Question

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

Answer 1

Answer:

$\huge{\bold☘}\mathfrak\pink{\bold{\underline{{ ℘ɧεŋσɱεŋศɭ}}}}{\bold☘}$

$\huge\tt\red{\bold{\underline{\underline{❥Question᎓}}}}$$\huge\tt\bold{\frac{x - 1}{x - 2} + \frac{x - 3}{x - 4} = 3 \frac{1}{3} }$

$\huge\tt\mathbb{\boxed{\overbrace{\underbrace{\blue{Answer</p><p> }}}}}$

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

$⟹\bold{\frac{(x - 1)(x - 4) + (x - 2)(x - 3)}{(x - 2)(x - 4)} = \frac{10}{3}}$

$⟹\bold{\frac{ {x}^{2} - x - 4x + 4 + {x}^{2} - 2x - 3x + 6 }{ {x}^{2} - 2x - 4x + 8} = \frac{10}{3} }$

$⟹\bold{\frac{2 {x}^{2} - 10x + 10}{ {x}^{2} - 6x + 8 } = \frac{10}{3}}$

$⟹\bold{6 {x}^{2} - 30x + 30 = 10 {x}^{2} - 60x + 80}$

$⟹\bold{4 {x}^{2} - 30x + 50 = 0}$

$⟹\bold{2 {x}^{2} - 15x + 25 = 0}$

$⟹\bold{2 {x}^{2} - 10x - 5x + 25 = 0}$

$⟹\bold{2x(x - 5) - 5(x - 5) = 0}$

$⟹\bold{(x - 5)(2x - 5) = 0}$

$⟹\bold{x - 5 = 0 \: or \: 2x - 5 = 0}$

$⟹\bold{x = 5 \: or \: x = \frac{5}{2}}$

$\bold{∴5\: and\: 5/2 \:are \:two \:solutions \:of \:given\: equation}$

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Answer 2

According to the question

$⟹\bold{\frac{(x - 1)(x - 4) + (x - 2)(x - 3)}{(x - 2)(x - 4)} = \frac{10}{3}} \\ \\ </p><p> </p><p></p><p> </p><p> </p><p></p><p>⟹\bold{\frac{ {x}^{2} - x - 4x + 4 + {x}^{2} - 2x - 3x + 6 }{ {x}^{2} - 2x - 4x + 8} = \frac{10}{3} } \\ \\ </p><p> </p><p> </p><p></p><p>⟹\bold{\frac{2 {x}^{2} - 10x + 10}{ {x}^{2} - 6x + 8 } = \frac{10}{3}} \\ \\ </p><p></p><p>⟹\bold{6 {x}^{2} - 30x + 30 = 10 {x}^{2} - 60x + 80} \\ \\ </p><p>⟹\bold{2 {x}^{2} - 15x + 25 = 0} \\ \\ </p><p></p><p>⟹\bold{2 {x}^{2} - 10x - 5x + 25 = 0} \\ \\ </p><p></p><p>⟹\bold{2x(x - 5) - 5(x - 5) = 0} \\ \\ </p><p></p><p>⟹\bold{(x - 5)(2x - 5) = 0} \\ \\ </p><p></p><p>⟹\bold{x - 5 = 0 \: or \: 2x - 5 = 0} \\ \\ </p><p></p><p>⟹\bold{x = 5 \: or \: x = \frac{5}{2}} \\ \\ </p><p>\bold{∴5\: and\: 5/2 \:are \:two \:solutions \:of \:given\: equation}$