algebra math word problem pls hrlp

Question

2. A radiator holds 4 gallons of fluid. If it is full with a 20% antifreeze solution, how much fluid should be drained and replaced with a 70% antifreeze mixture to result in a 50% mixture of antifreeze? Report the exact* answer in a sentence.

Answer 1

My primary strategy for these "% solution" problems is to create an equation that represents how much "pure" antifreeze (or whatever) we have after we're done draining and refilling the radiator.

So let's let X be the amount of 20% solution remaining, and let Y be the amount of 70% solution added. Then the amount of "pure" antifreeze from the 20% solution will be X * 0.2 and the pure antifreeze from the 70% solution is Y * 0.7. And the whole point is to end up with a 50% solution in our 4 gallon radiator. This gives us our main "pure-substance" equation:

X*0.2 + Y * 0.7 = 4 * 0.5 (Equation A)

And of course our 2nd equation is X + Y = 4 total gallons. (Equation B)

You know how to do the rest, I think. You should get Y = exactly 2.4 gallons to be drained and replaced by 70% solution.

Answer 2

Use the following equation. .2(4-x)+.7x =.5(4) where x= the amount to be removed. So .8 -.2x +.7x = 2 and .5x= 2-.8 =1.2 and x=2.4 gal to be removed.