How many gallons of 90% anti-freeze solution must be mixed with 100 gallons of 20% antifreeze to get a mixture that is 80% antifreeze

Answer 1

It is helpful to write 2 aligned lines like this:

X + 100 = X+100

90% 20% 80%

The first line adds up volumes of the mixed parts.

Now "multiply" first line by the second, so to speak:

X*0.9 + 100*0.2 = (X+100)*0.8

This equation represent the total amount of pure antifreeze.

It is essentially the same as in the previous expert answer.

Solve it for x: x=600

Answer 2

x=# gallons of 90% antifreeze needed

90%*x+20%*100=80%*(100+x)

0.9x+20=0.8(100+x)

0.9x+20=80+0.8x

0.9x-0.8x=80-20

0.1x=60

x=600 gallons