Question

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${\textsf{\textbf{Question\::}}}$
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$\tt{\dfrac{x+3}{x-2}-\dfrac{1-x}{x}=4\dfrac{1}{4}}$​

Answer 1

Answer:

Hope my answer will help you

Answer 2

$\large{\underline{\underline{\pmb{\frak{ Given :- }}}}}$

$\sf \dashrightarrow \dfrac{x+3}{x-2} - \dfrac{1-x}{x} = 4 \dfrac{1}{4}$  

$\large{\underline{\underline{\pmb{\frak{ Solution :- }}}}}$

$\sf \dashrightarrow \dfrac{x+3}{x-2} - \dfrac{1-x}{x} = 4 \dfrac{1}{4}$

$\sf \dashrightarrow \dfrac{x+3}{x-2} - \dfrac{1-x}{x} = \dfrac{17}{4}$

~Here, LCM of denominators of fractions of this equation is 4x( x - 2 ) . So, we’ll multiply it on both sides of the equation.  

$\sf \dashrightarrow 4x( x + 3 ) - ( 4x -8)(1-x) = 17x(x-2)$

$\sf \dashrightarrow (4x^{2} + 12x) -(12x-4x^{2} -8) = 17x^{2} -34x$

$\sf \dashrightarrow 4x^{2} +12x -12x+4x^{2}+8 = 17x^{2} -34x$

$\sf \dashrightarrow 8x^{2}+8 = 17x^{2}-34x$

$\sf \dashrightarrow ( 8x^{2} -17x^{2} ) + 8 = -34x$

$\sf \dashrightarrow -9x^{2} + 8x +34x = 0$

$\sf \dashrightarrow -9x^{2} +36x-2x+8 = 0$

$\sf \dashrightarrow 9x( -x+4) +2(-x+4) = 0$

$\sf \dashrightarrow (-x+4)(9x+2) = 0$

Solving first bracket :

$\sf \leadsto ( -x+4 ) = 0$

$\sf \leadsto -x = -4$

$\bigstar \: \underline{\boxed{\bf{x=4}}}$

Solving second bracket :

$\sf \leadsto (9x+2) = 0$

$\sf \leadsto 9x = -2$

$\bigstar \: \underline{\boxed{\bf{ x = \dfrac{-2}{9}}}}$

Hence,  

• Value of x can be 4 or -2/9