You play racquetball at a community club. You have two options for paying
for court time.
Option A: You pay $12 for court time each time you play.
Option B: You buy a club membership for $120 and then pay $2 each
time you play.
a. Write and solve a linear system to determine the number of visits for
which the cost would be the same for each option.
b. For what numbers of visits is option B less expensive?
the two equations are
y = 12x
y = 120+2x
You need a graph for this
To determine the number of visits for which the cost would be the same for each option, we need to create a linear system of equations.
Let's assume "n" represents the number of visits:
Option A cost: $12 per visit
Option B cost: $120 (membership fee) + $2 per visit
Now we can create the equations:
Equation 1: Cost for Option A = Cost for Option B
12n = 120 + 2n
To solve this equation, we want to isolate the variable "n":
12n - 2n = 120
10n = 120
n = 12
So, for 12 visits, both options will cost the same.
To determine the number of visits for which Option B is less expensive, we need to compare the costs of both options for different values of "n".
Let's compare the costs for Option A and Option B for different values of "n":
For n = 1:
Option A cost: $12 per visit
Option B cost: $120 (membership fee) + $2 per visit
Option A cost: $12
Option B cost: $120 + $2 = $122
In this case, Option A is less expensive.
For n = 2:
Option A cost: $12 per visit
Option B cost: $120 (membership fee) + $2 per visit
Option A cost: $24
Option B cost: $120 + $4 = $124
Again, Option A is less expensive.
We can continue this process for different values of "n" to determine the point at which Option B becomes less expensive.
For n = 12:
Option A cost: $12 * 12 = $144
Option B cost: $120 + $2 * 12 = $144
Here, both options cost the same.
For n > 12:
Option A cost > Option B cost
Therefore, for any number of visits greater than 12, Option B is less expensive.
In summary, for any number of visits greater than 12, Option B is less expensive.