Factorise $\frac{3}{x} -1+\frac{4}{x} -3\ =\ \frac{15}{x} +3$

Given:

$\frac{3}{x} -1+\frac{4}{x} -3\ =\ \frac{15}{x} +3$

To do: Factorize

Solution:

$\frac{3}{x-1}+\frac{4}{x-3}=\frac{15}{x+3}$

$\frac{3x - 9 + 4x - 4 }{x - 1}\times(x - 3) = \frac{15}{x+3}$

$\frac{7x - 13}{x - 1}\times(x - 3) = \frac{15}{x+3}$

$a\frac{7x - 13}{x+3} = 15\frac{x - 1}{x - 3}$ 

$7x^{2} - 13x + 21x - 39 = 15x^{2} - 60x + 45$

$8x^{2} - 68x + 84 = 0$

$2x^{2} - 17x + 21 = 0$

$2x^{2} - 14x -3x + 21 = 0$

$2x(x - 7) -3(x - 7)$

$(2x - 3)(x - 7) = 0$

$\frac{3}{x-1}+\frac{4}{x-3}=\frac{15}{x+3}$ factorized as

=$ (2x - 3)(x - 7)$