Question

solve for x:


$\frac{x - 1}{x - 2} + \frac{x - 3}{x - 4} = 3 \frac{1}{3} \: \: \: \: \: (x≠2,4). \:$

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Answer 1

$\huge\underline{\underline{\bf{ \purple{Solution:-}}}}$

$\tt \: The \: given \: equation \: is$

$\tt \frac{x- 1 }{x - 2} + \frac{x - 3}{x - 4 } = 3\frac{1}{3} (x \: ≠ \: 2, \: 4)$

$\tt \: Taking \: L.C.M.$

$\tt \frac{(x - 1)(x - 4) + (x - 3)(x - 2)}{(x - 2)(x - 4)} = 3 + \frac{1}{3}$

$\tt \implies \: \: \: \: \frac{x {}^{2} - 4x - x + 4 + x {}^{2} - 2x - 3x + 6}{ x{}^{2} - 4x - 2x + 8} = \frac{10}{3}$

$\tt \: Cross - \: multiplying$

$\tt \: 3(2x {}^{2} - 10x + 10) = 10(x {}^{2} - 6x + 8)$

$\implies \: \: \: \: \: \: \: \tt \: 6x {}^{2} - 30x + 30 = 10x {}^{2} - 60x + 80$

$\implies \: \: \: \: \tt \: - 4 x{}^{2} + 30x - 50 = 0$

$\tt \: Multiplying \: by \: - 1,$

$\tt \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 4x {}^{2} - 30x + 50 = 0$

$\tt \: Compairing \: with \: ax {}^{2} + bx + c = 0,$

$\tt \: We \: have$

$\tt \: a = 4, \: b = - 30, \: c = 50$

$\tt \: So, \: D = b {}^{2} - 4ac = ( - 30) {}^{2} - 4(4)(50)$

$\: \: \: \: \: \: \: \: \: \: \tt = 900 - 800 = 100$

$\tt \: Using \: Quadratic \: formula,$

$\tt \: We \: have$

$\: \: \: \: \: \: \: \: \: \: \tt x = \frac{ - b \: ± \: \sqrt{D} }{2a}$

$\: \: \: \: \: \: \: \: \: \tt \: = \frac{ - ( - 30) \: \: ±\sqrt{100} }{8}$

$\tt = \: \: \: \: \: \: \: \: \: \: \frac{ 30 \: ± \: 10}{8}$

$\implies \: \: \: \: \: \: \: \: \: \: \tt x = \frac{30 - 10}{8} = \frac{20}{8}$

$\implies \: \: \: \: \: \: \: \: \: \: \: \tt \: x = 5 \: \frac{5}{2}$

$\tt \: Hence, \: The \: Solution \: of \: the \: given \: equation \: \\ \tt \: are \: \red{5} \: and \red{ \frac{5}{2}}$