Its a word problem that has to do with money

Question

There were 26 nickels, dimes, and quarters in all, and their value was $2.25. How many coins of each type were there if there were 10 times as many nickels as quarters?

 

I've gotten to 10x+x

Answer 1

 

  N + D + Q = 26

 

 5N + 10D + 25Q = 225 <--- amount in pennies to avoid the decimals

 

 N = 10Q

 

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Substitutes the third equation into the first and second equations:

 

10Q + D + Q = 26

 

50Q + 10D + 25Q = 225

 

combines like terms:

 

11q + d = 26  --- > d = 26 - 11q

75q + 10d = 225

 

 

75q + 10( 26 - 11q) = 225

 

75q + 260 - 110q = 225

 

-35q + 260 = 225

 

-35q = 225 - 260

 

-35q = -35

 

q = 1

 

1 quarter

10 nickes

15 dimes

 

25 + 10*5 + 15*10 = 25 + 50 + 150 = 225

Answer 2

n = number of nickles

d= number of dimes

q= number of quarters

 

n+d+q = 26

 

There are 10 times as many nickles as there are quarters:

10q=n

 

And the total value is given by this equation:

 

0.05n + 0.1d + 0.25q = 2.25

 

Replace n in the first equation with 10q from the second equation:

n+d+q = 26

10q + d + q = 26

11q + d =26

Now solve for d:

d = 26-11q

 

Replace d in the 3rd equation with 26-11q, and replace n in the 3rd equation with 10q. Then solve for q:

0.05n + 0.1d + 0.25q = 2.25

0.05(10q) + 0.1(26-11q) + 0.25q = 2.25

0.5q + 2.6 - 1.1q + 0.25q =2.25

-0.35q = -0.35

q = 1

 

Find n from the second equation:

10q = n

10(1) = n

n= 10

 

And find d from the first equation:

n+d+q = 26
10 + d + 1 = 26

11 + d = 26

d = 15

 

There are 15 dimes, 10 nickles, and 1 quarter