SOLUTION: How many liters each of a 15% anrifreeze solution and a 30% antifreeze solution must be mixed to make 6 liters of a 20% antifreeze solution?

Question 795333: How many liters each of a 15% anrifreeze solution and a 30% antifreeze solution must be mixed to make 6 liters of a 20% antifreeze solution?
Answer by wilft1(217) (Show Source): You can put this solution on YOUR website!
this problem will be solved with 2 equations, using the subtraction method...
x + y = 6
30x + 15y = 20
top equation is two different quanties making 6 liters,
and the second line is how much of each solution to make the correct solution...
start off with multiplying 6 by 20, that gets us 120, substitute the 20 for that to make our new equations...
x + y = 6
30x + 15y = 120
now, we want to get rid of either x or y, y is the smallest of the two so we'll multiply the top line by -15, then add the two lines together.
-15x - 15y = -90
30x + 15y = 120
......................
15x = 30
divide both sides by 15
x = 2
now we know the value of x, lets plug that number into the original equation to find the value of y
x + y = 6
30(2) + 15y = 120
....................
x + y = 6
60 + 15y = 120
subtract 60 from both sides
x + y = 6
15y = 60
divide both sides by 15
y = 4
now we know both values, x is 2 and y is 4, means that you will need 2 liters of the 30 percent solution and 4 liters of the 15 percent solution to make 6 liters of 20 percent solution
hope this helps :)
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