There are 21 quarters and dimes in a can whose value adds to $3.60. How many of each coin are in the can?

N=number of nickels; D=number of dimes

N+D=21 D=21-N Use this to substitute for D

$0.05N+$0.10D=$31.60 Substitute for D. (This converts number of coins to value) $0.05N+$0.10(21-N)=$3.60 $0.05N+$2.10-$0.10N=$3.60 Subtract $2.10 from each side. -$0.05N -0.10N= $1.50

You take it up from there

May 16th, 2015

There are 21 quarters and dimes in a can whose value adds to $3.60. How many of each coin are in the can?

Q=number of quarters; D=number of dimes

Q+D=21 D=21-q Use this to substitute for D

$0.25Q+$0.10D=$3.60 Substitute for D. (This converts number of coins to value) $0.25Q+$0.10(21-Q)=$3.60 $0.25Q+$2.10-$0.10Q=$3.60 Subtract $2.10 from each side. $0.25-0.10Q= $1.50

-0.15Q = $1.50

Q= 10

So Number of dimes = 21-Q

So number of dimes = 11

May 16th, 2015

The second one is correct answer Sorry There was a confusion