SOLUTION: Joe has a collection of nickels and dimes that is worth $6.00. If the number of dimes were doubled and the number of nickels were increased by 6, the value of the coins would be $9

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Question 173613This question is from textbook Introductory Algebra
: Joe has a collection of nickels and dimes that is worth $6.00. If the number of dimes were doubled and the number of nickels were increased by 6, the value of the coins would be $9.90. How many dimes does he have? This question is from textbook Introductory Algebra

Answer by gonzo(654) About Me  (Show Source):
You can put this solution on YOUR website!
let n = number of nickels
let d = number of dimes
first equation is:
.05*n + .1*d = 6.00
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second equation is:
.05*(n+6) + .1*d*2 = 9.9
expanding the second equation, we get:
.05*n + .05*6 + .2*d = 9.9
which becomes:
.05*n + .3 + .2*d = 9.9
subtract .3 from both sides of equation to get:
.05*n + .2*d = 9.6
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2 equations to solve simultaneously are:
.05*n + .1*d = 6.00
.05*n + .2*d = 9.60
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subtract second equation from first to get:
.1*d = 3.60
d = 36
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number of dimes is 36
substitute in first equation to get:
.05*n + .1*(36) = 6
.05*n + 3.6 = 6
subtract 3.6 from both sides of equation to get:
.05*n = 2.4
divide both sides of equation by .05 to get:
= 2.4/.05 = 48
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we have:
number of dimes = 36
number of nickels = 48
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substitute these values in second equation to get:
.05*(48+6) + .1*36*2 = 9.9
.05*(54) + 72*.1 = 9.9
2.7 + 7.2 = 9.9
9.9 = 9.9
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equations are good.
answer to the question is:
he has 36 dimes.
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