Question

Solving Applied Problems: Two Equations? problem 1 The St.mark's Community bbq served 250 dinners. A child's plate cost $3.50 and an adult's plate cost $7.00. A total of $1347.50 was collected. How many of each type of plate was served?

The St.mark's Community bbq served 250 dinners. A child's plate cost $3.50 and an adult's plate cost $7.00. A total of $1347.50 was collected. How many of each type of plate was served?

Answer 1

Yes, you can build two equations here.

`c` = amount of child's plates
`a` = amount of adult's plates

What do you know?
1) you know that in total, 250 diners were served.
So, `c + a = 250`

What else do you know?
2) The costs for each plates and the total cost. This can be expressed as the following equation:
`3.5 c + 7 a = 1347.5`

Now, to solve the linear equation system, I would solve the first one for `c` or `a` - your choice - and plug it in the second.

For example, you can solve the first equation for `c`:
`c = 250 - a`

Pluging this in the second equation gives you:
`3.5 * ( 250 - a) + 7 a = 1347.5`
`875 - 3.5 a + 7 a = 1347.5`
`3.5 a = 472.5`
`a = 135`

This means that there were `135` adult's plates. The only thing left to do is to compute the amount of child's plates:
`c = 250 - a = 250 - 135 = 115`

Result: `135` adult's plates, `115` child's plates.