Question

2 upon (x-3)+x upon (x square-9) =4 upon (x + 3)​

Answer 1

Question :-

Solve for x

$\to \sf \frac{2}{x - 3} + \frac{x}{ {x}^{2} - 9} = \frac{4}{x + 3}$

Answer :-

→ x = 18

Formula used :-

→ a² - b² = ( a+b)(a - b)

Explanation :-

we have ,

$\to \sf \frac{2}{x - 3} + \frac{x}{ {x}^{2} - 9} = \frac{4}{x + 3} \\ \\ \to \sf \frac{2}{x - 3} + \frac{x}{ {x}^{2} - {3}^{2} } = \frac{4}{x + 3} \\ \\ \to \sf \frac{2}{x - 3} + \frac{x}{ (x - 3)(x + 3)} = \frac{4}{x + 3} \\ \: \\ \to \sf \frac{2(x + 3) + x}{(x - 3)(x + 3)} = \frac{4}{x + 3} \\ \\ \to \sf \frac{2x + 6 + x}{(x - 3)} = 4 \\ \\ \to \sf 3x + 6 = 4x - 12 \\ \\ \to \sf 4x - 3x = 6 + 12 \\ \\ \to \sf \boxed{ x = 18}$