Infinitely Many Solutions: Proving or Disproving a Linear System

Click For Summary

Homework Help Overview

The discussion revolves around the nature of solutions in linear systems, particularly focusing on the claim that a linear system with more unknowns than equations must have infinitely many solutions. Participants are examining specific examples of equations to explore this assertion.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants present various systems of equations with differing numbers of unknowns and equations, questioning the validity of the original claim. They explore specific cases, such as two equations with three unknowns, and consider the implications of subtracting equations to analyze the number of solutions.

Discussion Status

The discussion is active, with participants offering different perspectives on the nature of solutions in linear systems. Some participants express uncertainty about the original claim, while others provide examples to challenge or support it. There is no explicit consensus, but several lines of reasoning are being explored.

Contextual Notes

Participants note the lack of specification in the original problem and discuss the implications of specific examples, such as the equations x+y+z=1 and x+y+z=2, which lead to contradictions. The exploration of these examples raises questions about the assumptions underlying the claim regarding the number of solutions.

Mic :)
Messages
48
Reaction score
0
Member notified on the mandatory use of the homework template.
If a linear system has more unknowns than equations, then it must have infinitely many solutions.

Prove or disprove.

I'm not at all sure what to do with this one.

Thank you very much for any help. :)
 
Physics news on Phys.org
How many are the solutions of the following system of equations? Two equations with three unknowns...

x+y+z = 1
x+y+z = 2
 
ehild said:
How many are the solutions of the following system of equations? Two equations with three unknowns...

x+y+z = 1
x+y+z = 2

It is not specified.
I'm guessing any combination, unless there is one that disproves it.
I'm under the impression that the statement is true.
 
ehild said:
How many are the solutions of the following system of equations? Two equations with three unknowns...

x+y+z = 1
x+y+z = 2

If your'e able to show me how to do it with 2 equations and 3 unknowns, please go ahead so I can get and idea as to what's going on :)
 
Mic :) said:
It is not specified.
I'm guessing any combination, unless there is one that disproves it.
I'm under the impression that the statement is true.
Can you say a single solution?
Subtract one of the equations from the other one. What do you get? can it be true?
 
ehild said:
Can you say a single solution?
Subtract one of the equations from the other one. What do you get? can it be true?

Is there a way to determine the number of solutions by subtracting one from the other?
 
Subtracting two equations is legal. It serves to cancel some of the variables out.

How would you solve the following equations: x+y=5, x-y=1?
What do you get if you add or subtract them?
 
ehild said:
Subtracting two equations is legal. It serves to cancel some of the variables out.

How would you solve the following equations: x+y=5, x-y=1?
What do you get if you add or subtract them?

Add -----> 2x=6 , x=3

Sub. -----> 2y=4, y =2
 
And x=3, y=2 are solutions, as x+y=3+2=5 and x-y=3-2=1.

If you have two equations,
x+y+z=2 and
x+y+z=1,
these are two equations with three unknowns. You state that there must be infinitely many solutions.
Subtracting the second equation from the first (it is allowed) you get the equation x+y+z - (x+y+z) = 1,
The right-hand side is zero. So you get 0=1 which is a false statement. x+y+z can not be 1 and 2 at the same time. So there are no solutions.

Is it true then that all linear system has infinitely many solutions if the number of unknowns is more than the number of equations?
 
  • #10
Mic :) said:
It is not specified.
I'm guessing any combination, unless there is one that disproves it.
I'm under the impression that the statement is true.

Look at the system
X = 1
X = 2
Does this system have (a) no; (b) one; or (c) infinitely many solutions? Now what happens if you set x+y+z=X?
 

Similar threads

  • · Replies 30 ·
2
Replies
30
Views
2K
Replies
12
Views
2K
  • · Replies 18 ·
Replies
18
Views
2K
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
1
Views
2K