Sue has $1.90 in dimes and nickels. If she has 10 more dimes than nickels, how many of each coin does she have?

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This is an algebra problem. Let's write down what we know.

Let's call x=nickels and y= dimes.

We know that in total, we have 1.90 dollars worth of nickles and dimes.

nickles are worth .05 and dimes are worth .10

.05x+.10y=1.90 (equation 1)

We know she has 10 more dimes than nickles.

Thus y=x+10

Thus, we substitute y=x+10 into equation 1, and solve for x

.05x+.10(x+10)=1.90

.05x+.10x+1=1.90

.15x+1=1.90

.15x=.90

x=6.

Thus she has 6 nickles.

To find how many dimes we have, we substitute it into y=x+10

y=6+10=16.

Thus she has 16 dimes and 6 nickles.

CHECK your work

16*.10+6 *.05=1.6+.3=1.9

Thus we have the correct answer!

16 dimes and 6 nickles.

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