Question

If $\frac{x-1}{x-2}+\frac{x-3}{x-4}=3\frac{1}{3}$ and $(x\ne 2,4)$, then value of $x$ is

• A. $5,\frac{5}{2}$
• B. $0$
• C. $2$
• D. $-5,\frac{5}{2}$

Answer 1

The correct option is A $5,\frac{5}{2}$

Given equation is: $\frac{x-1}{x-2}+\frac{x-3}{x-4}=\frac{10}{3}$

Multiplying both sides by $3(x-2)(x-4)$, we get

$3(x-1)(x-4)+3(x-3)(x-2)$

$=10(x-2)(x-4)$

$3(x^2-5x+4)+3(x^2-5x+6)=10(x^2-6x+8)$

$3x^2-15x+12+3x^2-15x+18=10x^2-60x+80$

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

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

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

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

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

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

$\Rightarrow x-5=0$ and $2x-5=0$

$\Rightarrow x=5$ and $x=\frac{5}{2}$

Hence, $x=5,\frac{5}{2}$ are required solutions.