Question 313058
:
I have nickles, dimes, and quarters.
n, d, q
:
Write an equation for each statement:
:
 There are a total of 24 coins
n + d + q = 24
:
and their value is $3.00.
.05n + .10d + .25q = 3.00
:
 There is one nickle more than dimes.
n = d + 1
:
Replace n with (d+1) in the 1st equation
(d+1) + d + q = 24
2d + q = 24 - 1
2d + q = 23
:
Replace n with (d+1) in the 2nd equation
.05(d+1) + .10d + .25q = 3.00
.05d + .05 + .10d + .25q = 3.00
.15d + .25q = 3.00 - .05
.15d + .25q = 2.95
:
Multiply the above equation by 4 and subtract from 2d + q = 23
 2d + q = 23
.6d + q = 11.8
--------------
1.4d = 11.2
d = {{{11.2/1.4}}}
d = 8 dimes
:
Find q using eq: 2d + q = 23
2(8) + q = 23
q = 23-16
q = 7 quarters
:
Use the 1st equation to find n
n + d + q = 24
n + 8 + 7 = 24
n = 24 - 15
n = 9 nickels
:
:
Check solution in the total$ equation
.05(9) + .10(8) + .25(7) = 
.45 + .80 + 1.75 = 3.00

: