Question

Need help and explaining with some rational polynomials:
Solve 6/(x - 3) = 3/ x for x and determine if the solution is extraneous or not.

Answer 1

@SolomonZelman

Answer 2

Is it:
\(\Large\frac{6}{x - 3} = \frac{3}{ x}\)
(notice how parentheses are not required in the denominator, but are needed when we write the expression on a single line)

In this case, we cross multiply and end up with a linear equation, so there is no question of extraneous roots:
\(\Large 6x=3(x-3)\)
Simplify, distribute, isolate x and solve for x.