Question 83526
Let x=amount of 40% antifreeze needed
Then 20-x=amount of pure antifreeze needed

Now we know that the amount of pure antifreeze in the 40% solution (0.40x) plus the amount of pure antifreeze (20-x) must equal the amount of pure antifreeze in the final mixture 0.50(20).  So our equation to solve is:

0.40x+(20-x)=0.50(20)  get rid of parens

0.40x+20-x=10  subtract 20 from both sides

0.40x+20-20-x=10-20  collect like terms

-0.60x=-10  divide both sides by -0.60

x=16.666 gal --------------------------amount of 40% antifreeze

20-x=20-16.666=3.333 gal-------------------amount of pure antifreeze

CK
3.333+0.40(16.666)=0.50(20)
3.333+6.664=10
10=10


Hope this helps---ptaylor