18. Jace has $2.15 in nickels and dimes. If he has 7 more nickels than dimes, how many of each kind of coin does he have?

There are 12 dimes and 19 nickels

Since Jace has $2.15 in nickels and dimes.

We know that 1 nickel = $ 0.05 and 1 dime = $0.10

Let

We have that

0.05n + 0.10n' = 2.15   (1)

Since he has 7 more nickels than dimes, we have that

n = n' + 7       (2)

Substituting equation (2) into (1), we have

0.05n + 0.10n' = 2.15   (1)

0.05(n' + 7) + 0.10n' = 2.15  

0.05n' + 0.35 + 0.10n' = 2.15  

0.15n' + 0.35 = 2.15  

0.15n' = 2.15 - 0.35

0.15n' = 1.8

n' = 1.8/0.15

n' = 12

Since the number of nickels, n = n' + 7

Substituting n = 12 into the equation, we have

n = n' + 7

n = 12 + 7

n = 19

So, there are 12 dimes and 19 nickels

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