Question

Solve x:
x-1÷x-2+x-3÷x-4 =31÷3

Answer 1

Answer :

Now,

$\bold{\frac{x-1}{x-2} + \frac{x-3}{x-4} = \frac{31}{3}}$

$\to \bold{\frac{(x-1)(x-4)+(x-3)(x-2)}{(x-2)(x-4)} = \frac{31}{3}}$

$\to \bold{3 \{(x-1)(x-4) + (x-3)(x-2)\} = 31(x-2)(x-4)}$

$\to \bold{3 ({x}^{2} - 5x + 4 + {x}^{2} - 5x + 6) = 31 ({x}^{2} - 6x + 8)}$

$\to \bold{3 (2 {x}^{2} - 10x + 10) = 31 {x}^{2} - 186x + 248}$

$\to \bold{6 {x}^{2} - 30x + 30 = 31 {x}^{2} - 186x + 248}$

$\to \bold{(31 - 6) {x}^{2} + (-186+30)x + (248 - 30) = 0}$

$\to \bold{25 {x}^{2} - 156x + 218 = 0}$

Using Sridhar Acharya's theorem, we get

$\bold{x = \frac{ - ( - 156) \pm \sqrt{ {( - 156)}^{2} - 4(25)(218)} }{2 \times 25}}$

$= \bold{\frac{156 \pm \sqrt{24336 - 21800}}{50}}$

$= \bold{\frac{156 \pm \sqrt{2536} }{50}}$

$= \bold{\frac{(2 \times 78)\pm2 \sqrt{634} }{2 \times 25}}$

$= \bold{\frac{78 \pm \sqrt{634} }{25}}$

∴ the required solution be

$\boxed{\bold{x = \frac{78 + \sqrt{634} }{25}, \frac{78 - \sqrt{634} }{25}}}$

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