Question

Plzz guys answer me

Answer 1

$\frac{x+1}{x+2} - \frac{x+1}{x-2} = \frac{4}{3}$

$\frac{x+1}{x+2} * 3(x+2)(x-2) - \frac{x + 1}{x-2} * 3(x + 2)(x - 2) = \frac{4}{3} * 3(x+2)(x-2)$

3(x + 1)(x - 2) - 3(x + 1)(x + 2) = 4(x + 2)(x - 2)

3x^2 - 3x - 6 - 3x^2 - 9x - 6 = 4x^2 - 16

-12x - 12 = 4x^2 - 16

4x^2 - 4 = -12x

4x^2 + 12x - 4 = 0

x^2 + 3x - 1 = 0

We know that when an equation is in the form of ax^2 + bx + c = 0, its solutions are:

(1)

$x = \frac{-b + \sqrt{b^2 - 4ac} }{2a}$

$\frac{-3 + \sqrt{3^2 - 4 * 1 * (-1)} }{2 * 1}$

$\frac{-3 + \sqrt{3^2 + 4} }{2}$

$\frac{-3 + \sqrt{13} }{2}$



(2)

$\frac{-b - \sqrt{b^2 - 4ac} }{2a}$

$\frac{-3 - \sqrt{3^2 - 4 * 1 * (-1)} }{2 * 1}$

$\frac{-3 - \sqrt{3^2 + 4} }{2}$

$\frac{-3 - \sqrt{13} }{2}$


Therefore the solutions are:

$\frac{-3 + \sqrt{13} }{2}, \frac{-3 - \sqrt{13} }{2}$


Hope this helps!