Step-by-step explanation:
There are some equations for which there is no value of x that can make both of them true at once.
Take [tex]\frac{1}{x+3} +\frac{1}{x-3} =\frac{6}{x^2-9}[/tex] as an example. Let us see what happens when we try to solve for x.
[tex]\frac{1}{x+3} +\frac{1}{x-3} =\frac{6}{x^2-9} \\\\\frac{x-3}{x^2-9} +\frac{x+3}{x^2-9} =\frac{6}{x^2-9} \\\\\frac{2x-6}{x^2-9} =0\\\\\frac{2(x-3)}{(x+3)(x-3)} =0\\\\\frac{2}{x+3} =0\\\\2=0(x+3)\\\\2=0[/tex]
As you can see, the end of our "solution" gave us 2=0, which is not true. This means that there are no solutions to this equation. Therefore, there are no values of x to make this equation true.
