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Algebraic Solution of Inequality of One Variable and their Graphical Representation

Let us consider the inequality stands for the number of pens that can be bought for Rs. 100/- given each pen costs Rs. 6/-

Obviously cannot be negative integer or a fraction.

0, 1, 2...16 satisfy the given inequality, but 17 onwards, the values do not satisfy the inequality.
Solution set is {0, 1, 2, 3 ...16}

Thus, any solution of an inequality in one variable is a value of the variable which makes it a true statement.
But it is not always to go on trying the solution one by one. We shall use some rules for solving the inequalities.

  1. Equal numbers can be added (or subtracted from) both sides of an equation.
  2. Both sides of an equation can be multiplied or divided by the same non-zero number.
The same rules are applicable in the case of inequalities except that in rule (2) we have to observe that if we multiply (or divide) both sides by a negative number, then the inequality sign changes.

Example to illustrate the above point

(Change of sign on both sides changes the inequality)
-6<-4, Multiply by -3 both sides, we get (-6)(-3)>(-4)(-3)
18>12 (which is true)

Let us solve some inequalities

Example 1:
Solve: is an integer is a real number


Note: When '-2' is not included in the solution set use simple bracket at that end.

Example 2:


Note: When (-3) is also included in the solution set as in the above case, use []-square brackets, indicating it is a closed interval, where the end-point is also included.


​Example 3:


Example 4:


Example 5:


The sum could also be done differently from step (1)
Changing signs both sides & changing inequality,

Example 6:
Solve graphically (Solution on number line) for real values of satisfying.


Note: As 3 is excluded so we put a circle round 3

Example 7:
Solve graphically (Solution on number line) for real values of satisfying


Put a dark circle around '-1'

[-1 is also included in the solution]

Solving word problems using inequalities

Example: 8
Find all pairs of consecutive odd pairs of positive integers which are smaller than 10 such that their sum is more than 11.

Let be the consecutive odd numbers.
given, and

The pairs of numbers are (5,7) and (7,9).

Example: 9
The longest side of a triangle is 3 times the shortest side and the third side is 2cm. Shorter than the longest side. If the perimeter of the triangle is at least 61 cm, find the minimum length of the shortest side.


Let the shortest side be cm.
Then the longest side is cm.
Third side = cm.
The perimeter =

(Minimum value is 9)
 Minimum length of the shortest side = 9 cm

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