The Good Seed Flower Shop sells its own brand of fertilizer. It mixes one brand of fertilizer that contains 7% nitrogen with another than contains 8% nitrogen. The resulting mixture contains 7.7% nitrogen. How many pounds of the 8% fertilizer must be used to make 200 pounds of the 7.7% mixture?

Question

whpalmer4 · Answer

Let's call the variable representing the 7% nitrogen $s$
Let's call the variable representing the 8% nitrogen $e$

With mixture problems, it is often advantageous to work in terms of the amount of the thing of interest.  Here, that would be the nitrogen.

$0.07s$ gives the number of pounds of nitrogen in $s$ pounds of 7% fertilizer
$0.08e$ gives the number of pounds of nitrogen in $e$ pounds of 8% fertilizer

We are going to mix $s$ pounds of 7% with $e$ pounds of 8% to make 200 pounds of the 7.7% mixture, so we can write$$s+e=200$$

The amount of nitrogen in our mixture is going to be the amount of nitrogen in 200 pounds of 7.7% mixture, so we can also write
$$0.07s+0.08e = 200*0.077$$

Now you have a system of two equations in two unknowns, which can be easily solved by substitution.