SOLUTION: The manager of a candy shop wishes to mix candy worth $4 per pound, $6 per pound and $10 per pound to get 100 pounds of a mixture worth $7.60 per pound. The amount of $10 candy mus

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: The manager of a candy shop wishes to mix candy worth $4 per pound, $6 per pound and $10 per pound to get 100 pounds of a mixture worth $7.60 per pound. The amount of $10 candy mus      Log On


   



Question 448512: The manager of a candy shop wishes to mix candy worth $4 per pound, $6 per pound and $10 per pound to get 100 pounds of a mixture worth $7.60 per pound. The amount of $10 candy must equal the total amounts of the $4 and $6 candy. how many ounds of each must he use?
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let a = pounds of $4 candy needed
Let b = pounds of $6 candy needed
Let c = pounds of $10 candy needed
given:
(1) +a+%2B+b+%2B+c+=+100+
(2) +4a+%2B+6b+%2B+10c+=+7.6%2A100+
(2) +4a+%2B+6b+%2B+10c+=+760+
(2) +2a+%2B+3b+%2B+5c+=+380+
(3) +c+=+a+%2B+b+
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There are 3 equations and 3 unknowns, so it's solvable
(3) +c+=+a+%2B+b+
(3) +a+%2B+b+-+c+=+0+
Add (3) and (1)
(1) +a+%2B+b+%2B+c+=+100+
(3) +a+%2B+b+-+c+=+0+
+2a+%2B+2b+=+100+
+a+%2B+b+=+50+
From (3),
+c+=+50+
Substitute this into (2)
(2) +2a+%2B+3b+%2B+5%2A50+=+380+
(2) +2a+%2B+3b+%2B+250+=+380+
(2) +2a+%2B+3b+=+130+
Multiply both sides of (3) by 2
and subtract (3) from (2)
(2) +2a+%2B+3b+=+130+
(3) +-2a+-+2b+=+-100+
+b+=+30+
and, since
+a%2B+b+=+50+
+a+%2B+30+=+50+
+a+=+20+
20 pounds of $4 candy are needed
30 pounds of $6 candy are needed
50 pounds of $10 candy are needed
check answer:
(2) +2a+%2B+3b+%2B+5c+=+380+
(2) +2%2A20+%2B+3%2A30+%2B+5%2A50+=+380+
(2) +40+%2B+90+%2B+250+=+380+
(2) +380+=+380+
OK