[tex] \dfrac{5}{x^2 - 4} + \dfrac{2}{x} = \dfrac{2}{x - 2} [/tex]
[tex] \dfrac{5}{(x + 2)(x - 2)} + \dfrac{2}{x} = \dfrac{2}{x - 2} [/tex]
[tex] \dfrac{5}{(x + 2)(x - 2)} \times x(x + 2)(x - 2) + \dfrac{2}{x} \times x(x + 2)(x - 2) = \dfrac{2}{x - 2} \times x(x + 2)(x - 2) [/tex]
[tex] 5x + 2(x + 2)(x - 2) = 2x(x + 2) [/tex]
[tex] 5x + 2(x^2 - 4) = 2x^2 + 4x [/tex]
[tex] 5x + 2x^2 - 8 = 2x^2 + 4x [/tex]
[tex] 5x - 8 = 4x [/tex]
[tex] x - 8 = 0 [/tex]
[tex] x = 8 [/tex]
Now we look at the common denominator.
It is x(x + 2)(x - 2).
x cannot be zero, -2 or 2 because that would cause a zero in the denominator.
Since we get x = 8, and x = 8 does not have to be excluded from the domain, the answer is x = 8.
Answer: x = 8
