Question

1/x-3 + 1/x+3 = 1/3, find the value of x​

Answer 1

Answer:

x = ±3

Step-by-step explanation:

$\large \text{Given \dfrac{1}{x-3}+\dfrac{1}{x+3}=\dfrac{1}{3} }\\\\\\\large \text{we have to find value of x }\\\\\\\large \text{using identity (a+b)(a-b)=a^2-b^2 here}\\\\\\\large \dfrac{x+3+x-3}{(x+3)(x-3)}=\dfrac{1}{3}\\\\\\\large \dfrac{2x}{(x^2-9)}=\dfrac{1}{3}$

$\large 2x\times3=(x^2-9)\\\\\\\large x^2-6x-9=0\\\\\\\large \text{Rewrite 9 as (-3)^2 and -6x \ as \ x\times-3\times2 }\\\\\\\large (-x)^2-(x\times-3\times2)+(-3)^2=0\\\\\\\large \text{using identity (a^2-2ab+b^2)=(a+b)^2 }\\\\\\\large ((-x)+(-3))^2=0\\\\\\\large -x-3=0$

Thus we get answer x = ±3