SOLUTION: Ellen wishes to mix candy worth $1.23 per pound with candy worth $3.36 per pound to form 15 pounds of a mixture worth $2.37 per pound. How many pounds of the more expensive candy s

Algebra ->  Customizable Word Problem Solvers  -> Coins -> SOLUTION: Ellen wishes to mix candy worth $1.23 per pound with candy worth $3.36 per pound to form 15 pounds of a mixture worth $2.37 per pound. How many pounds of the more expensive candy s      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


   



Question 1021743: Ellen wishes to mix candy worth $1.23 per pound with candy worth $3.36 per pound to form 15 pounds of a mixture worth $2.37 per pound. How many pounds of the more expensive candy should she use?
Answer by macston(5194) About Me  (Show Source):
You can put this solution on YOUR website!
.
x=pounds of less expensive candy; y=pounds of more expensive candy
.
x+y=15
x=15-y . Use this to substitute for x below.
.
$1.23x+$3.36y=$2.37(15) . Substitute for x from above.
$1.23(15-y)+$3.36y=$35.55
$18.45-$1.23y+$3.36y=$35.55
$2.13y=$17.10
y=8.03
.
x=15-y=15-8.03=6.97
.
ANSWER: She should use 6.97 pounds of the less expensive candy and 8.03 pounds of the more expensive candy.
.
CHECK:
.
$1.23x+$3.36y=$2.37(15)
$1.23x+$3.36y=$35.55
$1.23(6.97)+$3.36(8.03)=$35.55
$8.57+$26.98=$35.55
$35.55=$35.55
.