SOLUTION: harry has $2.25 in nickles, dimes, and quarters. If he had twice as many nickels, half as many dimes and the same number of quarters, he would have $2.50. If he has 27 coins altoge

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Question 821805: harry has $2.25 in nickles, dimes, and quarters. If he had twice as many nickels, half as many dimes and the same number of quarters, he would have $2.50. If he has 27 coins altogether, then how many of each does he have?
Found 2 solutions by TimothyLamb, jsmallt9:
Answer by TimothyLamb(4379)   (Show Source): You can put this solution on YOUR website!
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5x + 10y + 25z = 225
x + y + z = 27
5(2)x + 10(1/2)y + 25z = 250
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put the system of linear equations into standard form
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5x + 10y + 25z = 225
x + y + z = 27
10x + 5y + 25z = 250
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copy and paste the above standard form linear equations in to this solver:
https://sooeet.com/math/system-of-linear-equations-solver.php
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answer:
x= nickels = 15
y= dimes = 10
z= quarters = 2
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Answer by jsmallt9(3758)   (Show Source): You can put this solution on YOUR website!
Let n = the number of nickels (at the start)
Let d = the number of dimes (at the start)
Let q = the number of quarters (at the start)

Since each nickel has a value of 5 cents, n nickels will have a value of 5n.
Since each dime has a value of 10 cents, d dimes will have a value of 10d.
Since each quarter has a value of 25 cents, q quarters will have a value of 25q.

From "harry has $2.25 in nickles, dimes, and quarters." we can write the equation:
5n + 10d + 25q = 225 (Note that we are using cents as our unit of money.)
From "If he had twice as many nickels, half as many dimes and the same number of quarters, he would have $2.50." we can write the equation:
5(2n) + 10(1/2)d + 25q = 250
which simplifies to:
10n + 5d + 25q = 250
From "he has 27 coins altogether" we can write the equation:
n + d + q = 27

So we have a system of three equations in three variables:
5n + 10d + 25q = 225
10n + 5d + 25q = 250
n + d + q = 27

I don't know what methods you may have learned which could be used to solve such a system. I am going to use a matrix and Gaussian elimination. (Note: algebra.com doesn't do matrices very well. So I am only going to show the rectangular array of numbers (without the usual brackets to indicate a matrix).)
     5    10    25   225
    10     5    25   250
     1     1     1    27
Replaced row 1 with 1/5 times row 1.
     1     2     5    45
    10     5    25   250
     1     1     1    27
Added -10 times row 1 to row 2.
      1      2      5     45
      0    -15    -25   -200
      1      1      1     27
Added -1 times row 1 to row 3.
      1      2      5     45
      0    -15    -25   -200
      0     -1     -4    -18
Replaced row 2 with -1/15 times row 2.
      1      2      5     45
      0      1    5/3   40/3
      0     -1     -4    -18
Added -2 times row 2 to row 1.
      1      0    5/3   55/3
      0      1    5/3   40/3
      0     -1     -4    -18
Added 1 times row 2 to row 3.
       1       0     5/3    55/3
       0       1     5/3    40/3
       0       0    -7/3   -14/3
Replaced row 3 with -3/7 times row 3.
      1      0    5/3   55/3
      0      1    5/3   40/3
      0      0      1      2
Added -5/3 times row 3 to row 1.
      1      0      0     15
      0      1    5/3   40/3
      0      0      1      2
Added -5/3 times row 3 to row 2.
    1    0    0   15
    0    1    0   10
    0    0    1    2
which translates into:
n = 15
d = 10
q = 2
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