# 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?

Nov 6, 2015

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:
$3.5 \cdot \left(250 - a\right) + 7 a = 1347.5$
$875 - 3.5 a + 7 a = 1347.5$
$3.5 a = 472.5$
$a = 135$

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.