SOLUTION: a coffee shop manager has coffee for $6 per pound and another type for $15 per pound. He wants to make a mixture that will cost $12 per pound. If he take 30lbs of the $15 coffee ho

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: a coffee shop manager has coffee for $6 per pound and another type for $15 per pound. He wants to make a mixture that will cost $12 per pound. If he take 30lbs of the $15 coffee ho      Log On


   



Question 1053917: a coffee shop manager has coffee for $6 per pound and another type for $15 per pound. He wants to make a mixture that will cost $12 per pound. If he take 30lbs of the $15 coffee how many pounds of the $6 coffee will he need to make a mixture that costs $12 per pound?
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +x+ = pounds of the $6 coffee needed
+15%2A30+=+450+ = cost of 30 pounds of $15 used
+6x+ = cost of the $6 coffee used
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+%28+6x+%2B+450+%29+%2F+%28+x+%2B+30+%29+=+12+
( this is [ cost/lb ] = [ cost/lb ] )
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+6x+%2B+450+=+12%2A%28+x+%2B+30+%29+
+6x+%2B+450+=+12x+%2B+360+
+6x+=+90+
+x+=+15+
15 pounds of the $6 coffee are needed
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check:
+%28+6x+%2B+450+%29+%2F+%28+x+%2B+30+%29+=+12+
+%28+6%2A15+%2B+450+%29+%2F+%28+15+%2B+30+%29+=+12+
+%28+90+%2B+450+%29+%2F+45+=+12+
+540+%2F+45+=+12+
+540+=+540+
OK