You can put this solution on YOUR website! How much pure acid should be mixed with 7 gallons of a 50% acid solution in order to get an 80% acid solution?
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let x = amt of pure acid
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A typical mixture equation
x + .50(7) = .8(x+7)
x + 3.5 = .8x + 5.6
x - .8x = 5.6 - 3.5
.2x = 2.1
x = 2.1/.2
x = 10.5 gallons of pure acid required:
:
:
confirm this in the original equation
10.5 + .5(7) = .8(10.5+7)
10.5 + 3.5 = .8(17.5)