Question 187524
oe has a collection of nickels and dimes that is worth $6.00.
 If the number of dimes was doubled and the number of nickels was increased by 6,
 the value of the coins would be $9.90.
How many dimes does he have?
:
Let n = no. of nickels; Let d = no. of dimes (originally)
:
"a collection of nickels and dimes that is worth $6.00."
.05n + .10d = 6.00
or 
.05n = 6.00 - .10d
n = {{{6/.05}}} - {{{.10/.05}}}d
n = 120 - 2d
:
"If the number of dimes was doubled and the number of nickels was increased by 6, the value of the coins would be $9.90."
.05(n+6) + 2(.10d) =  9.90
.05n + .3 + .20d = 9.90
.05n + .20d = 9.90 - .3
.05n + .20d = 9.60
:
Substitute (120-2d) for n in the above equation:
.05(120-2d) + .20d = 9.60
6 - .10d + .20d = 9.60
-.10d + .20d = 9.60 - 6
.10d = 3.60
d = {{{3.6/.1}}}
d = 36 dimes
:
:
Find the no. of nickels to check our solution
n = 120 - 2d
n = 120 - 2(36)
n = 120 - 72
n = 48 nickels
;
.05(48) + .10(36) = 
2.40 + 3.60 = 6.00; confirms our solution