Is y = (x^2−1) / (x−1) , x≠1 belong to two-variables linear equation?

Yoseph

New member
Is y = (x^2−1) / (x−1) , x≠1 belong to two-variables linear equation?

I have an equation as below:

[FONT=MathJax_Math]y[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main],[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]≠[/FONT][FONT=MathJax_Main]1[/FONT]​
Question: Does it belong to two-variables linear equation?
My effort: The equation can be simplified
[FONT=MathJax_Math]y[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main],[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]≠[/FONT][FONT=MathJax_Main]1[/FONT]​
[FONT=MathJax_Math]y[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main],[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]≠[/FONT][FONT=MathJax_Main]1[/FONT]​
The equation above in the form y = mx + c but I think it is not belong to two-variables linear equation because there is a hole (1,2) in line graph. Is it true?

Dr.Peterson

Elite Member
I have an equation as below:

[FONT=MathJax_Math]y[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Math](x[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]1)/([/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]1)[/FONT][FONT=MathJax_Main],[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]≠[/FONT][FONT=MathJax_Main]1[/FONT]
Question: Does it belong to two-variables linear equation?
My effort: The equation can be simplified
[FONT=MathJax_Math]y[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main])/[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main],[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]≠[/FONT][FONT=MathJax_Main]1[/FONT]
[FONT=MathJax_Math]y[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main],[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]≠[/FONT][FONT=MathJax_Main]1[/FONT]​
The equation above in the form y = mx + c but I think it is not belong to two-variables linear equation because there is a hole (1,2) in line graph. Is it true?
I would agree. It is a linear equation with an added restriction, not a true linear equation.