Question 238454
d = no. of dimes
n = no. of nickels
:
Just write an equation for each statement
:
daniel has 19 coins in dimes and nickles.
d + n = 19
or
n = (19-d); use this for substitution
: 
after spending 85 cents. he has 4 dimes and 2 nickles left..
.10d + .05n - .85 = .10(4) + .05(2)
.10d + .05n - .85 = .40 + .10
.10d + .05n = .50 + .85
.10d + .05n = 1.35
Replace n with (19-d)
.10d + .05(19-d) = 1.35
.10d + .95 - .05d = 1.35
.10d - .05d = 1.35 - .95
.05d = .40
d = {{{.40/.05}}}
d = 8 dimes
then
n = 19 - 8 
n = 11 nickels
:
I don't agree with any of these answers. He started out with 11 nickels
Check this out
.10(8) + .05(11) = 
.80 + .55 = 1.35
Subtract .85 and you .50 which is 4 dimes and 2 nickels