A chemist has three acid solutions. The first solution contains 15% acid, the second contains 35% and the third contains 65%. He wants to use all three solutions to obtain a mixture of 228 liters containing 25% acid, using 2 times as much of the 65% solution as the 35% solution. How many liters of each solution should be used?

Answer:171 liters of 15% acid19 liters of 35% acid38 liters of 65% acidStep-by-step explanation:If x is the liters of 15% acid, y is the liters of 35% acid, and z is the liters of 65% acid, then:x + y + z = 2280.15x + 0.35y + 0.65z = 0.25(228)z = 2ySolve the system of equations using elimination or substitution.  Using substitution:x + y + 2y = 228x + 3y = 228x = 228 − 3y0.15(228 − 3y) + 0.35y + 0.65(2y) = 0.25(228)34.2 − 0.45y + 0.35y + 1.3y = 571.2y = 22.8y = 19x = 228 − 3y = 171z = 2y = 38The chemist needs 171 liters of 15% acid, 19 liters of 35% acid, and 38 liters of 65% acid.