SOLUTION: explain how to solve using system of equations: Dr.combin-it wants to make 720 mL of a 40% acid solution. However, in his laboratory, he has only a 30% acid solution and a 60% acid

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Question 528497: explain how to solve using system of equations: Dr.combin-it wants to make 720 mL of a 40% acid solution. However, in his laboratory, he has only a 30% acid solution and a 60% acid solution. How much of each solution should he mix together to obtain his desired mixture?
Answer by oberobic(2304) About Me  (Show Source):
You can put this solution on YOUR website!
With mixture problems, a key strategy is to keep track of how much 'pure' stuff you have or need.
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He wants to make 720 mL of a 40% acid solution.
So, it will contain .4 * 720 = 288 mL of pure acid.
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He will mix 'x' mL of 30% acid and '720-x' mL of 60% acid.
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.3*x + .6(720-x) = .4(720)
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Multiply both sides by 10 to eliminate the decimals.
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3x + 6(720-x) = 4(720)
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3x + 4320 -6x = 2880
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-3x = 2880 - 4320
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-3x = -1440
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x = -1440/-3 = 1440/3 = 480
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x = 480 mL of 30% acid solution
and
(720-x) = 240 mL of 60% acid solution
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Does that work?
Check the amount of pure acid.
.3*480 = 144 mL
.6*240 = 144 mL
144 + 144 = 288 mL
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How much to you need?
288 mL, as shown above.
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So that's the correct answer.
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Answer: Mix 480 mL of 30% acid with 240 mL of 60% acid to make 720 mL of 40% acid.
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Done.