Question 1162779
let n = no. of nickels
let d = no. of dimes
let q = no. of quarters
:
Write an equation for each statement
:
 A collection of nickels, dimes, and quarters consist of 8 coins
n + d + q = 8
 with a total of $1.30. 
.05 + .10d + .25 = 1.30
If the number of dimes is equal to the number of nickels,
d = n
:
multiply the 2nd equation by 4 and subtract from the 1st equation
 1n + 1d +  1q = 8
.2n + .4d + 1q = 5.20
----------------------subtraction eliminates q
.8n + .6d = 2.80
d = n, therefore
.8n + .6n = 2.80
1.4n = 2.80
n = 2.8/1.4
n = 2 nickels
then since d = n
d = 2 dimes
and
q = 8 - 2 - 2
q = 4 quarters
:
:
See if that checks out
.05(2) + .10(2) + .25(4) =
.10 + .20 + 1.00 = 1.30