SOLUTION: Here is a problem I can't figure out for the lifeof me. Museum charges $10 for regular admission, $7 for members, and $5 for students. Saturday, 750 tickets were sold for $5400. If

Question 13556: Here is a problem I can't figure out for the lifeof me. Museum charges $10 for regular admission, $7 for members, and $5 for students. Saturday, 750 tickets were sold for $5400. If 20 more student tickets than regular tickets were sold, how many of each type of ticket were sold?
This is what I was trying to do:
10x+7y+5s+(x+20)=750
I am just not sure if you would have to figure out, oh never mind, I am not understanding what to do. Please help me. Thank you!!!!!!!! :(
Answer by LilSkittleMd(119) (Show Source): You can put this solution on YOUR website!
Let r=regular
s=student
m=member
There are 3 different equations in this problem.
r+m+s=750 (equation 1)
r+0m-s=-20 (equation 2)
10r+7m+5s=5400 (equation 3)
First take the first two equations, use the elimination method to eliminate one variable. Eliminate r
-10r-10m-10s=-7500
10r+7m+5s=5400 (equation 4)
Now take the first and third equations and use the same method and eliminate the same variable
-r-m-s=-750
r+0m-s=-20
-m-2s=-770 (equation 5)
Take equation 4 and equation 5, use the same method, and eliminate one of the remaining variables
3m+6s=2310
-3m-5s=-2100
s=210
plug in the value of s into either equation 4 or 5
-m-2(210)=-770
-m-420=-770
-m=-350
m=350
Now take the m and s values and plug them into one of the 3 original equations
r+350+210=750
r+560=720
r=190
There were 190 regular tickets sold,350 member tickets sold,and 210 student tickets sold

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