SOLUTION: 1. Joe has a collection of nickels and dimes that is worth $6.65. If the number of dimes were doubled and the number of nickels were increased by 11, the value of the coins would

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Question 275466: 1. Joe has a collection of nickels and dimes that is worth $6.65. If the number of dimes were doubled and the number of nickels were increased by 11, the value of the coins would be $11.60. How many dimes does he have?
2. Ellen wishes to mix candy worth $1.98 per pound with candy worth $3.96 per pound to form 20 pounds of a mixture worth $2.87 per pound. How many pounds of the more expensive candy should she use?
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Joe has a collection of nickels and dimes that is worth $6.65. If the number of dimes were doubled and the number of nickels were increased by 11, the value of the coins would be $11.60. How many dimes does he have?

let the number of nickel coins be x
let the number of dimes be y
5x+10y =665 cents
nickels were increase by 11 so it becomes x+11
Dimes were doubled so it beomes 2y
5(x+11)+10*2y = 1160 cents
5x+55+20y= 1160
5x+20y = 1160-55
5x+20y= 1105
5x+20y= 1105----------1
5x+10y =665 ----------2
subtract equation 2 from equation1
10y= 440
y= 44 = Number of dimes.
Number of nickels will be
5x+10y =665 put the value of y in the equation
5x+440 = 665
5x= 665-440
5x=225
x=45 number of nickels
2. Ellen wishes to mix candy worth $1.98 per pound with candy worth $3.96 per pound to form 20 pounds of a mixture worth $2.87 per pound. How many pounds of the more expensive candy should she use?
$1.98 candy--------------- $3.96 candy------------- $2.87 candy mix
x------------------------- 20-x----------------------20 pounds
The value of these two added should be equal to the value of the mixed candy
1.98x+3.96(20-x) = 20*2.87
1.98x+79.20 - 3.96x= 57.4
-1.98x= 57.4-79.2
-1.98x=-21.8
x= -21.8/-1.98
x=11 pounds which is $1.98 candy
$3.96 candy added will be 20-11=9 pounds