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# INEQUATIONS(Inequalities) Part 1

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INEQUATIONS (INEQUALITIES): Prologue

During GMAT exam, where time is precious, complicated inequalities often create a difficult situation. Purpose of this article is to narrate the paradigms of inequalities and to state the methods to solve the inequalities.

This Article is actually a digital version of my handwritten notes collected from various sources. Most of the input is from Indian School Text Books on Maths(CBSE Board) and remaining is from GMAT Math Strategy Guides of Elite Test Prep Companies and Guides for Indian CAT.

INEQUATIONS : Definition and Characterictics

The term Equation is defined as a statement involving variable(s) and the sign of equality(=)
Similarly, we can define the term Inequation as a statement involving variable(s) and the sign of inequality viz, >, ≥, <, or ≤
An Inequation may contain one or more variables. Also, it may be linear or quadratic or cubic etc.

Some Examples of Inequations

5x – 3 > 0

$$x^2$$ - 5x + 4 ≤ 0

$$ax^2$$ + $$by^2$$ < 1

$$x^3$$ + $$6x^2$$ + 11x + 6 ≤ 0

Linear Inequations with one variable

Let a be a non-zero real number and x be a variable.

Then Inequations of the form ax + b < 0, ax + b ≤ 0, ax + b > 0, ax + b ≥ 0 are known as linear Inequations in one variable

Examples : 9x – 15 > 0, 2x – 3 ≤ 0

Linear Inequations with two variables

Let a, b be non-zero real numbers and x, y be a variables.

Then Inequations of the form ax + by < c, ax + by ≤ c, ax + by > c, ax + by ≥ c are known as linear Inequations in two variables x and y

Examples : 2x + 3y ≤ 6, 2x + y ≥ 6

Let a be a non-zero real number.

Then Inequations of the form $$ax^2$$ + bx + c < 0, $$ax^2$$ + bx + c ≤ 0, $$ax^2$$ + bx + c > 0, $$ax^2$$ + bx + c ≥ 0 are known as Quadratic Inequations with one variable.

Examples : $$x^2$$ + x - 6 < 0, $$2x^2$$ + 3x + 1 > 0

The equation $$ax^2$$ + $$by^2$$ < 1 is a sample quadric Inequation with two variables.

Solution of an Inequation

Definition :- A Solution of an Inequation is the value(s) or the range of values of the variable(s) that makes the inequation a true statement

Consider the Inequation $$\frac{(3-2x)}{5}$$ < $$\frac{x}{3}$$ – 4

For x = 9, we have, LHS = $$\frac{(3-2*9)}{5}$$ = -3 and, RHS = $$\frac{9}{3}$$ - 4 = -1. Clearly -3 < -1 that means LHS < RHS, which is true. So x = 9 is a solution of the given inequation

For x = 6, LHS = $$\frac{(-9)}{5}$$ and, RHS = -2 ---------> LHS > RHS ---------> So x = 6 is not a solution of the given Inequation.

We can verify that any real number greater than 7 is a solution of the given Inequation.

Let us now consider the Inequation $$x^2$$ + 1 < 0

We know that $$x^2$$ ≥ 0 for all real values of x

$$x^2$$ +1 ≥ 1 for all real values of x--------------(Added 1 to both sides of ineqation)

So, there is no real value of x which makes the Inequation $$x^2$$ + 1 < 0 a true statement. Hence, it has no solution.

Solving an Inequation

It is the process of obtaining all possible solutions of an Inequation. Substitution by numbers is a lengthy process, and it is an impossible process for complicated quadratic equations. Best way to solve Inequations is shown in the following section.

Solution Set

The set of all possible solutions of an Inequation is known as its solution set.

For Example
The solution set of the Inequation $$x^2$$ + 1 ≥ 0 is the set of all real numbers
The solution set of the Inequation $$x^2$$ + 1 < 0 is empty/null set

Solving Linear Inequation with one variable

As mentioned earlier that solving an Inequation is the process of obtaining all possible solutions of the variable.
In the process of solving an Inequation, we use mathematical simplifications which are governed by the following rules

Rule 1 :- Same Number may be added to (or subtracted from) both sides of an Inequation without changing the sign of inequality. Example: x > 5 ---------> x + 2 > 5 + 2

Rule 2 :- Both sides of an Inequation can be multiplied (or divided) by the same positive real number without changing the sign of inequality. However, the sign of inequality is reversed when both sides of an Inequation are multiplied or divided by a negative number. Examples: x > 5 ----> 3x > 15; x > 1 ----> -x < -1

Rule 3 :- Any term of an Inequation may be taken to the other side with its sign changed without affecting the sign of inequality. Example: $$\frac{x}{3}$$ - 2 < $$\frac{x}{6}$$ -----> $$\frac{x}{3}$$ - $$\frac{x}{6}$$ < 2

Rule 4 :-
In inequalities never multiply for a term unless you know it does not equal 0 AND you know its sign.

Type I :- Algorithm to solve the Inequations of the form ax + b < 0, ax + b ≤ 0, ax + b > 0, ax + b ≥ 0

Step I :- Obtain the linear Inequation.
Step II :- Collect all terms involving the variable on one side of the Inequation and the constant terms on other side
Step III :- Simplify both sides of inequality in their simplest forms to reduce the Inequation in the form ax < b, or ax ≤ b, or ax > b, or ax ≥ b
Step IV :- Solve the Inequation obtained in step III by dividing both sides of the Inequation by the coefficient of the variable (of x)
Step V :- Write the solution set obtained in step IV in the form of an interval on the real line.

Example 1 :- Solve the Inequation 2x – 4 ≤ 0

Step 1) 2x – 4 ≤ 0
Step 2) (2x – 4) + 4 ≤ 0 + 4 --------------------------[ Adding 4 on both sides ]
Step 3) 2x ≤ 4
Step 4) $$\frac{2x}{2}$$ ≤ $$\frac{4}{2}$$
Step 5) x ≤ 2

Any real number less than or equal to 2 is a solution of the given Inequation.

Hence the solution set of the given Inequation is (-∞, 2]

NOTE :- The Interval [x, y] stands for the values between x and y, both inclusive, where as The interval (x, y) stands for the values between x and y, both exclusive

Example 2 :- -3x + 12 < 0

Step 1) -3x + 12 < 0

Step 2) -3x < -12 --------------------------[ Transposing 12 on right side ]

Step 3) $$\frac{-3x}{-3}$$ > $$\frac{-12}{-3}$$ --------------------------[ Dividing both sides by -3, and hence the sign of inequality changes ]

Step 4) x > 4

Any real number greater than 4 is a solution of the given Inequation.

Hence the solution set of the given Inequation is (4,∞)

Example 3 :- $$\frac{1}{(x-2)}$$ < 0

$$\frac{1}{(x-2)}$$ < 0 ---------> x-2 < 0 -----------------[ If $$\frac{a}{b}$$ < 0 and a > 0 then b < 0]

x < 2 ----------> Solution set is ( -∞, 2)

Type II :- Algorithm to solve the Inequations of the form $$\frac{(ax+b)}{(cx+d)}$$ < k, $$\frac{(ax+b)}{(cx+d)}$$ ≤ k, $$\frac{(ax+b)}{(cx+d)}$$ > k, $$\frac{(ax+b)}{(cx+d)}$$ ≥ k

Step I :- Obtain the Inequation.

Step II :- Transpose all terms on LHS

Step III :- Simplify LHS of the equation obtained in step II to obtain an Inequation of the form $$\frac{(px+q)}{(rx+s)}$$ < 0, $$\frac{(px+q)}{(rx+s)}$$ ≤ 0, $$\frac{(px+q)}{(rx+s)}$$ > 0, $$\frac{(px+q)}{(rx+s)}$$ ≥ 0

Step IV :- Make Coefficient of x positive in numerator and denominator if they are not.

Step V :- Equate Numerator and denominator separately to zero and obtain the values of x. These values of x are generally called critical points.

Step VI :-
Plot the critical points obtained in step V on real line. These points will divide the real line In three regions.

Step VII :-
In the rightmost region the expression on LHS of the Inequation obtained in step IV will be Positive and in other regions it will be alternatively negative and positive. So, mark positive Sign in the right most region and then mark alternatively negative and positive signs in Other regions.

Step VIII :-
Select Appropriate region on the basis of the sign of the information obtained in step IV. For +ve inequality signs (>, ≥), the range will be the region with +ve signs, i.e., the rightmost segment and leftmost segment. For -ve inequality signs (<, ≤), the range will be the region with -ve signs, i.e., the range between critical points. Write these regions in the form of intervals to obtain the desired solution sets of the given Inequation.

Example 1 :- $$\frac{(x-2)}{(x+5)}$$ > 2

Step I :- $$\frac{(x-2)}{(x+5)}$$ > 2

Step II :- $$\frac{(x-2)}{(x+5)}$$ - 2 > 0

Step III :- $$\frac{(-x-12)}{(x+5)}$$ > 0

Step IV :- $$\frac{(x+12)}{(x+5)}$$ < 0 ---------------[Multiplying by -1 to make coefficient of x positive in the expression in numerator]

Step V :- x+12 < 0 -----> x < -12 and x+5 < 0 --------> x < -5

Therefore x = -12, -5 are the critical points.

Step VI and Step VII :-

Step VIII:- The Real line is divided in to three regions and the signs of LHS of Inequation (step IV) marked. Since the Inequation in (step IV) possesses less than sign which means that LHS of the Inequation is negative. So, the solution set of the given Inequation is the union of the regions containing negative sign.

Hence the solution set of given Inequation is -12 < x < -5

Example 2 :- $$\frac{(x-3)}{(x-5)}$$ > 0

$$\frac{(x-3)}{(x-5)}$$ > 0

x-3 > 0 ------> x > 3 --------------------[ Equating x-3 to zero]

x-5 > 0 ------> x > 5 --------------------[ Equating x-5 to zero]

Upon equating x-3 and x-5 separately to zero, we obtain x = 3, 5 as critical points.

Plot these points on Real Line

The Real line is divided into three regions. Since the Inequation possesses greater than sign which means that LHS of the Inequation is positive. So, the solution set of the given Inequation is the union of the regions Containing positive sign.

Hence x > 5 or x < 3 -----------> (-∞, 3) U (5, ∞)

Example 3 :- $$\frac{(2x+4)}{(x-1)}$$ ≥ 5

$$\frac{(2x+4)}{(x-1)}$$ ≥ 5

$$\frac{(2x+4)}{(x-1)}$$ - 5 ≥ 0

$$\frac{((2x+4)-5(x-1))}{(x-1)}$$ ≥ 0

$$\frac{(-3x+9)}{(x-1)}$$ ≥ 0

$$\frac{(3x-9)}{(x-1)}$$ ≤ 0 ----------------[ Multiplying both sides by -1]

$$\frac{(3(x-3))}{(x-1)}$$ ≤ 0

$$\frac{(x-3)}{(x-1)}$$ ≤ 0 -----------------[ dividing both sides on Inequation by 3]

x = 3, 1

Sign of original Inequation obtained in Step 4 is negative so the solution set will be union of the regions containing negative sign.

Therefore Solution Set is 1 < x ≤ 3

NOTE :-
When x=1, the Inequation becomes $$\frac{6}{0}$$ ≥ 5 and we know that division by zero is not defined, hence we need to exclude x=1 from solution set.

Example 4 :- $$\frac{(x+3)}{(x-2)}$$ ≤ 2

$$\frac{(x+3)}{(x-2)}$$ ≤ 2

$$\frac{(x+3)}{(x-2)}$$ - 2 ≤ 0

$$\frac{(x+3-2x+4)}{(x-2)}$$ ≤ 0

$$\frac{(-x+7)}{(x-2)}$$ ≤ 0

$$\frac{(x-7)}{(x-2)}$$ ≥ 0 -------------------[Multiplying both sides by -1]

x-7 ≥ 0 ---------> x ≥ 7

x-2 ≥ 0 ---------> x ≥ 2

x = 7 and 2 are the critical points.

The Real line is divided into three regions. Since the Inequation (obtained in step 4) possesses greater than sign which means that LHS of the Inequation is positive. So, the solution set of the given Inequation is the union of the regions containing positive sign.

Hence x ≥ 7 or x ≤ 2 -----------> (-∞, 2] U [7, ∞)

In order this Article not be dense with too much information, we have decided to split the same in two parts. This is the first Part of Article. Second Part contains two remaining concepts which are 1) Modules - Absolute Value 2) Quadratic Inequalities.
Here is the link to Second Part of the Article inequations-inequalities-part-154738.html#p1238488.

Courtesy for the Information

Prof. Dr. R. D. Sharma
B.Sc. (Hons) (Gold Medalist), M.Sc. (Gold Medalist), Ph.D. in Mathematics
Author of CBSE Math Books

Mr. Arun Sharma
Aluminus – IIM Bangalore

Special Thanks to

BUNUEL, doe007, Zarrolou

Regards,

Narenn

---------- End of First Part ------------
Attachments

INEQUATIONS Part 1.pdf [160.45 KiB]

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Last edited by Bunuel on 02 Jul 2013, 12:20, edited 3 times in total.
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01 Jul 2013, 08:42
Narenn wrote:
Rule 4 :- In inequalities never multiply for a term unless you know it does not equal 0 AND you know its sign.

This is really important and it is a common mistake of test takers, especially in data sufficiency. Here is a really simple problem to demonstrate this:

Is x > 3?
1) xy > 3y
2) y > 1

If a test-taker divides each side of the inequality in statement 1 by y, then the result is x > 3. This appears to be sufficient to answer the question, but it is not because of your rule 4. y may not be a positive number, in which case we would have to switch the sign if we divide both sides by y. Instead, one should perform a bifurcated analysis in working statement 1, as below.

Statement 1:
Case 1: y > 0
This results in x > 3 because we can divide both sides by a non-zero y and we don't have to change the sign because y is positive

Case 2: y < 0
This results in x < 3 because we can divide both sides by a non-zero y and we have to change the sign because y is negative

Since x > 3 and x < 3 are each possible, this is enough to prove the insufficiency of statement 1, so we don't have to try the case of y = 0. In fact, y = 0 is not possible because then we would have 0 > 0

Statement 2 forces y to be a positive number. This is not enough by itself because we don't know x.

When we combine both statements, we are only left with case 1 (y > 0). This results in x > 3. Therefore, the answer is C.

Another way to avoid this trap is to choose numbers, put them into the equation, and then observe the outcome. When working with inequalities, be especially prone to try 0, 1, 2, -1, -2, 1/2, -1/2 as these will cover most outcomes. Choosing y = 1 and y = -1 disproves statement 1 by itself.
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12 Aug 2013, 15:49
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"Hence x ≥ 7 or x ≤ 2 -----------> (-∞, 2] U [7, ∞)"

x ≤ 2 should be x < 2, not equal to 2 else denom will become 0.
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21 Jul 2014, 13:17
Narenn wrote:

Example 3 :- $$\frac{(2x+4)}{(x-1)}$$ ≥ 5

$$\frac{(2x+4)}{(x-1)}$$ ≥ 5

$$\frac{(2x+4)}{(x-1)}$$ - 5 ≥ 0

$$\frac{((2x+4)-5(x-1))}{(x-1)}$$ ≥ 0

$$\frac{(-3x+9)}{(x-1)}$$ ≥ 0

$$\frac{(3x-9)}{(x-1)}$$ ≤ 0 ----------------[ Multiplying both sides by -1]

$$\frac{(3(x-3))}{(x-1)}$$ ≤ 0

$$\frac{(x-3)}{(x-1)}$$ ≤ 0 -----------------[ dividing both sides on Inequation by 3]

x = 3, 1

Sign of original Inequation obtained in Step 4 is negative so the solution set will be union of the regions containing negative sign.

Therefore Solution Set is 1 < x ≤ 3

NOTE :-
When x=1, the Inequation becomes $$\frac{6}{0}$$ ≥ 5 and we know that division by zero is not defined, hence we need to exclude x=1 from solution set.

Great work!

Shouldn't the highlighted portion be 1 ≤ x ≤ 3
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22 Jul 2014, 00:40
Game wrote:
Example 3 :- $$\frac{(2x+4)}{(x-1)}$$ ≥ 5

$$\frac{(2x+4)}{(x-1)}$$ ≥ 5

Great work!

Shouldn't the highlighted portion be 1 ≤ x ≤ 3

No, because (x-1) in denominator will yield zero at X=1 and answer undefined.
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08 Nov 2014, 19:36
PiyushK wrote:
Game wrote:
Example 3 :- $$\frac{(2x+4)}{(x-1)}$$ ≥ 5

$$\frac{(2x+4)}{(x-1)}$$ ≥ 5

Great work!

Shouldn't the highlighted portion be 1 ≤ x ≤ 3

No, because (x-1) in denominator will yield zero at X=1 and answer undefined.

Example 4 :- \frac{(x+3)}{(x-2)} ≤ 2

Why based on the same concept, 2 has not been excluded in example 4? In this example also, 2 in the denominator will result the expression to undefined
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25 Jan 2015, 22:44
PiyushK wrote:
Game wrote:
Example 3 :- $$\frac{(2x+4)}{(x-1)}$$ ≥ 5

$$\frac{(2x+4)}{(x-1)}$$ ≥ 5

Great work!

Shouldn't the highlighted portion be 1 ≤ x ≤ 3

No, because (x-1) in denominator will yield zero at X=1 and answer undefined.

How come when you multiply the equation by (x-1), you don't eliminate it from the denominator?
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