Question

Solve the following quadratic equation by factorization :
Solve for $x:\frac{x-1}{x-2}+\frac{x-3}{x-4}=3\frac{1}{3},x\ne 2,4$

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

Answer 1

Answer: (B)

$5$, $\frac{5}{2}$

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

$\Rightarrow$ $\frac{(x-1)(x-4)+(x-3)(x-2)}{(x-2)(x-4)}=\frac{10}{3}$

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

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

$\Rightarrow$ $\frac{2(x^2-5x+5)}{x^2-6x+8}=\frac{10}{3}$

$\Rightarrow$ $\frac{x^2-5x+5}{x^2-6x+8}=\frac{5}{3}$

$\Rightarrow$ $3x^2-15x+15=5x^2-30x+40$

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

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

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

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

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

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