Infinitely Many Solutions: Proving or Disproving a Linear System

[Mic :)]

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. :)

 

[ehild]

How many are the solutions of the following system of equations? Two equations with three unknowns...

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

 

[Mic :)]

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.

 

[Mic :)]

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 :)

 

[ehild]

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?

 

[Mic :)]

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?

 

[ehild]

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?

 

[Mic :)]

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

 

[ehild]

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?

 

[Ray Vickson]

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?