Question

Solve $\frac{x-3}{x-4}+\frac{x-5}{x-6}=0$

Answer 1

We have,

$\frac{x-3}{x-4}+\frac{x-5}{x-6}=0$

$\Rightarrow \frac{(x-3)(x-6)+(x-5)(x-4)}{(x-4)(x-6)}=0$

$\Rightarrow \frac{x^2-3x-6x+18+x^2-5x-4x+20}{(x-4)(x-6)}=0$

$\Rightarrow x^2-9x+18+x^2-9x+20=0$

$\Rightarrow 2x^2-18x+38=0$

$\Rightarrow x^2-9x+19=0$

Comparing that, $a{x}^2+bx+c=0$ and we get,

$a=1,b=-9,c=19$

Using quadratic equation formula

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$x=\frac{-(-9)\pm \sqrt{81-4\times 1\times 19}}{2\times 1}$

$x=\frac{9\pm \sqrt{81-76}}{2}$

$x=\frac{9\pm \sqrt{5}}{2}$

Hence, this is the answer.