Question

mathematics rd sharma​

Answer 1

Question :-

Solve for 'x'.

$\dfrac{x-1}{x-2}+\dfrac{x-3}{x-4}=3\dfrac{1}{3};\:x\neq2,4$

Solution :-

$\dfrac{x-1}{x-2}+\dfrac{x-3}{x-4}=3\dfrac{1}{3}$

$\dfrac{x-1}{x-2}+\dfrac{x-3}{x-4}=\dfrac{(3\times3)+1}{3}$

$\dfrac{(x-1)(x-4)+(x-3)(x-2)}{(x-2)(x-4)}=\dfrac{9+1}{3}$

$\dfrac{x^2-x-4x+4+x^2-3x-2x+6}{x^2-2x-4x+8}=\dfrac{10}{3}$

$\dfrac{x^2+x^2-x-4x-3x-2x+6+4}{x^2-2x-4x+8}=\dfrac{10}{3}$

$\dfrac{2x^2-10x+10}{x^2-6x+8}=\dfrac{10}{3}$

Cross Multiplying,

$3(2x^2-10x+10)=10(x^2-6x+8)$

$6x^2-30x+30=10x^2-60x+80$

$10x^2-6x^2-60x+30x+80-30=0$

$4x^2-30x+50=0$

$2(2x^2-15x+25)=0$

$2x^2-15x+25=0$

Factorising it using Splitting method,

$2x^2-10x-5x+25=0$

$2x(x-5)-5(x-5)=0$

$(x-5)(2x-5)=0$

So,

$x-5=0$

$x=5$

and

$2x-5=0$

$2x=5$

$x=\dfrac{5}{2}$

Answer :-

$So,value\:of\:\bold{x=5, \dfrac{5}{2}}.$