Mathematics

Inequalities

Posted by Asok K Nadhani on

Inequality

Inequalities or inequations are statements which shows an unequal relationship between any two or more given quantities

\displaystyle Ex.\text{ }4x+15\text{ }\le \text{ }9,\text{ }3x+12y\ge 5.

Inequalities are often applied in business to solve problems where there are upper & lower limits (like Demand, Supply, Production), within the prevailing constraints.

Example : A Company produces 2 types goods. Grade 1 of X tons & Grade 2 of Y tons. However, the Company cannot produce more than 10 tons of goods. This can be represented as \displaystyle x+y\le 10.

Rules of Linear Inequalities

The rules for inequations are similar to those for equations except for multiplying or dividing by a negative number, using one or more of the following methods

  • add the same number or expression to both sides.
  • deduct the same number or expression from both sides.
  • multiply  both sides by the same positive number.
  • divide both sides by the same positive number.
  • multiply or divide by the same negative number, and change the inequation direction (the inequality symbol (<) changes to (>) and vice versa)

Ex.

  • If x < 3, then – x > –3                               (Multiplying by – 1 )
  • \displaystyle 3x1\ge 15,\text{ }then\text{ }4\left( {3x1} \right)~\le 60~~~      (Multiplying by – 4 )
  • \displaystyle \text{ }6x\le 18,\text{ }then\text{ }x\ge 3~~~                           (Dividing by – 6 )

So, symbol of an inequation should be reversed, when multiplying or dividing by a negative number.

Number Systems

The Number System includes Natural Numbers, Whole Numbers and Integers

Natural Numbers

Natural Numbers (denoted by a symbol ‘N’) are obtained by adding one to the previous number. So, it is a set of infinite  number of them.

It is normally expressed as N = {1, 2, 3,………..}.

Whole Numbers

The set of Natural numbers do not include zero. The set of whole number contains zero also.

The set of whole numbers is denoted by ‘W’ and is expressed as:   W = {0, 1, 2, 3, ………….}                                                                        

Integers

The set of whole numbers does not include negative numbers. The set of Integers, denoted by the symbol ‘I’. contains negative whole numbers also and is represented as

 I = {…..−3, −2, −1, 0, 1, 2, 3, …….}

This set of Integers (I) is super set of Natural (N) and Whole (W) numbers, as it covers more number elements. So set of Integers include the set of Natural Numbers and Whole Numbers.

Inequality : Problems

Inequality  Problems and solutions

Ex.1 : Solve the inequality \displaystyle 4-2x\ge x12\text{ },x\in N.

\displaystyle 42x\ge x12~

Step 1: \displaystyle 42x4\ge x124\text{ }[\text{ }x\in N\text{ }Means\text{ }x\text{ }is\text{ }natural\text{ }number\text{ }\left( {Subtract\text{ }4\text{ }} \right)]

Step 2 : – 2x\displaystyle \ge   x – 16 

Step3 : 1–3x \displaystyle \ge – 16 ((Subtract x )

Step 4 : x<16/3 (Divide by – 3 and reverse the symbol)

As \displaystyle x\in N, the solution set is { 1, 2, 3, 4, 5 } [ because 5 is less than 16/3, satisfying the condition, but the number element 6 …onwards does not include in the set, as it is greater than 16/3)

Ex 2: Solve 25 – 3(2a – 5 ) < 21 where a is  a positive integer.

25 – 3(2a – 5 ) < 21

Step 1: 25 – 6a + 15 < 21   (simplifying the expression)

Step 2: 40 – 6a  < 21                            

Step 3: 40 – 6a – 40 < 21 – 40    (eliminate constant values from LHS to solve a)

Step 4: – 6a <  – 19

a>19/6 (divide by -1 and change the symbol)

 As a is a positive integer, the solution set is { 4, 5, 6, ……. }. [ because 3 is less than 19/3, not satisfying the condition, but the number element 4 …onwards satisfy the condition. So included in the set)

Ex. 3 : Solve 14 + 2a > 4 where a is a negative integer.

14 + 2a > 4

Step 1: 14 + 2a – 14 > 4 –14 (subtract 14)

Step 2: 2a > – 10

a > – 5  (Divide by 2)

 As a is a negative integer, the solution set is  { – 4, -3, – 2, – 1 }. [ because -4 is greater than -5, satisfying the condition, a belongs to the set of negative integers for -5 onwards to -1)

Ex. 4 : Solve the inequation \displaystyle \frac{{\left( {\mathbf{2a}\mathbf{1}} \right)}}{4}\le \mathbf{5},\text{ }\mathbf{a}\in \mathbf{W}.

\displaystyle \frac{{\left( {\mathbf{2a}\mathbf{1}} \right)}}{4}\le \mathbf{5}

Step 1: 2a – 1 \displaystyle \le 20 (multiply by 4)

Step 2: 2a – 1 + 1 \displaystyle \le 20 + 1   (add 1)

2a  \displaystyle \le 21

a \displaystyle \le 10.5

As a  \displaystyle \in W, the solution set is { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 }.

As a is Whole Number (a \displaystyle \in W.), the set contains whole numbers upto 10 (which is less than 10.5)

Graphical Solution of linear inequation

Steps of  graphical solution of an inequation of the form ax + by + c \displaystyle \ge 0 or ax +by + c \displaystyle \le 0

  • Replace the inequality sign ‘\displaystyle \ge ‘ or ‘ \displaystyle \le ‘ by equality to get the equation ax + by =    c,    which represents a straight line in xy plane.
  • Draw the graph of the straight line ax + by =c. finding any two  points on the line and drawing a straight line through them. Preferably, choose the two points on an axis, i.e. y = 0 in the equation to get the   point,   where the line meets the x –axis and put x = 0 in the equation to get the point,  where the   line meets the y – axis.
  • If the inequality is ‘\displaystyle \le ‘ or ‘ \displaystyle \ge ‘ the points on this line are included and the line drawn is thick.
  • If the inequality is ‘<‘ or ‘ > ‘ the points on this line are excluded and the line drawn is  dotted.
  • The line divides the XOY plane in two half planes or regions. To determine    which    region satisfies the inequation :

1. Select an arbitrary point and plot it. It will be in either of the regions. 2. If the point satisfies the given inequation, the region where the point lies is the desired region. 3. If the point does not satisfy the given inequation, the region where the point lies is the desired region. 4. Shade the relevant region.

If the graph of the corresponding equation does not pass through the origin, the  origin (0, 0) is the most convenient point to select.

Linear inequation – Problems

Linear Inequation-   Graphical solutions

Ex. Solve  3a + 5b < 15 graphically.

Step 1 : 3a + 5b = 15 (Replacing the inequality sign by equality)

Step 2 : 5b = 15 (put a=0). Or b=3. Put b=0, you get 3a=15, or a=5. So the line passes through the points (0, 3) and (5, 0 )

The graph of 3a + 5b < 15 is the line AB.Step 3 : Putting  a= 0, b = 0 in 3a + 5b <15, we get 0 <15, which is true.

Region Containing the point (0, 0 ) is the solution set of the given inequation. Shade the portion which contain the origin, i.e. the point (0, 0).

All points in the shaded region satisfy the given inequation and form the solution set.  But the points on the graph line are not included  in the solution set. That is why the graph line is shown dotted.

Linear inequation – Problems

Linear Inequation-   Graphical solutions

Ex. Solve graphically :  \displaystyle \mathbf{3a}+\mathbf{4b}\ge \mathbf{12},~\mathbf{a}\ge \mathbf{1},\text{ }\mathbf{b}\ge \mathbf{2}

Step 1 : 3a + 4b = 12 (Replacing the inequality sign by equality)
Step 2 : 4b = 12 (put a=0). Or b=3. Put b=0, you get 3a=12, or a=4. So the line passes through the points A (0, 3) and B (4, 0 )
The graph of 3a + 4b > 12 is the line AB.

Step 3 : Putting  a= 0, b = 0. The graph of 3a + 4b =12 is the line AB which passes trough A (0, 3) and B (4, 0). The graph of a = 1 is a line a parallel to y – axis and the graph of b =2 is a line parallel to x axis.

Putting a = 0, b = 0 in 3a + 4b \displaystyle \ge 12, we get 0 \displaystyle \ge 12, which is false. Hence, the solution set of 3a + 4b \displaystyle \ge 12 is non – origin side of the line including the points on AB.

Step 4 : A \displaystyle \ge 1 represents the region on the right side of the line a =1 including the points on the line a = 1

Step 5  : b\displaystyle \ge 2 represents the region on and above the line b =2.

The common region is the shaded region. Any point in this region is required solution of the system of linear inequations.

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