Achemical company makes two brands of antifreeze. the first brand is 65% pure antifreeze, and the second brand is 90% pure antifreeze. in order to obtain 40 gallons of a mixture that contains 80% pure antifreeze, how many gallons of each brand of antifreeze must be used?

it should be used 16 gallons of 65% pue antifreeze and 24 gallons of 80% antifreeze.

Step-by-step explanation:

Let 'x' = amount of 65% antifreeze.

Let 'y' = amount of 90% antifreeze.

We need to obtain 40 gallons of mixture, then:

x + y = 40 gallons [1]

Also we know that the mixture should contain 80% pure antifreeze, then:

0.65x + 0.9y = 0.80(x+y) →  0.15x = 0.1y → y = 1.5x  [2]

Now, subtituting the value of 'y' into [1]:

x + y = 40 gallons → x + 1.5x = 40 gallons → 2.5x = 40 gallons

⇒ x = 16 gallons.

Then: y = 40 - 16 = 24 gallons.

Now, it should be used 16 gallons of 65% pue antifreeze and 24 gallons of 80% antifreeze.

set each factor equal to 0

4x-3 = 0

2x-1 = 0

solve for x.

4x-3=0

4x=3

x= 3/4

solve for x. 

2x-1= 0

2x= 1

x= 1/2

your answer should be 1/2 is less than or equal to x and 3/4 is greater than or equal to x