Solving a percent mixture problem using a system of linear equations

Question

A chemical company mixes pure water with their premium antifreeze solution to create an inexpensive antifreeze mixture. The premium antifreeze solution contains 60% pure antifreeze. The company wants to obtain 180 gallons of mixture that contains 20% pure antifreeze. How many gallons of water and how many gallons of the premium antifreeze solution must be mixed?

Water: gallons

Premium antifreeze: gallons

Answer 1

x+y = 180

.6x +.0y = .2(280)

.6x =36

x =36/.6 = 60 gal premium

y= 180-60= 120 gal water

.6x + 0(180-x) = .2(180)

.6x = 36

x =36/.6 =60 gallons of 60% premium

mixed with 180-60= 120 gallons of water

yield 180 gallons of 20%

60-20 = 40

20-0 = 20

40/20 = 2

you need twice as much water as premium

because 0% is twice as close to 20% as 60% is

120/60= 2