Question

Where is the mistake ?my answer is not coming.​

Answer 1

Answer:

Step-by-step explanation

I advice you to try futher sum by delta method :)

Answer 2

$\large\underline\purple{\bold{Solution :- }}$

$\tt \:  \longrightarrow \: \dfrac{x + 3}{x - 2} - \dfrac{1 - x}{x} = \dfrac{17}{4}$

$\tt \:  \longrightarrow \: \dfrac{(x + 3)x - (x - 2)(1 - x)}{(x - 2)x} = \dfrac{17}{4}$

$\tt \:  \longrightarrow \: \dfrac{ {x}^{2} + 3x - (x - {x}^{2} - 2 + 2x) }{ {x}^{2} - 2x} = \dfrac{17}{4}$

$\tt \:  \longrightarrow \: \dfrac{ {x}^{2} + 3x - 3x + 2 + {x}^{2} }{ {x}^{2} - 2x} = \dfrac{17}{4}$

$\tt \:  \longrightarrow \: \dfrac{2 {x}^{2} + 2}{ {x}^{2} - 2x} = \dfrac{17}{4}$

$\tt \:  \longrightarrow \: {8x}^{2} + 8 = {17x}^{2} - 34x$

$\tt \:  \longrightarrow \: {9x}^{2} - 34x - 8 = 0$

$\tt \:  \longrightarrow \: {9x}^{2} - 36x + 2x - 8 = 0$

$\tt \:  \longrightarrow \: 9x(x - 4) + 2(x - 4) = 0$

$\tt \:  \longrightarrow \: (x - 4)(9x + 2) = 0$

$\tt\implies \: \boxed{ \red {\tt \:  x \: = 4 \: or \: x \: = \: - \: \dfrac{2}{9} }}$