Question

Solve the given quadratic equation by factorisation method: $\frac{x-1}{x-2}+\frac{x-3}{x-4}=3\frac{1}{3}$

Answer 1

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

$⟹\frac{x-1}{x-2}+\frac{x-3}{x-4}=\frac{10}{3}$

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

$⟹x^2-5x+4+x^2-5x+6=\frac{10}{3}(x^2-6x+8)$

$⟹2x^2-10x+10=\frac{10}{3}(x^2-6x+8)$

$⟹3x^2-15x+15=5x^2-30x+40$

$⟹2x^2-15x+25=0$

$⟹2x^2-10x-5x+25=0$

$⟹2x(x-5)-5(x-5)=0$

$⟹(2x-5)(x-5)=0$

$⟹x=\frac{5}{2},5$