<PRE><B>Question 164965</B>
Let x=amount that needs to be drained off and replaced with pure antifreeze
Now we know that the amount of pure antifreeze left after x amount is drained off (0.20(14-x)) plus the amount of pure antifreeze added(x) has to equal the amount of pure antifreeze in the final mixture (0.40(14)).  So, our equation to solve is:

0.20(14-x) + x=0.40*14  get rid of parens (distributive)
2.8-0.20x+x=5.6  subtract 2.8 from each side
2.8 - 2.8-0.20x+x=5.6-2.8 collect like terms

0.80x=2.8  divide each side by 0.80
x=3.5 qts-----------------------------------ans
CK
0.20*10.5+3.5=5.6
2.1+3.5=5.6
5.6=5.6

Hope this helps---ptaylor


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