Question

Solve for X and give correct answer with correct explanation.
$\frac{x - 1}{2x + 1} + \frac{2x + 1}{x - 1} = \frac{5}{2}$
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Answer 1

➫QuesTion :-

$\tt \: solve \: \: for \: \: x \\ \to \: \tt \frac{x - 1}{2x + 1} + \frac{2x + 1}{x - 1} = \frac{5}{2}$

➫Answer :-

→ x = -1

➫SoluTion :-

We have ,

$\to \: \tt \frac{x - 1}{2x + 1} + \frac{2x + 1}{x - 1} = \frac{5}{2} \: \\ \\ \mathbb{Tak \tt{ing} \: \: \mathbb{LCM \: }} \\ \\ \to \tt \: \frac{ {(x - 1)}^{2} + {(2x + 1)}^{2} }{(x - 1)(2x + 1)} = \frac{5}{2} \\ \\ \to \: \tt \frac{ {x}^{2} + 1 - 2x + 4 {x}^{2} + 1 + 4x }{(x - 1)(2x + 1)} = \frac{5}{2} \\ \\ \to \: \tt \frac{5 {x}^{2} + 2x + 2}{(x - 1)(2x + 1)} = \frac{5}{2} \\ \\ \red{\bf Cross \: multiplication } \\ \\ \to \tt 2(5 {x}^{2} + 2x + 2) = 5(x - 1)(2x + 1)</p><p> \: \\ \\ \to \tt10 {x}^{2} + 4x + 4 = 5 \{2 {x}^{2} + x - 2x - 1 \} \\ \\ \to \tt10 {x}^{2} + 4x + 4 = 10 {x}^{2} - 5x - 5 \\ \\ \green{\bf Seperating \: one \: side } \\ \\ \to \tt10 {x}^{2} + 4x + 4 - 10 {x}^{2} + 5x + 5 = 0 \\ \\ \to \tt 9x + 9 = 0 \\ \\ \to \tt 9x = - 9 \\ \\ \to \boxed{ \tt \: x \: = - 1}$

➫Verification :-

$\to \: \tt \frac{x - 1}{2x + 1} + \frac{2x + 1}{x - 1} = \frac{5}{2} \: \\ \\ \tt \: put \: \: \: x \: = - 1 \\ \\ \to \tt \frac{ - 1 - 1}{2( - 1) + 1} + \frac{2( - 1) + 1}{ - 1 - 1} = \frac{5}{2} \\ \\ \to \tt \frac{ - 2}{ - 1} + \frac{( - 1)}{( - 2)} = \frac{5}{2} \\ \\ \to \tt \: \frac{2}{1} + \frac{1}{2} = \frac{5}{2} \\ \\ \to \tt \: \frac{2 \times 2 + 1}{2} = \frac{5}{2} \\ \\ \to \tt \frac{5}{2} = \frac{5}{2}$

Hence verified .