# How do you solve for x: x/(x-3) = 3/(x-3) + 3 ?

Aug 4, 2016

$x \in \emptyset$

#### Explanation:

The first thing to do here is get rid of the denominators. Note that a possible solution to the original equation must satisfy the condition

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{x - 3 \ne 0 \implies x \ne 3} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

So, to get rid of the denominators, multiply the last term by $1 = \frac{x - 3}{x - 3}$. This will get you

$\frac{x}{x - 3} = \frac{3}{x - 3} + 3 \cdot \frac{x - 3}{x - 3}$

You can now focus exclusively on the numerators

$x = 3 + 3 \cdot \left(x - 3\right)$

Rearrange to get $x$ alone on one side of the equation

$x = 3 + 3 x - 9$

$x = 3 x - 6$

$2 x = 6 \implies x = \frac{6}{2} = 3$

Notice that $x$ came out to be equal to the one value that it cannot take, i.e. $x = 3$.

This means that your original equation has no solution, $x \in \emptyset$.