Question

(x-3/x+3)-(x+3/x-3)=48/7​

Answer 1

$\huge\bold{\gray{\sf{Answer:}}}$

$\bold{Explanation:}$

$= > ( \dfrac{x - 3}{x + 3} ) - ( \dfrac{x + 3}{x - 3} ) = \dfrac{48}{7}$

$= > \dfrac{(x - 3)(x - 3) - (x + 3)(x + 3)}{(x + 3)(x - 3)} = \dfrac{48}{7}$

Here, identity is used:

$\bold{\boxed{ {(x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy}}$

$\bold{\boxed{ {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy}}$

$\bold{\boxed{(x + y)(x - y) = {x}^{2} - {y}^{2}}}$

$= > \dfrac{ {(x - 3)}^{2} - {(x + 3)}^{2} }{ {x}^{2} - {3}^{2} } = \dfrac{48}{7}$

$= > \dfrac{ {x}^{2} + {y}^{2} - 2(x)(3) - [{x}^{2} + {y}^{2} + 2(x)(3)] }{ {x}^{2} - {3}^{2} } = \dfrac{48}{7}$

$= > \dfrac{ {x}^{2} + {y}^{2} - 6x - {x}^{2} - {y}^{2} - 6x }{ {x}^{2} - {3}^{2} } = \dfrac{48}{7}$

$= > \dfrac{ - 12x}{ {x}^{2} - {3}^{2} } = \dfrac{48}{7}$

$= > - 84x = 48( {x}^{2} - {3}^{2} )$

$= > - 84x = 48 {x}^{2} - 432$

$= > 48 {x}^{2} + 84x - 432 = 0$

Taking 4 common from the equation:-

$= > 4(12 {x}^{2} + 21x - 108) = 0$

$= > 12 {x}^{2} + 21x - 108 = 0$

Taking 3 common from the equation:-

$= > 4 {x}^{2} + 7x - 36 = 0$

$= > 4 {x}^{2} + 16x - 9x - 36 = 0$

$= > 4x(x + 4) - 9(x + 4) = 0$

$= > (4x - 9)(x + 4) = 0$

$= > 4x - 9 = 0$

$\bold{\red{x = \dfrac{9}{4}}}$

$= > x + 4 = 0$

$\bold{\red{x = - 4}}$

$\therefore$x=-4 and x=9/4

Answer 2

Step-by-step explanation:

$\huge\rm{Solution:-} \\ \\ \rm→ \frac{(x-3)²-(x+3)²}{(x+3)(x-3)}= \frac{48}{7} \\ \\ \rm→ \frac{x²+9-6x-(x²+9+6x)}{x²-3x+3x-9} = \frac{48}{7} \\ \\ \rm→ \frac{x²+9-6x-x²-9-6x}{x²-9} = \frac{48}{7} \\ \\ \rm→ \frac{-12x}{x²-9} \neq \frac{48}{7} \\ \\ \rm→-84x=48x²-48(9) \\ \\ →12(-7)=12\rm(4x²-36) \\ \\ \rm→0=4x²+7x-36 \\ \\ \rm→4x²+16x-9x-36=0 \\ \\ \rm→4x(x+4)-9(x+4)=0 \\ \\ \rm→(x+4)(4x-9)=0 \\ \\ \rm{\fbox{→x=-4}} \: \: \: |\: \: \: 4x-9=0 \\ \\ \rm→ \: \: \: \: \: \: \: \: \: \: \: 4x=9 \\ \\ \rm \: Answer→\: \: \: \: \: \: x= \frac{9}{4}$

x = -4; x = 9/4