# Homework Help: If a linear system has more unknowns than equations, then it must have inifnitely many solutions

1. Nov 17, 2014

### Mic :)

• 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. :)

2. Nov 17, 2014

### 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

3. Nov 17, 2014

### Mic :)

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.

4. Nov 17, 2014

### Mic :)

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 :)

5. Nov 17, 2014

### ehild

Can you say a single solution?
Subtract one of the equations from the other one. What do you get? can it be true?

6. Nov 17, 2014

### Mic :)

Is there a way to determine the number of solutions by subtracting one from the other?

7. Nov 17, 2014

### 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?

8. Nov 17, 2014

### Mic :)

Sub. -----> 2y=4, y =2

9. Nov 17, 2014

### 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?

10. Nov 17, 2014

### Ray Vickson

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?