Question 288228
A radiator contains 6 liters of a 25% anti-freeze solution.
 How much should be drained and replaced with pure anti-freeze to produce a 33% solution?
:
You almost got it:
:
Let x = amt of 25% to be drained, and 
x = amt of pure antifreeze to be added 
:
.25(6-x) + x = .33(6)
1.5 - .25x + x = 1.98
-.25x + x = 1.98 - 1.5
.75x = .48
x = {{{.48/.75}}}
x = .64 liters of 25% removed and .64 liters of pure antifreeze to be added