First, find the LCM of the denominators (x-2) and 3. It is 3(x-2).
Multiply both sides of the equation by the LCM to cancel out all the denominators. Then, solve for x.
[tex]\begin{gathered} \frac{x}{x-2}=\frac{2}{x-2}-\frac{2}{3} \\ \Rightarrow3(x-2)\mleft(\frac{x}{x-2}\mright)=3(x-2)\mleft(\frac{2}{x-2}-\frac{2}{3}\mright) \end{gathered}[/tex]
Expand the parenthesis on the right member of the equation:
[tex]\Rightarrow3(x-2)\mleft(\frac{x}{x-2}\mright)=3(x-2)\mleft(\frac{2}{x-2}\mright)-3(x-2)\mleft(\frac{2}{3}\mright)[/tex]
Cancel out the factors that appear on the denominator in each term:
[tex]\Rightarrow3(x)=3(2)-(x-2)(2)[/tex]
Expand the parenthesis on the right member of the equation:
[tex]\Rightarrow3(x)=3(2)-(x)(2)+(2)(2)[/tex]
Expand all the parentheses and solve for x:
[tex]\begin{gathered} \Rightarrow3x=6-2x+4 \\ \Rightarrow3x+2x=6+4 \\ \Rightarrow5x=6+4 \\ \Rightarrow5x=10 \\ \Rightarrow x=\frac{10}{5} \\ \therefore x=2 \end{gathered}[/tex]
Notice that the expression x-2 appears in the denominator. Nevertheless, since the value of x that would make the equation true is 2, then the expression x-2 is equal to 0. On the other hand, the denominator cannot be equal to 0 because division over 0 is not defined.
Therefore, this equation has no solution (the solution set is the empty set).
