SOLUTION: SOLVE USING DETERMINANTS and CRAMER'S RULE
A coin collector has 41 coins consisting of nickels, dimes and quarters and they are worth a total of $4.00. If the number of dimes pl
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Question 208212: SOLVE USING DETERMINANTS and CRAMER'S RULE
A coin collector has 41 coins consisting of nickels, dimes and quarters and they are worth a total of $4.00. If the number of dimes plus the number of quarters is one more than the number of nickels, then how many of each does he have ?
REQUIRED: Find the number of quarters, dimes and nickels.
Answer by pjalgeb(9) (Show Source): You can put this solution on YOUR website!
SOLUTION:
Let x = number of nickels
y = number of dimes
z = number of quarters
Write equations from the information given in the word problem:
a. A coin collector has 41 coins consisting of nickels, dimes and quarters...:
x + y + z = 41 -->Equation 1
b. ...and they are worth a total of $4.00:
0.05x + 0.1y + 0.25z = 4.00 -->Equation 2
c. ...the number of dimes plus the number of quarters is one more than the number of nickels...:
y + z = x + 1
x - y - z = -1 -->Equation 3
Now we have a system of 3 linear equations with 3 unknowns:
x + y + z = 41 -->Equation 1
0.05x + 0.1y + 0.25z = 4 -->Equation 2
x - y - z = -1 -->Equation 3
The augmented 3x3 matrix for the above system of equations is:
| 1 1 1 | 41 |
| .05 .1 .25 | 4 |
| 1 -1 -1 | -1 |
Using Cramer's Rule:
First we solve for D, Dx ,Dy and Dz:
D = | 1 1 1 |
| .05 .1 .25 |
| 1 -1 -1 |
D = 1(-.1+.25) - .05(-1+1) + 1(.25-.1)
D = 1(.15) - .05(0) + 1(.15)
D = .15 - 0 - .15
D = 0
We cannot solve using Determinants because D = 0 so try other methods.
Multiply Equation 2 by 4 and add to Equation 3 to eliminate z:
[ 0.05x + 0.1y + 0.25z = 4 ] * 4 -->4*Equation 2
.2x +.4y + z = 16 -->Equation 4
x - y - z = -1 -->Equation 3
___________________
1.2x - .6y = 15 -->Equation 5
Equation 1 + Equation 3:
x + y + z = 41 -->Equation 1
x - y - z = -1 -->Equation 3
_______________________
2x = 40 -->Equation 6
2x = 40
Substitute x = 20 in Equation 5:
1.2(20) - .6y = 15
24 - .6y = 15
24 - 15 = .6y
9 = .6y
y = 9/.6
Substitute and in Equation 1:
20 + 15 + z = 41
35 + z = 41
z = 41-35
ANSWER: The collector has 20 nickels, 15 dimes and 6 quarters.
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