Question

Kinda irritating math problem

person spent $14.00 on postage staps
the stamps were $0.15, $0.20, and $0.35
they bought 45 stamps

use the gauss-jordan elimination method (matrices) to solve and SHOW ALL WORK

Answer 1

x=number of stamps at a price of $0.15
y=number of stamps at a price of $0.20
z=number of stamps at a price of $0.35

We have a system of two equations with three unknowns

x+y+z=45
0.15x+0.20y+0.35 z=14

We solve this system of equations by Gauss-Jordan elimination method.

1 1 1 45
0.15 0.2 0.35 14


1 1 1 45
15 20 35 1400 (100R₂)



1 1 1 45
0 5 20 725 (R₂-15R₁)



1 1 1 45
0 1 4 145 (R₂/5)


Z=λ; λ∈N; λ∈[0,45]
y+4λ=145 ⇒ y=145-4λ

x+(145-4λ)+λ=45
x=45-λ+4λ-145
x=-100+3λ

The solution is:
x=-100+3λ
y=145-4λ
z=λ λ∈[0.45]

Now, we calculate the possible solutions:
x=-100+3λ
x≥0
-100+3λ≥0
λ=≥33.33333....⇒λ can be: 34,35,36,....45;

y=145-4λ
y≥0
145-4λ≥0
-4λ≥-145
-λ≥-145/4
λ≤36.25 ⇒λ can be (34,35 or 36)

if λ=34
z=34
y=145-4λ=145-136=9
x=-100+3 λ=-100+102=2

if λ=35
z=35
y=145-4λ=145-140=5
x=-100+3λ=-100+105=5

if λ=36
z=36
y=145-4λ=145-144=1
x=-100+3λ=-100+108=8

Answer:
We have threes possible solutions:
solution 1:
2 stamps at a price of $0.15
9 stamps at a price of $0.2
34 stamps at a price of $0.35


solution 2:
5 stamps at a price of $0.15
5 stamps at a price of $0.2
35 stamps at a price of $0.35


solution 3:
8 stamps at a price of $0.15
1 stamps at a price of $0.2
36 stamps at a price of $0.35

Answer 2

Solutions

To solve this problem lets use the variables >> "x,y,and z"

x=number of stamps at a price of $0.15
y=number of stamps at a price of $0.20
z=number of stamps at a price of $0.35

We have a system of two equations as the three unknowns

x+y+z=45

0.15x+0.20y+0.35 z=14

We solve this system of equations by Gauss-Jordan elimination method.The Gauss-Jordan elimination method is used to solve a system of linear equations.


`\left[\begin{array}{ccc}1&1&1\\0.15&0.2&0.35\\45&14&0\end{array}\right]`


`\left[\begin{array}{ccc}1&1&1\\15&20&35\\45&1400&0\end{array}\right]`

(100R₂)


`\left[\begin{array}{ccc}1&1&1\\0&5&20\\45&725&0\end{array}\right]`

(R₂-15R₁)


`image`

(R₂/5)

Calculations

Z=λ; λ∈N; λ∈[0,45]
y+4λ=145 ⇒ y=145-4λ

x+(145-4λ)+λ=45
x=45-λ+4λ-145
x=-100+3λ

The solution is:
x=-100+3λ
y=145-4λ
z=λ ≡ λ∈[0.45]

Our next step is to calculate all the possible solutions.

x=-100+3λ
x≥0
-100+3λ≥0
λ=≥33.33333....⇒λ can be numbers: 34,35,36,....45;

y=145-4λ
y≥0
145-4λ≥0
-4λ≥-145
-λ≥-145/4
λ≤36.25 ⇒λ can be the numbers (34,35 or 36)

if λ=34
z=34
y=145-4λ=145-136=9
x=-100+3 λ=-100+102=2

if λ=35
z=35
y=145-4λ=145-140=5
x=-100+3λ=-100+105=5

if λ=36
z=36
y=145-4λ=145-144=1
x=-100+3λ=-100+108=8

Simplify

We have three solutions

⇒solution 1:

2 stamps at a price of $0.15
9 stamps at a price of $0.2
34 stamps at a price of $0.35


⇒solution 2:

5 stamps at a price of $0.15
5 stamps at a price of $0.2
35 stamps at a price of $0.35


⇒solution 3:

8 stamps at a price of $0.15
1 stamps at a price of $0.2
36 stamps at a price of $0.35

Note - λ∈N = λ belong to natural numbers (1,2,3,4....)