solve the equation: $\frac{x - 4}{x - 1} + \frac{x - 7}{x - 3} + \frac{x - 2}{x - 9} = 3$ (Algebra)

Created: 1 year ago | Updated: 1 year ago

Updated: 1 year ago

$\frac{x - 4}{x - 1} + \frac{x - 7}{x - 3} + \frac{x - 2}{x - 9} = 3 \Rightarrow \frac{x - 4}{x - 1} + \frac{x - 7}{x - 3} = 3 - \frac{x - 2}{x - 9} \Rightarrow \frac{(x - 4) (x - 3) + (x - 7) (x - 1)}{(x - 1) (x - 3)} = \frac{3 (x - 9) - (x - 2)}{x - 9} \Rightarrow \frac{x^{2} - 3 z - 4 z + 12 + x^{2} - x - 7 x + 7}{x^{2} - 3 x - x + 3} = \frac{3 x - 27 - x + 2}{x - 9} \Rightarrow \frac{2 x^{2} - 15 x + 19}{x^{2} - 4 x + 3} = \frac{2 x - 25}{x - 9} \Rightarrow (2 x^{2} - 15 x + 19) (x - 9) = (x^{2} - 4 x + 3) (2 x - 25) \Rightarrow 2 x^{3} - 15 x^{2} + 19 x - 18 x^{2} + 135 x - 171 = 2 x^{3} - 8 x^{2} + 6 x - 25 x^{2} + 100 x - 75 \Rightarrow - 33 x^{2} + 154 x - 171 = - 33 x^{2} + 106 x - 75 \Rightarrow - 33 x^{2} + 33 x^{2} + 154 x - 106 x = - 75 + 171 \Rightarrow 48 x = 96 ∴ x = \frac{96}{48} = 2$