Greg the trainer has two solo workout plans that he offers his clients: plan a and plan b. each client does either one or the other (not both). on wednesday there were 3 clients who did plan a and 5 who did plan b. on thursday there were 9 clients who did plan a and 7 who did plan b. greg trained his wednesday clients for a total of 6 hours and his thursday clients for a total of 12 hours. how long does each of the workout plans last?

This problem can be solved by algebraic method.
Let
x = the total time spent of all clients in Plan A
y = the total time spent of all clients in Plan B
We represent two variables x and y because there are two plans that won't be happened simultaneously.
On Wednesday, the two workout plans have the total time of 6 hours. We equate
3x + 5y = 6
While on Thursday, the total time is 12 hours. We also equate
9x + 7y = 12

To find x and y, we can use the substitution method. For the first equation, we arrange it in terms of y, that is
5y = 6 - 3x
y = (6 - 3x)/5

Substitute it to the second equation:
9x + (7/5)(6 - 3x) = 12
9x + (42/5) - (21/5)x = 12
Multiply the equation by 5 to cancel the denominator:
45x + 42 - 21x = 60
45x - 21x = 60 - 42
24x = 18
x = 18/24 = 3/4 hours

For y:
3(3/4) + 5y = 6
9/4 + 5y = 6
Multiply the equation by 4 to cancel the denominator:
9 + 20y = 24
20y = 24 - 9
20y = 15
y = 15/20 = 3/4 hours

Hence, each workout plans are done within 3/4 hours (or 45 minutes).

Ithink the answer would be knew