SOLVING RATIONAL INEQUALITY EXAMPLES
About "Solving rational inequality examples"
Solving rational inequality examples :
Here we are going to see how to solve rational inequalities.
(i) (ax+b)/cx + d) > 0
(ii) (ax+b)/cx + d) < 0
(iii) (ax+b)/cx + d) ≥ 0
(iv) (ax+b)/cx + d) ≤ 0
If the given question is in one of the above forms, we may follow the steps given below to solve the inequality.
If we have any numerical value other than zero on the right, we have to subtract that particular number on both sides in order to get zero on the right side.
Step 1 :
Make sign of coefficient of x as positive, if they are not.
Step 2 :
Equate the numerator and denominator to zero and find the values of x. These values are known as critical number.
Step 3 :
Draw the number line and plot critical numbers.
Step 4 :
These critical numbers will divide the number line into three regions.
Step 5 :
In the right most region the expression on L.H.S of the equation obtained in the step 4 will be positive and other regions will be alternatively negative and positive in other regions.
Step 6 :
- If the given question has the inequalities < or ≤, we have to choose the intervals in the negative region.
- If the given question has the inequalities > or ≥, we have to choose the interval in the positive region.
Let us look into some example problems to understand the above concept.
Example 1 :
Solve the following linear inequations
(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).
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