SOLUTION: Write a system of equations in 2 or 3 variables. Then use one of the methods using matrices to solve the system and state the solution. A newsboy had $2.65, consisting of nicke

Algebra ->  Matrices-and-determiminant -> SOLUTION: Write a system of equations in 2 or 3 variables. Then use one of the methods using matrices to solve the system and state the solution. A newsboy had $2.65, consisting of nicke      Log On


   

Question 1113204: Write a system of equations in 2 or 3 variables. Then use one of the methods using matrices to solve the system and state the solution.
A newsboy had $2.65, consisting of nickels, dimes, and quarters. If the number of dimes exceeded the number of nickels by 1, and the number of quarters was equal to the number of nickels decreased by one, what was the number of each kind of coin?
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
this can be solved easily without using matrices, so i'll do that first to see what the answer should be.

n = number of nickels.
d = number of dimes.
q = number of quarters.

$2.65 * 100 = 265 cents.
1 * n = 5 cents
1 * d = 10 cents
1 * = 25 cents

number of dimes exceeds the number of nickels by 1, so you get d = n + 1

number of quarters is 1 less than the number of quarters so you get q = n - 1

your 3 equations are:

5n + 10d + 25q =265
d = n + 1
q = n - 1

in the first equation, replace d with n + 1 and q with n - 1 to get:

5n + 10 * (n + 1) + 25 * (n - 1) = 265

simplify to get:

5n + 10n + 10 + 25n - 25 = 265

combine like terms to get:

40n -15 = 265

add 15 to both sides to get:

40n = 280

divide both sides by 40 to get:

n = 7

since d = n + 1, then d = 8

since q = n - 1, then q = 6

your solution is:

n = 7
d = 8
q = 6

7 * n = 7 * 5 = 35 cents
8 * d = 8 * 10 = 80 cents
6 * q = 6 * 25 = 150 cents.

add them together and your total is 265 cents.

divide that by 100 and your total is $2.65.

solution is confirmed to be good.

to solve this using matrices, you would do the following:

first you want to get the constants on the right side of the equations and the variables on the left side of the equations.

the matrix columns will be in the following order: 

       n    d    q    c


your first equation stays the same.

your second equation is d = n + 1
subtract n from both sides to get d - n = 1
reorder the terms and provide 0 coefficient for the missing terms to get:
-1n + 1d + 0q = 1

your third equation is q = n - 1
add 1 to both sides and subtract q from both sides to get 1 = n - q
flip sides to get n - q = 1
provide 0 coefficient for the missing terms to get:
1n - 0d - 1q = 1

now create your matrix using the coefficients of the variables and the constant term.

it will look like this: 

                 n          d         q        c
row 1            5          10        25       265
row 2           -1          1         0        1
row 3            1          0        -1        1


the first row is your heading to remind you what the coefficients and constants in each column are representing.
n is the number of nickels
d is then umber of dimes
q is the number of quarters
c is the number of cents

the first column just tells you what the row number is.

i did the rest in excel because keeping everything in the right position is difficult without using it.

the progression of the matrix looks like this:

$$$
$$$
$$$

your answer is the same as i got without using the matrix, as it should.

my way is only one way to get to the answer.
if you do the steps correctly, you should arrive at the same answer, even if you did the steps a different way.

a tutorial on row manipulation is shown below:

http://www.purplemath.com/modules/mtrxrows.htm