Solving for h in linear system

[cotfw]

1x + hx₂ = 4
3x + 6x₂ = 8


I don't know how to solve this. When I try replacement, interchange, and scaling, I still have the h value unsolved. I keep going around in circles

 

[cotfw]

X₂ is just another variable. You could call it Y or anything else. There's only the one h in the question

 

[pka]

1x + hy = 4
3x + 6y = 8

If you know anything about matrices then \(\displaystyle \left| {\begin{array}{*{20}{c}}1&h\\3&6\end{array}} \right| = (1)(6) - (3)(h) \ne 0\)
means the system has an unique solution.

What else?

 

[Ishuda]

1x + hx₂ = 4
3x + 6x₂ = 8


I don't know how to solve this. When I try replacement, interchange, and scaling, I still have the h value unsolved. I keep going around in circles

If all three of x, x₂, and h are unknown then there is no unique solution set. You can choose an independent variable and write the other quantities as functions of that variable. The set of equations does not always have solutions for arbitrary values of x, x₂, and h though, i.e. let x₂ = h/2.

EDIT: meant to say: The set of equations does not always have solutions for arbitrary values of x, x₂, and h though, i.e. let h = 2.

 

[pka]

If all three of x, x₂, and h are unknown then there is no unique solution set.

Really?? Did you bother to read the thread?

There's only the one h in the question

So if $\displaystyle h\ne 2$ then for each h there is an unique solution.
That solution is here.

 

[Ishuda]

If all three of x, x₂, and h are unknown then there is no unique solution set. You can choose an independent variable and write the other quantities as functions of that variable....

Really?? Did you bother to read the thread?


So if $\displaystyle h\ne 2$ then for each h there is an unique solution.
That solution is here.

Let's see. For each h there is a unique solution. That sounds a lot like "You can choose an independent variable and write the other quantities as functions of that variable." But if you think there is a unique solution why don't you tell me exactly what x, x₂, and h are given only
1x + hx₂ = 4
3x + 6x₂ = 8
with x, h, and x2 unknown. Even the link you provided picked an 'independent variable' and wrote the solution for the other two in terms of it.

Now no fair peaking to see what h has to be, remember all of them are supposed to be unknown.

EDIT: BTW pka, did you read the title of the thread?