SOLUTION: A coin collection consists of 14 coins with a value of $1.35. The coins are nickels, dimes, and quarters. The numbers of nickles is three less than twice the number of dimes. How m

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Let x=number of nickels
y=number of dimes
z=number of quarters
Lets deal in pennies; we are told that:
5x+10y+25z=135---------------------eq1
x+y+z=14-----------------------------eq2
and
x=2y-3-------------------------------eq3
multiply eq2 by 25 (we get: 25x+25y+25z=350) and subtract eq1 from it and we get:
20x+15y=215---------------------eq2a
substitute x=2y-3 from eq3 into eq2a:
20(2y-3)+15y=215 simplifying we get:
40y-60+15y=215 add 60 to each side
55y=275 divide each side by 5
y=5 -------------------------------------number of dimes
From eq3:
x=2y-3=2*5-3=7------------------------------number of nickels
substitute x=7 and y=5 into eq2 and we get:
7+5+z=14
z=2--------------------------------------------number of quarters
CK
7+5+2=14
14=14
and
7*5+10*5+2*25=135
35+50+50=135
135=135

Hope this helps---ptaylor