SOLUTION: Hello I need help with this problem please. Plato has 58 coins in nickels, dimes, and quarters. The number of nickels is three less than twice the number of dimes. The total val

Question 274770: Hello I need help with this problem please.
Plato has 58 coins in nickels, dimes, and quarters. The number of nickels is three less than twice the number of dimes. The total value of the coins is $7.40. How many of each type of coin does Plato have?
Answer by ptaylor(2198) (Show Source): You can put this solution on YOUR website!
OK. We can work this problem either using one unknown or two unknowns. I will work it using one unknown and then set it up for you to solve using two unknowns.
Let x=number of dimes
Then 2x-3=number of nickels
Number of nickels and dimes=x+2x-3=3x-3
So, number of quarters=58-(3x-3)=58-3x+3=61-3x
Now we are told that($ understood):
0.10x+0.05(2x-3)+0.25(61-3x)=7.40 get rid of parens
0.10x+0.10x-0.15+15.25-0.75x=7.40 simplify by combining like terms
-0.55x+15.10=7.40 subtract 15.10 from each side
-0.55x+15.10-15.10=7.40-15.10 combine like terms again
-0.55x=-7.70 divide each side by -0.55
x=14--------------------number of dimes
2x-3=2*14-3=25----------number of nickels
61-3x=61-42=19-----------number of quarters
CK
14+25+19=58
58=58
and
0.10*14+0.05*25+0.25*19=7.40
1.40+1.25+4.75=7.40
7.40=7.40
Now using 2 unknowns:
Let x=number of dimes
then 2x-3=number of nickels
and let y=number of quarters
x+(2x-3)+y=58 or
3x-3+y=58 add three to both sides
3x+y=61----------------------------eq1
and
0.10x+0.05(2x-3)+0.25y=7.40 simplify
0.20x-0.15+0.25y=7.40 add 0.15 to each side
0.20x+0.25y=7.55-----------------------------eq2
Now just solve the two equations (hint: multiply eq2 by 4 and then subtract from eq1)
Hope this helps---ptaylor

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