# A girl scout troop sells 62 tickets to their mother and daughter dinner, for a total of \$216.If the tickets cost \$4.00 for mothers and \$3.00 for daughters how many of each ticket did they sell?

Ok so you need 2 equations. I like to use variables that go with the problem
Ok so the first equation is the quantity equation. How many of each ticket was sold for a total of how many tickets? We don't know how many of each ticket was sold, so that's where the variables come in.
M+D=62
the second equation is the value equation. How much is each ticket worth? Then it equals the total cost of all the tickets
4M+3D=216
Ok so you have
M+D=62
4M+3D=216
You have to get one variable to eliminate it. This is appropriately named the elimination method:) personally, I like to multiply. I don't know why it just works easier for me. So I need to get one variable the same, so I'll solve for M, so I'll multiply equation 1 by 4.
4M+4D=248 I multiplied the entire thing by 4
4M+3D=216 now that you have a variable the same (4M) you can subtract the second equation from the first
4M-4M is 0, so that's gone and you're left with
D=32 plug 32 back into the very first equation
M+32=62 subtract 32 from 62
M=30
so 32 daughter tickets were sold and 30 mother tickets were sold:D
thanked the writer.
If all 62 tickets are sold for daughters, the revenue is \$186. The revenue is actually \$30 more. We know that selling a mother ticket instead of a daughter ticket increases revenue by \$1, so there must have been 30 mother tickets sold. The remaining 32 tickets were daughter tickets.
thanked the writer.
A systems problem. Change all money to pennies for convenience of operation.

M = moms

D = daughters

- 300 (M + D = 62)
400M + 300D = 21600

- 300M - 300D = - 18600
400M + 300D = 21600

100M = 3000

M (moms) = 30 tickets * \$4.00 = \$120.00
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D(daughters) = 32 tickets * \$3.00 = \$96.00 