SOLUTION: In changing a $5, Marge received 9 more dimes than nickels and 7 fewer quarters than dimes. How many of each coin did she receive? nickels=x dimes=9+x quarters= x-7 = 9+x-7 =

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Question 903361: In changing a $5, Marge received 9 more dimes than nickels and 7 fewer quarters than dimes. How many of each coin did she receive?
nickels=x
dimes=9+x
quarters= x-7 = 9+x-7 = x-2
x+9+x+x-2=?
3x+7=?
Not sure what the number should be that the equation equals. Would it be 500?
Found 3 solutions by jim_thompson5910, josgarithmetic, MathTherapy:
Answer by jim_thompson5910(35256)   (Show Source): You can put this solution on YOUR website!
n = # of nickels
d = # of dimes
q = # of quarters


"Marge received 9 more dimes than nickels" ---> d = n+9

"7 fewer quarters than dimes" ----> q = d - 7 = (n+9)-7 = n + 2

--------------------------------------------------------------------

She has n nickels, n+9 dimes, n+2 quarters

she has 0.05n dollars in just nickels
she has 0.10(n+9) dollars in just dimes
she has 0.25(n+2) dollars in just quarters

They must add to $5

So, 0.05n+0.10(n+9)+0.25(n+2) = 5

Now solve for n 

	0.05n+0.10(n+9)+0.25(n+2) = 5


	0.05n+0.10(n)+0.10*(9)+0.25(n)+0.25*(2) = 5


	0.05n+0.10n+0.9+0.25n+0.50 = 5


	0.40n+1.40 = 5


	0.40n = 5-1.40


	0.40n = 3.60


	n = 3.60/0.40


	n = 9

-------------------------------------------------------

Use n = 9 to find d

d = n+9
d = 9+9
d = 18

Then use d = 18 to find q.

q = d - 7
q = 18 - 7
q = 11


-------------------------------------------------------

Summary: n = 9, d = 18, q = 11

She has 9 nickels, 18 dimes, 11 quarters
Answer by josgarithmetic(39623)   (Show Source): You can put this solution on YOUR website!
Use more variables, and do not restrict assigning to x or y or z.

DIMES, NICKELS, QUARTERS -------
Try n, d, q for how many of each coin.

"9 more dimes than nickels", .

"7 fewer quarters than dimes", .

You need to think carefully about forming those algebraic sentences; mistakes in there can easily happen.

Another equation is for the "In changing $5...".
Account for the change:

Simplify this money count equation.


System of equations is


More than one way to solve is possible. Try substituting for q in the other two equations. Use q=d-7. (Cannot do that to the d-n=9 because it has no q in it);



Now use the d-n=9 equation to substitute for d, in the form of d=n+9:





-----ONE OF THE COIN, NINE NICKELS


because just found n=9,
-------HOW MANY DIMES



-------THE COUNT OF QUARTERS
Answer by MathTherapy(10555)   (Show Source): You can put this solution on YOUR website!
In changing a $5, Marge received 9 more dimes than nickels and 7 fewer quarters than dimes. How many of each coin did she receive?
nickels=x
dimes=9+x
quarters= x-7 = 9+x-7 = x-2
x+9+x+x-2=?
3x+7=?
Not sure what the number should be that the equation equals. Would it be 500?


You can use one variable, as you did. 

Number of nickels: x

Number of dimes: 9 + x

Number of quarters: 9 + x - 7, or x + 2, NOT x - 2 as you stated



Now, the equation becomes: , or

 -------- MULTIPLYING by 20 to get rid of decimals

The first equation would equal 500 if you multiply it by 100, but that's unnecessary

since multiplying by 20 does the job of getting rid of the decimals

Solve for x, the number of nickels.

Add 9 to x to get the number of dimes

Add 2 to x to get the number of quarters. 

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