- #1
AlphaNoodle
- 3
- 0
I encountered a system:
3x+5y+7z = 9
7x+3y-z=-5
12x+13y+14z=15
And the solution was infinite solutions.
However, when looking at each equation, the constants (including coefficients) increase/decrease by a constant amount.
3x+5y+7z = 9 (+2)
7x+3y-z=-5 (-4)
12x+13y+14z=15 (+1)
And I made other systems using that same format, where all the equations' constants increased/decreased arithmetically (by a constant)
And the solution was always infinity.
I am curious to why this is and is there any proof behind this?
I got as far as:
Ax+(A+m)y+(A+2m)z=(A+3m)
Bx+(B-n)y+(B-2n)z=(B-3n)
Cx+(C+k)y+(C+2k)z=(C+3k)
But I am clueless to why there is always infinite solutions to these types of systems. I always get to the point where a number = a number (using elimination) to solve the system, and form there the solutions are always infinite. I don't even know how to proceed with a proof, and I was hoping for some help. I am sure I am interpreting something wrong or missing something, but anything would be appreciated. Thanks in advanced.
3x+5y+7z = 9
7x+3y-z=-5
12x+13y+14z=15
And the solution was infinite solutions.
However, when looking at each equation, the constants (including coefficients) increase/decrease by a constant amount.
3x+5y+7z = 9 (+2)
7x+3y-z=-5 (-4)
12x+13y+14z=15 (+1)
And I made other systems using that same format, where all the equations' constants increased/decreased arithmetically (by a constant)
And the solution was always infinity.
I am curious to why this is and is there any proof behind this?
I got as far as:
Ax+(A+m)y+(A+2m)z=(A+3m)
Bx+(B-n)y+(B-2n)z=(B-3n)
Cx+(C+k)y+(C+2k)z=(C+3k)
But I am clueless to why there is always infinite solutions to these types of systems. I always get to the point where a number = a number (using elimination) to solve the system, and form there the solutions are always infinite. I don't even know how to proceed with a proof, and I was hoping for some help. I am sure I am interpreting something wrong or missing something, but anything would be appreciated. Thanks in advanced.