Question 219856
How many gallons of a 70% antifreeze solution must be mixed with 70 gallons of
 25% antifreeze to get a mixture that is 60% antifreeze?
 use the six step method
;
I'm not sure what the six step method, but this is how I would do it
:
Let x = amt of  70% solution will be required
:
A simple mixture equation
.70x + .25(70) = .60(x + 70)
:
.70x + 17.5 = .60x + 42
:
.70x - .60x = 42 - 17.5
:
.10x = 24.5
x = {{{24.5/.10}}}
x = 245 gal of 70% solution
:
:
Check solution in the original equation
.70(245) + .25(70) = .60(245 + 70)
171.5 + 17.5 = .60(315)
189 = 189