Hello Cammie,
Let's let $T$ be the total value of the two candies in dollars, and $x$ be the amount in pounds of the 50-cent candy that must be used to meet the stated goal.
Before mixing, we may state the combined value of the two candies as:
$T=20\cdot0.80+x\cdot0.50$
and we may arrange this as:
(1) $2T-x=32$
After mixing, we may state the total value of the mixture as:
$T=0.60(20+x)$
which we may arrange as:
(2) $5T-3x=60$
To eliminate $T$, we may multiply (1) by 5 and (2) by -2 and add them to get:
$x=40$
Thus, we have found the troop needs to add 40 lbs. of the 50-cent candy to the 20 lbs. of 80-cent candy to get a mixture that can be sold for 60 cents per pound without any gain or loss.
You may be interested to know that what we have done is equivalent to using a weighted average to solve the problem:
$\displaystyle \frac{20\cdot0.80+x\cdot0.50}{20+x}=0.60$
$\displaystyle 20\cdot0.80+x\cdot0.50=0.60(20+x)$
Do you see that this is $T=T$?
$\displaystyle 16+\frac{1}{2}x=12+\frac{3}{5}x$
$\displaystyle 4=\left(\frac{3}{5}-\frac{1}{2} \right)x=\frac{1}{10}x$
$\displaystyle x=40$