Question

x−1 / x + x /x−1 = 61 30 ; x ≠ 0, 1​​​

Answer 1

$\red{\bold{\underline{\underline{❥Question᎓}}}}$

$\sf \frac{x−1}{x} + \frac{x }{x−1} = \frac{61}{30}; x ≠ 0, 1$

$\huge\tt\underline\blue{Answer}$

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$\sf ➜ \: \frac{x-1}{x} + \frac{x}{x-1} = \frac{61}{30}$

$\sf ➜ \: \frac{(x-1)(x-1) + (x × x)}{x(x-1)} = \frac{61}{30}$

$\sf ➜ \: \frac{x(x-1) -1(x-1) + x²}{x²-x} = \frac{61}{30}$

$\sf ➜ \: \frac{x²-x -x+1 + x²}{x²-x} = \frac{61}{30}$

$\sf ➜ \: \frac{2x²-2x+1}{x²-x} = \frac{61}{30}$

$\sf ➜ \: (2x²-2x+1)30 = 61(x²-x)$

$\sf ➜ \: 60x²- 60x + 30 = 61x² - 61x$

$\sf ➜ \: 60x²- 60x + 30 - 61x² + 61x = 0$

$\sf ➜ \: -x² + x + 30 = 0$

$\sf ➜ \: -x² + x + 30 = 0$

$\sf ➜ \: -x² + 6x - 5x + 30 = 0$

$\sf ➜ \: -x(x - 6) - 5(x - 6) = 0$

$\sf ➜ \: (-x - 5)(x - 6) = 0$

___________________________

$\sf -x -5 = 0$

$\sf -x = 5$

$\boxed{\sf x = -5}$

___________________________

$\sf x -6 = 0$

$\boxed{\sf x = 6}$

___________________________

$\sf \red{Therefore, \: value \: of \: x \: is \: 6 \: or \: -5}$

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Answer 2

$\bf \pink{Question}$

$\sf ➜ \: \frac{(x-1)(x-1) + (x × x)}{x(x-1)} = \frac{61}{30}$

$\sf ➜ \: \frac{x(x-1) -1(x-1) + x²}{x²-x} = \frac{61}{30}$

$</p><p>\sf ➜ \: \frac{x²-x -x+1 + x²}{x²-x} = \frac{61}{30}$

$\sf ➜ \: \frac{2x²-2x+1}{x²-x} = \frac{61}{30}$

$\sf ➜ \: (2x²-2x+1)30 = 61(x²-x)$

$\sf ➜ \: 60x²- 60x + 30 = 61x² - 61x$

$\sf ➜ \: 60x²- 60x + 30 - 61x² + 61x = 0$

$\sf ➜ \: -x² + x + 30 = 0$

$\sf ➜ \: -x² + x + 30 = 0$

$\sf ➜ \: -x² + 6x - 5x + 30 = 0$

$\sf ➜ \: -x(x - 6) - 5(x - 6) = 0$

$\sf ➜ \: (-x - 5)(x - 6) = 0$

$___________________________$

$\sf -x -5 = 0$

$\sf -x = 5$

$\boxed{\sf \pink{ x = -5}}$

$___________________________</p><p></p><p>$

$</p><p>\sf x -6 = 0$

$\boxed{\sf x = 6}$

$___________________________$

$\sf \red{Therefore, \: value \: of \: x \: is \: 6 \: or \: -5}$