Answer:
x = 5,-2
Step-by-step explanation:
Multiply the whole equation by 3(x+1) to get rid of the denominator.
[tex] \large{ \frac{x + 2}{3} \times 3(x + 1) = \frac{2(x + 2)}{x + 1} \times 3(x + 1)} \\ \large{(x + 2)(x + 1) = 2(x + 2) \times 3} \\ \large{ {x}^{2} + 3x + 2 = 6(x + 2)} \\ \large{ {x}^{2} + 3x + 2 = 6x + 12}[/tex]
Then we solve the equation.
[tex] \large{ {x}^{2} + 3x + 2 - 6x - 12 = 0} \\ \large{ {x}^{2} - 3x - 10 = 0} \\ \large{(x - 5)(x + 2) = 0} \\ \large{x = 5, - 2}[/tex]
Since we are also solving rational equation, make sure that x ≠ -1 for this problem. Because if x = -1, the denominator is 0 which is undefined. Since our solutions are x = 5,-2 and not -1. Therefore the answer is x = 5,-2
