SOLUTION: I am having trouble figuring my math problem out. The problem is "Molly has $5.45 in coins. She has two more nickels than dimes. She has nine fewer quarters than dimes. How many of

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Question 997320: I am having trouble figuring my math problem out. The problem is "Molly has $5.45 in coins. She has two more nickels than dimes. She has nine fewer quarters than dimes. How many of each coin does she have?"
Found 3 solutions by josgarithmetic, KMST, MathTherapy:
Answer by josgarithmetic(39618) About Me  (Show Source):
You can put this solution on YOUR website!
n nickels, d dimes, q quarters;

system%280.05n%2B0.1d%2B0.25q=5.45%2Cn=d%2B2%2Cq=-9%2Bd%29

Divide the money equation by 0.05 and adjust the quarters equation.
system%28n%2B2d%2B5q=109%2Cn=d%2B2%2Cq=d-9%29
Notice that the n and the q equations both are in terms of d.
..
..
You see what you can do?

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
You have to assume that Molly has no pennies, half-dollar coins, or dollar coins.
The problem does not tell you that, but maybe it should.

IF YOU STUDIED SYSTEMS OF LINEAR EQUATIONS:
You start by defining three variables:
N= number of nickels Molly has,
D= number of dimes Molly has, and
Q= number of quarters Molly has.
So,
$0.05N= amount of money that Molly has as nickels.
$0.10D= amount of money that Molly has as dimes.
$0.25Q= amount of money that Molly has as quarters.
The half assumed fact that Molly has $5.45 in coins" between nickels dimes and quarters, with no pennies, half-dollars, or dollars, means that
0.05N%2B0.1D%2B0.25Q=5.45 is one of your equations.
"She has two more nickels than dimes" translates as the equation
N=D%2B2 .
"She has nine fewer quarters than dimes" translates as
Q=D-9 .
With all 3 equations, you have the system
system%280.05N%2B0.1D%2B0.25Q=5.45%2CN=D%2B2%2CQ=D-9%29 .
You can have a simpler system, if you simplify equation 0.05N%2B0.1D%2B0.25Q=5.45 by
dividing both sides of the equal sign by 0.05=1%2F20 , or
multiplying both sides of the equal sign times 20 (same thing, really).
then you get N%2B2D%2B5Q=109 , which is simpler.
The system system%280.05N%2B0.1D%2B0.25Q=5.45%2CN=D%2B2%2CQ=D-9%29 or system%28N%2B2D%2B5Q=109%2CN=D%2B2%2CQ=D-9%29 is easy to solve by substitution,
substituting D%2B2 for N%7D%7D+%2C+and+%7B%7B%7BD-9 for Q in the first equation:
system%28N%2B2D%2B5Q=109%2CN=D%2B2%2CQ=D-9%29-->system%28%28D%2B2%29%2B2D%2B5%28D-9%29=109%2CN=D%2B2%2CQ=D-9%29-->system%28D%2B2%2B2D%2B5D-45=109%2CN=D%2B2%2CQ=D-9%29-->system%288D-43=109%2CN=D%2B2%2CQ=D-9%29-->system%288D=43%2B109%2CN=D%2B2%2CQ=D-9%29-->system%288D=152%2CN=D%2B2%2CQ=D-9%29-->system%28D=152%2F8%2CN=D%2B2%2CQ=D-9%29-->system%28D=19%2CN=D%2B2%2CQ=D-9%29-->system%28D=19%2CN=19%2B2%2CQ=19-9%29-->highlight%28system%28D=19%2CN=21%2CQ=10%29%29

IF YOU HAVE NOT STUDIED SYSTEMS OF LINEAR EQUATIONS:
You carefully choose one variable to define, so that you can calculate everything from that variable:
D= number of dimes Molly has.
"She has two more nickels than dimes" means that
D%2B2= number of nickels Molly has, and
"She has nine fewer quarters than dimes" means that
D-9= number of quarters Molly has.
Then,
$0.05%28D%2B2%29= amount of money that Molly has as nickels,
$0.10D%7D%29%7D=+amount+of++money+that+Molly+has+as+dimes%2C%0D%0A%24%7B%7B%7B0.25%28D-9%29= amount of money that Molly has as quarters, and
$0.05%28D%2B2%29%2B0.10D%2B0.25%28D-9%29= $5.45= amount of money that Molly has in coins.
Then you solve the equation
0.05%28D%2B2%29%2B0.10D%2B0.25%28D-9%29=5.45 to find D .

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!

I am having trouble figuring my math problem out. The problem is "Molly has $5.45 in coins. She has two more nickels than dimes. She has nine fewer quarters than dimes. How many of each coin does she have?"
Let number of dimes be D
Then number of nickels = D + 2
Number of quarters = D - 9
We then get the value equation: .1D + .05(D + 2) + .25(D - 9) = 5.45
.1D + .05D + .1 + .25D - 2.25 = 5.45
.1D + .05D + .25D + .1 - 2.25 = 5.45
.4D - 2.15 = 5.45
.4D = 5.45 + 2.15
.4D = 7.6
D, or number of dimes = 7.6%2F.4, or highlight_green%2819%29
Number of nickels: 19 + 2, or highlight_green%2821%29
Number of quarters: = 19 - 9, or highlight_green%2810%29