SOLUTION: I don't understand how to set up this systems of equations. A coin bank contains 27 coins, all nickels, dimes, or quarters. The combined value of the nickels and dimes is $0.90. Th

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Question 1049416: I don't understand how to set up this systems of equations. A coin bank contains 27 coins, all nickels, dimes, or quarters. The combined value of the nickels and dimes is $0.90. The combined value of the dimes and quarters is $4. How many coins of each type are in the bank?
Use a matrix
Found 3 solutions by Edwin McCravy, ankor@dixie-net.com, AnlytcPhil:
Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!

Your problem is solved at the bottom by using a matrix, 
under my alias AnlytcPhil.  Scroll down.

Edwin

Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
n = no. of nickels
d = no. of dimes
q = no. of quarters
:
Write an equation for each statement,
:
A coin bank contains 27 coins, all nickels, dimes, or quarters.
n + d + q = 27
:
The combined value of the nickels and dimes is $0.90.
.05n + .10d = .90
:
The combined value of the dimes and quarters is $4.00
.10d + .25q = 4.0
:
How many coins of each type are in the bank?
Use elimination on the last two equations
0n + .10d + .25q = 4.00
.05n + .10d + 0q = 0.90
--------------------------Subtraction eliminates d
-.05n +.25q = 3.10
.25q = .05n + 3.10
multiply equation by 4
1q = .20n + 12.40
:
Substitute for q in the first equation
n + d + .2n + 12.40 = 27
1.2n + d = 14.60
d = -1.2n + 14.60
:
Substitute for d in the 2nd equation
.05n + .10(-1.2n+14.60) = .90
.05n - .12n - 1.46 = .90
-.07n = .90 - 1.46
-.07n = -.56
n = -.56/-.07
n = + 8 nickels
then
q = .20(8) + 12.40
q = 1.6 + 12.4
q = 14 quarters
and using the 1st equation
8 + d + 14 = 27
d = 27 - 22
d = 5 dimes
:
:
Check this in the 3rd equaton
.10(5) + .25(14) = 4
.50 + 3.50 = 4
:
:
A long journey but we finally got it. Hope this made sense to you. CK

Answer by AnlytcPhil(1807) About Me  (Show Source):
You can put this solution on YOUR website!

Hi!  I'm Edwin McCravy.  AnlytcPhil is my alias! :)

I promised you I'd solve it later.  I notice the other tutor
has already solved it correctly in the meantime, but not 
with a matrix.

A coin bank contains 27 coins, all nickels, dimes, or quarters. 
N + D + Q = 27

The combined value of the nickels and dimes is $0.90. 
0.05N + 0.10D = 0.90

The combined value of the dimes and quarters is $4. 
0.10D + 0.25Q = 4.00

The system of equations is:

    N +     D +     Q = 27
0.05N + 0.10D         = 0.90
        0.10D + 0.25Q = 4.00

The matrix is



Multiply the 2nd and 3rd rows each by 100 to clear
the decimals:

That's these row operations:

100R2->R2
100R3->R3



Divide the 2nd and 3rd rows each by 5:

R2/5->R2
R3/5->R3



To get a 0 where the 1 is on the 2nd row,
multiply the 1st row by -1 and add to the 2nd row:

-1R1+R2->R2



To get a 0 where the 2 is on the bottom row,
multiply the 2nd row by -2 and add to the bottom
row.

That's -2R2+R3->R3



Divide the bottom row through by 7 to
get a 1 where the 7 is:

R3/y->R3



That's the system:

x + y + z = 27
    y - z = -9
        z = 14

Substitute 14 for z in the 2nd equation:

    y - z = -9
   y - 14 = -9
        y = 5

Substitute 14 for z and 5 for y in the 1st equation 

 x + y + z = 27
x + 5 + 14 = 27
    x + 19 = 27
         x = 8

So there are 8 nickels, 5 dimes and 14 quarters.

Edwin