SOLUTION: Sam found a number of nickels, dimes, and quarters in his room. He found 6 more dimes than nickels but 3 times as many quarters as dimes. The total value of the coins was $11.40. H

Question 201312: Sam found a number of nickels, dimes, and quarters in his room. He found 6 more dimes than nickels but 3 times as many quarters as dimes. The total value of the coins was $11.40. How many coins of each type did Sam find?
6y=11.40
y=1.90
3z=11.40
z=3.80
3.80 + 1.90=5.70
11.40-5.70=5.70
Found 2 solutions by ptaylor, josmiceli:
Answer by ptaylor(1894) (Show Source):
You can put this solution on YOUR website!

Let x=number of nickels
Then x+6=number of dimes
And 3(x+6)=number of quarters
Now lets deal in pennies
5x+10(x+6)+25*3(x+6)=1140 get rid of parens and simplify
5x+10x+60+75x+450=1140 collect like terms
90x+510=1140 subtract 510 from each side
90x=1140-510=630 divide each side by 90
x=7-------------number of nickels
x+6=7+6=13-------------number of dimes
3(x+6)=3*13=39-----------number of quarters
CK
5*7+10*13+39*25=1140
35+130+975=1140
1140=1140
Hope this helps---ptaylor

Answer by josmiceli(6786) (Show Source):
You can put this solution on YOUR website!
Let = number of nickels
Let = number of dimes
Let = number of quarters
given:

(in pennies)
---------------------------

Then, by substituting,

And, since

Also,

Sam found 7 nickels, 13 dimes and 39 quarters
check answer:

OK