SOLUTION: You have one type of candy that sells for $1.50/lb and another type of candy that sells for $6.80/lb. You would like to have 31.8 lbs of a candy mixture that sells for $5.50/lb. Ho

Algebra ->  Finance -> SOLUTION: You have one type of candy that sells for $1.50/lb and another type of candy that sells for $6.80/lb. You would like to have 31.8 lbs of a candy mixture that sells for $5.50/lb. Ho      Log On


   



Question 1060013: You have one type of candy that sells for $1.50/lb and another type of candy that sells for $6.80/lb. You would like to have 31.8 lbs of a candy mixture that sells for $5.50/lb. How much of each candy will you need to obtain the desired mixture?

Thank you for your help

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +a+ = pounds of $1.50/lb candy needed
Let +b+ = pounds of $6.80/lb candy needed
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(1) +a+%2B+b+=+31.8+
(2) +%28+1.5a+%2B+6.8b+%29+%2F+31.8+=+5.5+
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(2) +1.5a+%2B+6.8b+=+5.5%2A31.8+
(2) +1.5a+%2B+6.8b+=+174.9+
(2) +15a+%2B+68b+=+1749+
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Multiply both sides of (1) by +1.5+
and subtract (1) from (2)
(1) +1.5a+%2B+1.5b+=+47.7+
(1) +15a+%2B+15b+=+477+
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(2) +15a+%2B+68b+=+1749+
(1) +-15a+-+15b+=+-477+
-----------------------------
+53b+=+1272+
+b+=+24+
and
(1) +a+%2B+24+=+31.8+
(1) +a+=+7.8+
--------------------------
7.8 pounds of $1.50/lb candy needed
24 pounds of $6.80/lb candy needed
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check:
(2) +%28+1.5a+%2B+6.8b+%29+%2F+31.8+=+5.5+
(2) +%28+1.5%2A7.8+%2B+6.8%2A24+%29+%2F+31.8+=+5.5+
(2) +%28+11.7+%2B+163.2++%29+%2F+31.8+=+5.5+
(2) +174.9+=+5.5%2A31.8+
(2) +174.9+=+174.9+
OK