If x is negative, what is the range of values of x within which $\frac{x + 1}{3}$ > $\frac{1}{X + 3}$

3 < x < 4

-4 < x < -3

-2 < x < -1

-3 < x < 0

Correct answer is B

$\frac{x + 1}{3}$ > $\frac{1}{X + 3}$ = $\frac{x + 1}{3}$ > $\frac{x + 3}{X + 3}$

= (x + 1)(x + 3)² > 3(x + 3) = (x + 1)[x² + 6x + 9] > 3(x + 3)

x³ + 7x² + 15x + 9 > 3x + 9 = x³ + 7x² + 12x > 0

= x(x + 3)9x + 4) > 0

Case 1 (+, +, +) = x > 0 , x + 3 > 0, x + 4 > 0

= x > -4 (solution only)

Case 2 (+, -, -) = x > 0, x + 4 < 0

= x > 0, x < -3, x < -4 = x < -3(solution only)

Case 3 (-, +, -) = x < 0, x > -3, x < -4 = x < -0, -4 < x < 3(solutions)

Case 4 (-, -, +) = x < 0, x + 3 < 0, x + 4 > 0

= x < 0, x < -5, x > -4 = x < -0, -4 < x < -3(solution)

combining the solutions -4 < x < -3

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