# EXAMPLES OF SOLVING RATIONAL INEQUALITIES

Solve the following rational inequalities.

Example 1 :

(x + 6) / (x² - 5x - 24) ≥ 0

Solution :

Let f(x) = (x + 6) / (x² - 5x - 24)

f(x) ≥ 0

(x + 6) / (x² - 5x - 24) ≥ 0

(x + 6) / (x - 8) (x + 3) ≥ 0

By equating the numerator and denominator to zero, we get

x + 6 = 0, x - 8 = 0 and x + 3 = 0

x = -6, x = 8 and x = -3

 Intervals choosing value       f(x) from intervals f(x) (-∞, -6] -7 Є (-∞, -6] f(x) ≤ 0 [-6, -3) -4 Є [-6, -3) f(x) ≥ 0 (-3, 8) 0 Є (-3, 8) f(x) ≤ 0 (8, ∞) 9 Є (8, ∞) f(x) ≥ 0

From the above, the possible values of x are

-6 ≤ x < -3 (or) x > 8

By writing it as interval notation, we get

[-6, -3) υ (8, ∞)

So, the required solution is [-6, -3) υ (8, ∞)

Example 2 :

-10/(x - 5) ≥ -11/(x - 6)

Solution :

-10/(x - 5) ≥ -11/(x - 6)

Add 11/(x - 6) on both sides,

-10/(x - 5) + 11/(x - 6) ≥ -11/(x - 6) + 11/(x - 6)

-10/(x - 5) + 11/(x - 6) ≥ 0

Let f(x) = -10/(x - 5) + 11/(x - 6)

f(x) ≥ 0

[-10(x - 6) + 11(x - 5)] / (x - 5)(x - 6) ≥ 0

[(-10x + 60) + (11x - 55)]/(x - 5)(x - 6) ≥ 0

(-10x + 60 + 11x - 55)/(x - 5)(x - 6) ≥ 0

(x + 5)/(x - 5)(x - 6) ≥ 0

By equating the numerator and denominator to zero, we get

x + 5 = 0, x - 5 = 0, and x - 6 = 0

x = -5, x = 5, and x = 6

 Intervals choosing value       f(x) from intervals f(x) (-∞, -5] -6 Є (-∞, -5] f(x) ≤ 0 [-5, 5) 0 Є [-5, 5) f(x) ≥ 0 (5, 6) 5.5 Є (5, 6) f(x) ≤ 0 (6, ∞) 7 Є (6, ∞) f(x) ≥ 0

From the above, the possible values of x are

- 5 ≤ x < 5 (or) x > 6

By writing it as interval notation, we get

[-5, 5) υ (6, ∞)

So, the required solution is - 5 ≤ x < 5 or x > 6.

Example 3:

-3/(x + 7) ≤ -4/(x + 8)

Solution:

-3/(x + 7) ≤ - 4/(x + 8)

Add 4/(x + 8) on both sides,

-3/(x + 7) + 4/(x + 8) ≤ - 4/(x + 8) + 4/(x + 8)

Let f(x) = -3/(x + 7) + 4/(x + 8)

-3/(x + 7) + 4/(x + 8) ≤ 0

[-3(x + 8) + 4(x + 7)] / (x + 7)(x + 8) ≤ 0

(-3x - 24 + 4x + 28) / (x + 7)(x + 8) ≤ 0

(x + 4) / (x + 7)(x + 8) ≤ 0

By equating the numerator and denominator to zero, we get

(x + 4) = 0, x + 7 = 0 and x + 8 = 0

x = -4, x = -7, and x = -8

 Intervals choosing value       f(x) from intervals f(x) (-∞, -8) -9 Є (-∞, -8) f(x) ≤ 0 (-8, -7) -7.5 Є (-8, -7) f(x) ≥ 0 (-7, -4] -5 Є (-7, -4] f(x) ≤ 0 [-4, ∞) 0 Є [-4, ∞) f(x) ≥ 0

From the above, the possible values of x are

x < -8 (or) -7 <  -4

By writing it as interval notation, we get

[-∞, -8) υ (-7, -4]

So, the required solution is [-∞, -8) υ (-7, -4]

Example 4 :

-7/(x + 5) ≤ -8/(x + 6)

Solution:

-7/(x + 5) ≤ -8/(x + 6)

Add 8/(x + 6) on both sides,

-7/(x + 5) + 8/(x + 6) ≤ -8/(x + 6) + 8/(x + 6)

-7/(x + 5) + 8/(x + 6) ≤ 0

Let f(x) = -7/(x + 5) + 8/(x + 6)

-7/(x + 5) + 8/(x + 6) ≤ 0

[-7(x + 6) + 8(x + 5)] / (x + 5)(x + 6) ≤ 0

(-7x - 42 + 8x + 40) / (x + 5)(x + 6) ≤ 0

(x - 2) / (x + 5)(x + 6) ≤ 0

By equating the numerator and denominator to zero, we get

x - 2 = 0, x + 5 = 0, and x + 6 = 0

x = 2, x = -5, and x = -6

 Intervals choosing value       f(x) from intervals f(x) (-∞, -6) -7 Є (-∞, -6) f(x) ≤ 0 (-6, -5) -5.5 Є (-6, -5) f(x) ≥ 0 (-5, 2] 0 Є (-5, 2] f(x) ≤ 0 [2, ∞) 3 Є [2, ∞) f(x) ≥ 0

From the above, the possible values of x are

x < -6 (or) -5 <  2

By writing it as interval notation, we get

(-∞, -6) υ (-5, 2]

So, the required solution is x < -6 or -5 <  2.

Example 5 :

(x + 7) (x - 3) / (x - 5)² > 0

Solution :

Let f(x) = (x + 7) (x - 3) / (x - 5)²

f(x) > 0

(x + 7) (x - 3) / (x - 5)² > 0

By equating the numerator and denominator to zero, we get

x + 7 = 0, x - 3 = 0, and (x - 5)2 = 0

x = -7, x = 3, and (x - 5)2 = 02

x = -7, x = 3, and x = 5

 Intervals choosing value       f(x) from intervals f(x) (-∞, -7) -8 Є (-∞, -7) f(x) > 0 (-7, 3) 0 Є (-7, 3) f(x) < 0 (3, 5) 4 Є (3, 5) f(x) > 0 (5, ∞) 6 Є [5, ∞) f(x) > 0

From the above, the possible values of x are

x < -7 (or) 3 < x < 5 (or) x > 5

By writing it as interval notation, we get

(-∞, -7) υ (3, 5) υ (5, ∞)

So, the required solution is (-∞, -7) υ (3, 5) υ (5, ∞)

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