Question 161968
A radiator contains 6 quarts of a 25% antifreeze solution. How much antifreeze
 should be drained and replaced with pure antifreeze to yield a 33% solution?
:
Let x = amt to be removed and amt to be added (1x = pure antifreeze)
:
a decimal equation of the amt of antifreeze
.25(6-x) + 1x = .33(6)
:
1.5 - .25x + 1x = 1.98
:
-.25x + 1x = 1.98 - 1.5
:
.75x = .48
x = {{{.48/.75}}}
x = .64 qts to replaced
:
:
Check solution in the original equation
.25(6 - .64) + 1(.64) = .33(6)
.25(5.36) + .64 = 1.98
1.34 + .64 = 1.98; confirms our solution