Question

Find the value of x and verify it :-
$\\ \sf \displaystyle \sf\frac{1}{x -3 } - \frac{x}{ {x}^{2} - 9} = \frac{2}{3 - x}$

Don't spam..!!​

Answer 1

Step-by-step explanation:

Answer refers to the attachment

Answer 2

Solution:

Given Equation:

$\rm\longrightarrow \dfrac{1}{x-3}-\dfrac{x}{x^{2}-9}=\dfrac{2}{3-x}$

Can be written as:

$\rm\longrightarrow \dfrac{1}{x-3}-\dfrac{x}{(x+3)(x-3)}=\dfrac{2}{3-x}$

$\rm\longrightarrow \dfrac{1}{x-3}\bigg(1-\dfrac{x}{x+3}\bigg)=\dfrac{2}{3-x}$

$\rm\longrightarrow \dfrac{1}{x-3}\bigg(1-\dfrac{x}{x+3}\bigg)=\dfrac{-2}{x-3}$

$\rm\longrightarrow \bigg(1-\dfrac{x}{x+3}\bigg)=-2\ \ [x\neq3]$

$\rm\longrightarrow \dfrac{-x}{x+3}=-2-1$

$\rm\longrightarrow \dfrac{-x}{x+3}=-3$

$\rm\longrightarrow \dfrac{x}{x+3}=3$

$\rm\longrightarrow x=3x+9$

$\rm\longrightarrow -2x=9$

$\rm\longrightarrow x=\dfrac{-9}{2}$

⊕ So, the value of x satifying the given equation is -9/2.

Answer:

• x = -9/2