Math question system of linear equations

Question

A chemist has three different acid solutions. The first acid solution contains 20% acid, the second contains 35% and the third contains 80%. They want to use all three solutions to obtain a mixture of 108 liters containing 40% acid, using 2 times as much of the 80% solution as the 35% solution. How many liters of each solution should be used?

Answer 1

Let X = liters of 20% solution

Let Y = liters of 35% solution

Let Z = liters of 80% solution

X + Y + Z = 108

.20X + .35Y + .80Z = .40(108)

Z = 2X

X + Y + 2X = 108

.20X + .35Y + .80(2X) = 43.2

3X + Y = 108

.20X + .35Y + 1.60X = 43.2

1.80X + .35Y = 43.2

Y = 108 - 3X

1.80X + .35(108 - 3x) = 43.2

1.80X + 37.80 - 1.05X = 43.2

.75X + 37.80 = 43.2

.75X = 5.40

X = 7.20

Z = 2(7.20) = 14.40

Y = 108 - 3(7.20) = 86.40

Checking --- .20(7.2) + .35(86.4) + .80(14.4) = 43.2

Use 7.2 liters of 20% solution and 86.4 liters of 35% solution and 14.4 liters of 80% solution.