Simulataneous linear equations

[TyErd]

The simultaneous linear equations
mx+12y=24---------------1
3x+my=m------------------2

have a unique solution for m= ?
The answer should be m Є R\{-6,6}


what I tried doing was substituting the 2nd equation into m, expanded, grouped them and factorised but I struck a dead end.

 

[berkeman]

The simultaneous linear equations
mx+12y=24---------------1
3x+my=m------------------2

have a unique solution for m= ?
The answer should be m Є R\{-6,6}


what I tried doing was substituting the 2nd equation into m, expanded, grouped them and factorised but I struck a dead end.

That's the whole question? Seems like you have 3 unknowns and 2 equations... ?

 

[TyErd]

Yes, that is the whole question.

 

[Mark44]

The problem isn't to solve for m, but to state conditions for which there will be a unique solution for m.

The two equations can be put in an augmented matrix form, like so.
[m 12 | 24]
[3 m | 12]

Do you know how to use matrix operations to row-reduce this augmented matrix?

 

[TyErd]

I've touched on a little bit of matrix operations.
shouldn't it be
[m 12 | 24]
[3 m | m] ?? and from there i wouldn't know where to go.

 

[Mark44]

Right. That 3rd entry in the 2nd row should be m, not 12 as I had.

The basic row operations are

• Exchange two rows
• Replace a row by a multiple of itself
• Add a multiple of one row to another row

I think this is the direction they want you to go in this problem.

 

[TyErd]

okkkk, but i am not very familiar with how to do that.

 

[Mark44]

So going back to the system of equations,

mx + 12y = 24
3x + my = m

is there some multiple of the first equation you could add to the second equation to eliminate a variable? The operations you can apply to your system of equations are exactly the same operations as I listed in my previous post