Answer:
given solution x =3 is extraneous.
Step-by-step explanation:
[tex]\frac{3}{x-3} =\frac{x}{x-3} -\frac{3}{2} \\ \frac{3}{x-3} =[tex]\frac{3}{x-3} =\frac{2x-3(x-3)}{2(x-3)}\\=\frac{2x-3x+9)}{2(x-3)}\\= \frac{9-x}{2(x-3)} \\\\\frac{3}{x-3} = \frac{9-x}{2(x-3)} \\\\ \\ 6(x-3) = (9-x)(x-3) \\6(x-3) - (9-x)(x-3) = 0 \\(x-3) (6-9+x) = 0 \\(x-3) (x-3) = 0x=3[/tex] [taking LCM right hand side] ,we get
gives x =3 as a solution which makes the original equation undefined
therefore the solution extraneous .
