Question 154118

Let x=amount of coolant that must be drained and replaced with water to yield a 40% solution.


Now we know that the amount of pure antifreeze in the radiator after we drain x amount out (0.80(3.8-x)) plus the amount of pure antifreeze in the water that we add (0) has to equal the amount of pure antifreeze in the final solution (0.40(3.8)).  So our equation to solve is:

0.80(3.8-x)=0.40*3.8  get rid of parens

3.04-0.80x=1.52  subtract 3.04 from each side

3.04-3.04-0.80x=1.52-3.04  collect like terms

-0.80x=-1.52 divide each side by -0.80

x=1.9 liters----------------amount that must be drained and replaced with pure water

CK
0.80(3.8-1.9)=0.40*3.8
0.80*1.9=0.40*3.8
1.52=1.52


Hope this helps--ptaylor