Out of school for 35 years and never taking algebra makes my kids homework a real challange, especially when my daughter is in honors math and looks to me for help. She works very hard to stay in class and I guess I have to keep up whith the times so I can be there for her and my son right behind her in school. After beating my head against the wall all week on this one problem I came acrossed this forum with a question. I can figure the answer the hard way but I am very curious about the formula. Question seems so simple:
The class made $1220.00 on a school play they sold 295 tickets, adult tickets were $5 and children were $2 how many Adult tickets were sold.
Thank You
Jim
I used elementary row operations to solve the matrices: addition/subtraction of rows, multiplying by a scalar, and both operations combined.
The first was multiple row 1 by 5 and subtract it from the second.
Then, I divide row 2 by -3.
This is a simultaneous equations question.
Let represent the number of adult tickets and represent the number of children's tickets.
There were tickets sold, so .
They made per adult ticket and per children's ticket, getting a total of . So .
Now you have a system of equations:
.
Solve them simultaneously for and .
Another way to do is to solve as a simple system:
Let 'a' be the number adult tickets sold, and 'c' be the number of children tickets.
a + c = 295
5a + 2c = 1220
Solve for 'c' in the first:
c = 295 - a
Sub 'c = 295 - a' into the second:
5a + 2(295 - a) = 1220
5a + 590 - 2a = 1220
3a = 630
a = 210
Which is the number of adult tickets sold. The number of children can be found by subbing back into the first equation with a = 210.
210 + c = 295
c = 85
Edit: Beaten!