SOLUTION: How much pure alcohol must be added to 10 gallons of 35% alcohol to have a solution that is 60% alcohol? Solve by substitution method

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Question 478960: How much pure alcohol must be added to 10 gallons of 35% alcohol to have a solution that is 60% alcohol? Solve by substitution method
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Let x = the number of gallons of pure alcohol to be added to the 10 gallons of 35% alcohol solution to obtain (10+x) gallons of 60% alcohol solution.
You can write the equation for the amount of alcohol as follows (after changing the percentages to their decimal equivalents):
0.35%2810%29%2Bx+=+%280.6%29%2810%2Bx%29 Simplify and solve for x.
0.35%2810%29%2Bx+=+6%2B0.6x
3.5%2Bx+=+6%2B0.6x Subtract 0.6x from both sides.
3.5%2B0.4x+=+6 Subtract 3.5 from both sides.
0.4x+=+2.5 Finally, didvide oth sides by 4.
x+=+6.25gallons of pure alcohol.