Question

1/(x-3)+1/(x-1)=5/x-1)-2(x-2),solve for x​

Answer 1

Step-by-step explanation:

$\frac{1}{x - 3} + \frac{1}{x - 1} = \frac{5}{x - 1} - 2(x - 2)$

$= > \frac{1}{x - 3} + 2 = \frac{4}{x - 1} \\ = > \frac{1 + 2x - 6}{x - 3} = \frac{4}{x - 1} \\ = > \frac{2x - 5}{x - 3} = \frac{4}{x - 1} \\ = > (2x - 5)(x - 1) = 4(x - 3) \\ = > 2 {x}^{2} - 2x - 5x + 5 = 4x - 12 \\ = > 2 {x}^{2} - 7x + 5 = 4x - 12 \\ = > 2 {x}^{2} - 11x + 17 = 0 \\ = > \frac{2 {x }^{2} - 11x + 17 }{2} = 0 \\ = > {x}^{2} - \frac{11}{2} x + \frac{17}{2} = 0 \\ = > {x}^{2} - \frac{11}{2} x = - \frac{17}{2} \\ = > {x}^{2} - 2 \times x \times \frac{11}{4} + ({ \frac{11}{4}) }^{2} - ( { \frac{11}{4}) }^{2} = - \frac{17}{2} \\ = > ( {x - \frac{11}{4} })^{2} = \frac{121}{16} - \frac{17}{2} \\ = > { (x - \frac{11}{4}) }^{2} = \frac{121 - 136}{16} \\ = > {(x - \frac{11}{4} })^{2} = \frac{ - 15}{16} \\ = > x - \frac{11}{4} = \frac{ \sqrt{ - 15} }{4} \\ = > x = \frac{11}{4} + \frac{ \sqrt{ - 15} }{4} \\ = \frac{11}{4 } + i \frac{ \sqrt{15} }{4}$

so here's your answer ....

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