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