SOLUTION: A solution if 52% fertilizer is to be mixed with a solution of 22% fertilizer to form 90 liters of a 42% solution. How many of the 52% solution must be used?

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Question 710351: A solution if 52% fertilizer is to be mixed with a solution of 22% fertilizer to form 90 liters of a 42% solution. How many of the 52% solution must be used?
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
This one is the same general problem as question #710342. Different materails, different mixture unit, different values; but the same problem.

VARIABLES
High concentration fertilizer percent, H =52%
Low concentration fertilizer percent, L =22%
Target percent wanted, T = 42%
Amount of High conce. to use, v
Amount of Low conce. to use, u
Amount of target mixture produced, M = 90 Liters

This system can be formed from the description and question
%28uL%2BvH%29%2FM=T
u%2Bv=M

If we use the second equation as v=M-u, we can substitute it into the rational equation and solve that one for u, giving us this:
highlight%28u=%28M%28T-H%29%29%2F%28L-H%29%29

Now, just substitute the values and compute u.
We can then go to our simpler-looking expression for v and find (compute) it, too. highlight%28v=M-u%29