SOLVING RATIONAL INEQUALITY EXAMPLES

Example 1 :

Solve the following linear inequalities. 

(x - 3)/(x - 5) > 0

Solution :

(x - 3)/(x - 5) > 0

By equating the numerator and denominator to zero, we get

x = 3 and x = 5

The critical number is dividing the number line into three regions.

Since we have greater than sign (>) in the given question, we have to choose only positive regions.

Hence, the solution set of given equation is

(-∞, 3) U (5, ∞)

Example 2 :

Solve the following linear inequations

(x - 2)/(x + 5) > 2

Solution :

(x - 2)/(x + 5) > 2

Subtract by 2 throughout the equation

[(x - 2)/(x + 5)] - 2 > 2 - 2

By taking L.C.M we get,

[(x - 2) - 2(x + 5)}/(x + 5) > 0

(x - 2 - 2x - 10)/(x + 5) > 0

-x - 12/(x + 5) > 0

- (x + 12)/(x + 5) > 0

In order to make the coefficient x as positive, we have to multiply by -1 through out the equation

(x + 12)/(x + 5) < 0

So the > has being converted into <.

Equate the numerator and denominator to zero, we get

x + 12 = 0    x + 5 = 0

x = -12 and x = -5 (Critical numbers)

The critical number is dividing the number line into three regions.

Since we have greater than sign (<) in the last step, we have to choose only negative regions.

Hence, the solution set of given equation is

(-12, -5)

Example 3 :

(x - 7)/(x - 1) < 0

Solution :

(x - 7)/(x - 1) < 0

By equating the numerator and denominator to zero, we get

x = 7 and x = 1

Plotting the numbers in the number line, 

(-∞, 1), (1, 7) and (7, ∞)

x = 0 ∈ (-∞, 1)

f(x) = (x - 7)/(x - 1) < 0

f(0) = (0 - 7)/(0 - 1) < 0

= 7 < 0

False

x = 2 ∈ (1, 7)

f(2) = (2 - 7)/(2 - 1) < 0

= -5/1 < 0

= -5 < 0

True

x = 8 ∈ (7, ∞)

f(8) = (8 - 7)/(8 - 1) < 0

= 1/7 < 0

False

For the interval (1, 7), the function is true. Then the solution is (1, 7).

Example 4 :

(x + 6)/(x² - 5x - 24) ≥ 0

Solution :

(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, x = 8 and x = -3

Plotting the numbers in the number line and decomposing into intervals, we get

(-∞, -6), (-6, -3) (-3, 8) and (8, ∞)

When x = -7 ∈ (-∞, -6)

(-7 + 6)/(-7 - 8)(-7 + 3) ≥ 0

(-1)/(-15)(-4) ≥ 0

-1/60 ≥ 0

False

When x = -5 ∈ (-6, -3)

(-5 + 6)/(-5 - 8)(-5 + 3) ≥ 0

1/(-13)(-2) ≥ 0

1/26 ≥ 0

True

When x = 0 ∈ (-3, 8)

(0 + 6)/(0 - 8)(0 + 3) ≥ 0

6/(-24) ≥ 0

-1/3 ≥ 0

False

When x = 9 ∈ (8, ∞)

(9 + 6)/(9 - 8)(9 + 3) ≥ 0

15/12 ≥ 0

True

So, the solution is (-6, -3) U (8, ∞).

Example 5 :

-10/(x - 5) ≥ -11/(x - 6)

Solution :

-10/(x - 5) ≥ -11/(x - 6)

-10/(x - 5) + 11/(x - 6) ≥ 0

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

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

(5 + x)/(x - 6)(x - 5) ≥ 0

Equating each factor to 0, we get

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

Intervals are (-∞, -5) (-5, 5), (5, 6) and (6, ∞).

When x = -6 ∈ (-∞, -5)

(5 - 6)/(-6 - 6)(-6 - 5) ≥ 0

-1/110 ≥ 0

False

When x = 0 ∈ (-5, 5)

(5 + 0)/(0 - 6)(0 - 5) ≥ 0

5/30 ≥ 0

True

When x = 5.2 ∈ (5, 6)

(5 + 5.2)/(5.2 - 6)(5.2 - 5) ≥ 0

10.2/(-0.8)(0.2) ≥ 0

False

When x = 7 ∈ (6, ∞)

(5 - 7)/(7 - 6)(7 - 5) ≥ 0

2/1(2) ≥ 0

True

So, the solutions are (-5, 5) and (6, ∞).

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