SOLUTION: A chemist wishes to strengthen a mixture from 10% alcohol to 30% alcohol. How much pure alcohol should be added to 7 liters of the 10% mixture?

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Question 124543This question is from textbook A Graphical Approach to College Algebra
: A chemist wishes to strengthen a mixture from 10% alcohol to 30% alcohol. How much pure alcohol should be added to 7 liters of the 10% mixture? This question is from textbook A Graphical Approach to College Algebra

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Let x = the required number of liters of pure alcohol to be added to the 7 liters of 10% alcohol solution to get (7+x) liters of 30% alcohol solution.
The 7 liters of 10% alcohol solution can be expressed as:
7%280.1%29 and we'll add x liters of 100% alcohol to this to get the (7+x)(0.3) liters.
So, here's the equation:
7%280.1%29%2Bx+=+%287%2Bx%29%280.3%29 Simplify and solve for x.
0.7%2Bx+=+2.1%2B0.3x Subtract 0.3x from both sides.
0.7%2B0.7x+=+2.1 Now subtract 0.7 from both sides.
0.7x+=+1.4 Finally, divide both sides by 0.7
x+=+2
The chemist will need to add 2 liters of pure alcohol to the 7 liters of the 10% alcohol solution to obtain 9 liters of 30% alcohol solution.
Check:
7%280.1%29%2B2+=+%287%2B2%29%280.3%29
0.7%2B2+=+9%280.3%29
2.7+=+2.7