For this case, the first thing we should do is define a variable.
We have then:
x: number of classes
We now write the equations that model the problem:
With membership:
members pay a $ 20 membership fee and $ 3 for each aerobics class:
\(3x + 20\)
Without membership:
Nonmembers pay $ 5 for each aerobics class:
\(5x\)
Then, by equating both equations we have:
\(3x + 20 = 5x\)
Clearing x we have:
\(5x-3x = 20\\2x = 20\\x = \frac {20} {2}\\x = 10\)
Answer:
The number of aerobics classes that the cost for members and nonmembers will be the same is:
\(x = 10\)
