# Thread: Worded problem.

1. ## Worded problem.

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 2. Originally Posted by Sven22 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.
Let a = adult tickets sold; then children tickets sold = 295 - a ; OK?

So 5a + 2(295 - a) = 1220 ; OK?

Can you solve that for a?

3. $\displaystyle 295=a+c$

$\displaystyle 5a+2c=1220$

$\displaystyle \displaystyle \begin{bmatrix} 1 & 1 & 295\\ 5 & 2 & 1220 \end{bmatrix}\Rightarrow \begin{bmatrix} 1 & 1 & 295\\ 0 & -3 & -255 \end{bmatrix}\Rightarrow\begin{bmatrix} 1 & 1 & 295\\ 0 & 1 & 85 \end{bmatrix}$

$\displaystyle c=85$

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.

4. This is a simultaneous equations question.

Let $\displaystyle \displaystyle a$ represent the number of adult tickets and $\displaystyle \displaystyle c$ represent the number of children's tickets.

There were $\displaystyle \displaystyle 295$ tickets sold, so $\displaystyle \displaystyle a+c = 295$.

They made $\displaystyle \displaystyle \$5$per adult ticket and$\displaystyle \displaystyle \$2$ per children's ticket, getting a total of $\displaystyle \displaystyle \$1220$. So$\displaystyle \displaystyle 5a + 2c = 1220$. Now you have a system of equations:$\displaystyle \displaystyle a+c = 295\displaystyle \displaystyle 5a+2c=1220$. Solve them simultaneously for$\displaystyle \displaystyle a$and$\displaystyle \displaystyle c\$.

5. 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!

6. so the algerbraic expression to solve this problem is 5a+2(295-a)=1220

7. Originally Posted by Sven22
so the algerbraic expression to solve this problem is 5a+2(295-a)=1220
Solve for a.

Plug the value of a into the equation a+c=295 to obtain c.