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)$
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