Achemical company makes two brands of antifreeze. the first brand is 55 % pure antifreeze, and the second brand is 80 % pure antifreeze. in order to obtain 90 gallons of a mixture that contains 75 % pure antifreeze, how many gallons of each brand of antifreeze must be used?

We have 55% and 80% antifreeze.  We need 90 gallons of 75%.  

F - 55% and E - 80%

A) F + E = 90

B) .55 F + .80 E = 67.50 (which is .75 * 90)  Multiplying A) by -.55

A) -.55 F -.55 E = -49.50 (which is -.55 * 90) then adding A) & B)

.25 E = 18

E = 72 gallons of 80% and

F = 18 gallons of 55%

DOUBLE CHECK

72 * .80 = 57.60  and 18 * .55 = 9.90

57.60 + 9.90 = 67.50 gallons of antifreeze and we have 90 total gallons

(67.50 / 90) * 100 = 75% correct!

Source: 1728.com/mixture.htm

126 maybe. sorry

c

step-by-step explanation: