Question

Please solve this as fast as possible and let me know

Answer 1

$let \ \frac{x-3}{x+3} = y \\ \\ So, \\ \\ \text{Equation will be } \\ \\ y - \frac{1}{y} = \frac{48}{7} \\ \\ \frac{y^{2} - 1}{y} = \frac{48}{7} \\ \\ 7y^{2} - 7 = 48y \\ \\ 7y^{2} - 48y - 7 = 0 \\ \\ 7y^{2} - 49y + y - 7 = 0 \\ \\ 7y(y-7) + 1(y - 7) = 0 \\ \\ (7y + 1)( y - 7) = 0 \\ \\ if \\ \\ y = 7 \\ \\ \frac{x-3}{x+3} = 7 \\ \\ x-3 = 7x + 21 \\ \\ -6x = 24 \\ \\ x = -4 \\ \\ \textbf{if } \\ \\ y = \frac{-1}{7} \\ \\ \frac{x- 3}{x+3} = \frac{-1}{7} \\ \\ 7x - 21 = -x - 3$

$7x - 21 = -x -3 \\ \\ 8x = 18 \\ \\ x = \frac{9}{4}$