ICSE Class 10 Maths Chapter 04 Linear Inequations

Unit 2: Algebra

Chapter 4: Linear Inequations (In One Variable)

4.1 Introduction

If x and y are two quantities; then both of these quantities will satisfy any one of the following four conditions (relations):

i.e. either (i) x > y (ii) x >= y (iii) x < y or (iv) x <= y

Each of the four conditions, given above, is an inequation.

In the same way, each of the following also represents an inequation:

x < 8, x >= 5, -x + 4 <= 3, x + 8 > 4, etc.

4.2 Linear Inequations In One Variable

If a, b and c are real numbers, then each of the following is called a linear inequation in one variable:

(i) ax + b > c. Read as: ax + b is greater than c.

(ii) ax + b < c. Read as: ax + b is less than c.

(iii) ax + b >= c. Read as: ax + b is greater than or equal to c.

(iv) ax + b <= c. Read as: ax + b is less than or equal to c.

In an inequation, the signs '>', '<', '>=' and '<=' are called signs of inequality.

4.3 Solving A Linear Inequation Algebraically

To solve a given linear inequation means to find the value or values of the variable used in it.

Thus; (i) to solve the inequation 3x + 5 > 8 means to find the variable x.

(ii) to solve the inequation 8 - 5y <= 3 means to find the variable y and so on

The following working rules must be adopted for solving a given linear inequation:

Rule 1: On transferring a positive term from one side of an inequation to its other side, the sign of the term becomes negative.

e.g. 2x + 3 > 7 - 2x > 7 - 3, 5x + 4 <= 15 - 5x <= 15 - 4, 23 >= 4x + 15 - 23 - 15 >= 4x and so on.

Rule 2: On transferring a negative term from one side of an inequation to its other side, the sign of the term becomes positive.

e.g. 2x - 3 > 7 - 2x > 7 + 3, 5x - 4 <= 15 - 5x <= 15 + 4, 23 >= 4x - 15 - 23 + 15 >= 4x and so on.

Rule 3: If each term of an inequation be multiplied or divided by the same positive number, the sign of inequality remains the same.

That is, if p is positive

(i) x < y - px < py and \(\frac{x}{p} < \frac{y}{p}\)

(ii) x > y - px > py and \(\frac{x}{p} > \frac{y}{p}\)

(iii) x <= y - px <= py and \(\frac{x}{p} <= \frac{y}{p}\)

and, (iv) x >= y - px >= py and \(\frac{x}{p} >= \frac{y}{p}\)

Thus, x <= 6 - 4x <= 4 x 6, x >= 5 - 3x >= 3 x 5, x <= 2 - \(\frac{x}{10} <= \frac{2}{10}\) and so on.

Rule 4: If each term of an inequation be multiplied or divided by the same negative number, the sign of inequality reverses.

That is, if p is negative

(i) x < y - px > py and \(\frac{x}{p} > \frac{y}{p}\)

(ii) x >= y - px <= py and \(\frac{x}{p} <= \frac{y}{p}\)

Thus, x <= 6 - -4x >= -4 x 6, x > 5 - -3x < -3 x 5 x >= y - \(\frac{x}{-2} <= \frac{y}{-2}\) and so on.

Rule 5: If sign of each term on both the sides of an inequation is changed, the sign of inequality gets reversed.

i.e. (i) -x > 5 - x < -5 (ii) 3y <= 15 - -3y >= -15

(iii) -2y < -7 - 2y > 7 and so on.

Rule 6: If both the sides of an inequation are positive or both are negative, then on taking their reciprocals, the sign of inequality reverses.

i.e. if x and y both are either positive or both are negative, then

(i) x > y - \(\frac{1}{x} < \frac{1}{y}\) (ii) x <= y - \(\frac{1}{x} >= \frac{1}{y}\)

(iii) x >= y - \(\frac{1}{x} <= \frac{1}{y}\) and so on.

4.4 Replacement Set And Solution Set

The set, from which the value of the variable x is to be chosen, is called replacement set and its subset, whose elements satisfy the given inequation, is called solution set.

e.g. Let the given inequation be x < 3, if:

(i) the replacement set = N, the set of natural numbers; the solution set = {1, 2}

(ii) the replacement set = W, the set of whole numbers; the solution set = {0, 1, 2}

(iii) the replacement set = Z or I, the set of integers; the solution set = {..........., -2, -1, 0, 1, 2}

But, if the replacement set is the set of real numbers, the solution set can only be described in set-builder form, i.e. {x : x in R and x < 3}.

Problem 1: If the replacement set is the set of natural numbers (N), find the solution set of: (i) 3x + 4 < 16 (ii) 8 - x <= 4x - 2.

Solution:

(i) 3x + 4 < 16

- 3x < 16 - 4 [Using rule 1]

- 3x < 12

- \(\frac{3x}{3} < \frac{12}{3}\) [Using rule 3]

i.e. x < 4

Since, the replacement set = N (set of natural numbers)

- Solution set = {1, 2, 3} Ans.

(ii) 8 - x <= 4x - 2 - -x - 4x <= -2 - 8 [Using rule 1]

- -5x <= -10

- \(\frac{-5x}{-5} >= \frac{-10}{-5}\) [Using rule 4]

i.e. x >= 2

Since, the replacement set = N

- Solution set = {2, 3, 4, 5, 6, ...........}

Alternative method:

8 - x <= 4x - 2 - 4x - 2 >= 8 - x [x <= y and y >= x mean the same]

- 4x + x >= 8 + 2 [Using rule 2]

- 5x >= 10

- x >= 2 and x in N

- Solution set = {2, 3, 4, 5, 6, ...........} Ans.

Teacher's Note

Linear inequations help us compare quantities in real situations like budgeting (spending less than available money) or setting score requirements (achieving at least a minimum grade). Understanding solution sets connects abstract math to practical decision-making.

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