[tex] \frac{4}{x-4} = \frac{x}{x-4}- \frac{4}{3} [/tex]
[tex] \frac{4}{x-4}- \frac{x}{x-4} =- \frac{4}{3} [/tex]
[tex] \frac{4-x}{x-4}=- \frac{4}{3} [/tex]
[tex]3(4-x)=-4(x-4)[/tex]
[tex]12-3x=-4x+16[/tex]
[tex]4x-3x=16-12[/tex]
[tex]x=4[/tex]
The solution is extraneous one because although it is a result of solving algebraically, it is not a valid solution. If we substitute x = 4 back into the equation, we will have two fractions with denominator of zero
[tex] \frac{4}{4-4}= \frac{x}{4-4}- \frac{4}{3} [/tex]
[tex] \frac{4}{0}= \frac{x}{0}- \frac{4}{3} [/tex]
A fraction with zero denominators is undefined, hence the solution is not valid.
Correct answer: A
