Joe has a collection of nickels and dimes that is worth $6.05.

If the number of dimes was doubled and the number of nickels was decreased by 10,

the value of the coins would be $9.85.

How many dimes does he have?

:

Write an equation for each statement:

:

"Joe has a collection of nickels and dimes that is worth $6.05."

.05n + .10d = 6.05

:

"If the number of dimes was doubled and the number of nickels was decreased by

10, the value of the coins would be $9.85."

.05(n-10) + .10(2d) = 9.85

.05n - .50 + .20d = 9.85

.05n + .20d = 9.85 + .50

.05n + .20d = 10.35

:

Subtract the 1st equation from the above:

.05n + .20d = 10.35

.05n + .10d = 6.05

---------------------Subtraction eliminates n find d

.10d = 4.30

d =

d = 43 dimes is what he has

;

:

That's the answer they want, but to check the problem we have to find n

.05n + .10(43) = 6.05

.05n + 4.30 = 6.05

.05n = 6.05 - 4.30

.05n = 1.75

n =

n = 35 dimes

:

Check solution in the 2nd equation

.05(n-10) + .10(2d) =

.05(25) + .10(86) =

1.25 + 8.60 = 9.85; confirms out solution of 43 dimes