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Ray Of Light [21]
3 years ago
8

You have 23 coins, including nickels, dimes, and quarters. if you have two more dimes than quarters, and the total value of the

coins is $2.50. how many of each kind of coin do you have?
Mathematics
2 answers:
olchik [2.2K]3 years ago
8 0

Hi there i hope this will help

n + d + q = 23 . . . . . . . there are 23 coins total

 0n +d -q = 2 . . . . . . . . .there are 2 more dimes than quarters

 5n +10d +25q = 250 . .the total value is $2.50

The collection includes 11 nickels, 7 dimes, and 5 quarters.

I used the matrix function of my calculator to solve these equations. You can find q by subtracting from the last equation five times the sum of the first two equations.

 (5n +10d +25q) -5((n +d +q) +(d -q)) = (250) -5(23 +2)

 25q = 125 . . . . . . . simplify

 q = 5

From the second equation,

 d = q +2 = 7

And from the first,

 n = 23 -5 -7 = 11

<h2>answer is 11</h2>
Tanzania [10]3 years ago
7 0
You can write three equations in the numbers of nickels (n), dime (d), and quarters (q).
  n + d + q = 23 . . . . . . . there are 23 coins total
  0n +d -q = 2 . . . . . . . . .there are 2 more dimes than quarters
  5n +10d +25q = 250 . .the total value is $2.50

The collection includes 11 nickels, 7 dimes, and 5 quarters.

_____
I used the matrix function of my calculator to solve these equations. You can find q by subtracting from the last equation five times the sum of the first two equations.
  (5n +10d +25q) -5((n +d +q) +(d -q)) = (250) -5(23 +2)
  25q = 125 . . . . . . . simplify
  q = 5
From the second equation,
  d = q +2 = 7
And from the first,
  n = 23 -5 -7 = 11
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