Math word problem

Question

A solution of 60% fertilizer is to be mixed with a solution of 21% fertilizer to form 234 liters of a 43% solution. How many liters of the 60% solution must be used?

Answer 1

Let x be the amount of the 60% solution and y be the 21% solution.

x+y = 234 --> y = 234-x

.6x + .21y = .43(234)

.6x + .21(234-x) = .43(234)

.6x + .21(234)-.21x = .43(234)

.39x = 234(.43-.21) = 234(.22)

x = 234(.22)/.39 = 132

132 liters of the 60% solution need to be used.