Question

\[\frac{ 3 }{ x-3}=\frac{ x }{ x-3 }-\frac{ 3 }{ 2 }x \]
\[\frac{ 3 }{ x-3 }=\frac{ 2x-3x(x-3) }{ 2\left( x-3 \right) }\]
multiply both sides by 2(x-3)
\[6=2x-3x^2+9x,3x^2-11x+6=0\]
3x^2-9x-2x+6=0
3x(x-3)-2(x-3)=0
(x-3)(3x-2)=0
\[x-3=0 ~or~ x=3~is~an~extraneous~ root.\]
or 3x-2=0
calculate x