solve for x:
4x = 10 + 2x
Combine like terms:
4x - 2x = 10
Simplify:
2x = 10
Divide both sides by 2:
x = 5
So, at 5 classes, the cost will be the same for both nonmembers and members. To find the cost, substitute x = 5 into either equation:
For nonmembers:
y = 4(5) = 20
For members:
y = 10 + 2(5) = 20
Therefore, at 5 classes, the cost will be $20 for both nonmembers and members.
c. To determine when being a member is cost-effective, we can compare the total cost for a certain number of classes for both nonmembers and members.
Let's say we want to find the number of classes at which being a member is cost-effective. We can set up an inequality to represent this:
10 + 2x < 4x
Simplify:
10 < 2x
Divide both sides by 2:
5 < x
So, being a member is cost-effective when the number of classes is greater than 5.