How many gallons of a 50% antifreeze solution must be mixed with 90 gallons of 10% antifreeze to get a mixture that is 40% antifreeze? Use the six step method.

I don't know what the "six-step method" is, but I do know how to solve the problem:

Let x = number of gallons of 50% solution

amount of antifreeze in 50% solution = 0.50x

amount of antifreeze in 10% solution = 0.10 (80) = 8

amount of 40% solution = 0.40 (x+80) = 0.40x + 32

Then, 0.50x + 8 = 0.40x + 32

0.10x = 24

x = 240 gal