Answer:[tex]\dfrac{6}{17}[/tex]
Step-by-step explanation:
Reciprocal of [tex]3-x\ \text{is}\ \dfrac{1}{3-x}[/tex]
Reciprocal of [tex]1-2x\ \text{is}\ \dfrac{1}{1-2x}[/tex]
According to question
[tex]\Rightarrow \dfrac{1}{3-x}+\dfrac{1}{1-2x}=6\times \dfrac{1}{3-4x}\\\\\Rightarrow \dfrac{(1-2x)+(3-x)}{(1-2x)(3-x)}=\dfrac{6}{3-4x}\\\\\Rightarrow \dfrac{4-3x}{(1-2x)(3-x)}=\dfrac{6}{3-4x}[/tex]
[tex]\Rightarrow (4-3x)(3-4x)=6(1-2x)(3-x)\\\Rightarrow 12-16x-9x+12x^2=18-6x-36x+12x^2\\\Rightarrow 12x^2-25x+12=18-42x+12x^2=0\\\Rightarrow 42x-25x=6\\\Rightarrow 17x=6\\\Rightarrow x=\dfrac{6}{17}[/tex]
