LINEAR INEQUATIONS (Including Number Lines) - Q&A
EXERCISE 15
1. Multiple Choice Type:
Choose the correct answer from the options given below.
(i) If 11 + 2x > 5 and x ∈ {negative integers}, then the set to which x belongs is:
(a) {-2, -1}
(b) {-3, -2, -1, .....}
(c) {-3, -2, -1}
(d) {1, 2, 3}
Solution:
Given: 11 + 2x > 5
Subtracting 11 from both sides:
2x > 5 - 11
2x > -6
Dividing by 2:
x > -3
Since x belongs to the set of negative integers, the values greater than -3 are -2, -1.
Solution set = {-2, -1}
Ans. (a) {-2, -1}
(ii) If 3x + 1 ≤ 16 and x ∈ {real numbers}, then the values of x represented on a number line are:
(a) Number line showing values ≤ 5
(b) Number line showing values ≥ 5
(c) ...
(d) ...
Solution:
Given: 3x + 1 ≤ 16
3x ≤ 16 - 1
3x ≤ 15
x ≤ 5
Since x is a real number, the solution represents all real numbers less than or equal to 5. On a number line, this is represented by a solid dot at 5 and a thick line extending towards the left.
Ans. (a) (Referring to the graph showing x ≤ 5)
(iii) If 3 - 2x ≥ x - 10, x ∈ W the solution set is:
(a) {1, 2, 3, 4, .......}
(b) {1, 2, 3, 4, 5}
(c) {0, 1, 2, 3, 4}
(d) {1, 2, 3, 4}
Solution:
Given: 3 - 2x ≥ x - 10
3 + 10 ≥ x + 2x
13 ≥ 3x
13/3 ≥ x
4.33 ≥ x or x ≤ 4.33
Since x ∈ W (Whole numbers), the values are 0, 1, 2, 3, 4.
Ans. (c) {0, 1, 2, 3, 4}
(iv) If x ∈ W and (2x - 1)/3 ≥ 4, the solution set is:
(a) {0, 1, 2, 3, 4, 5...}
(b) {0, 1, 2, 3, 4, 5, 6}
(c) {7, 8, 9, 10, .......}
(d) {7, 8, 9, 10}
Solution:
Given: (2x - 1)/3 ≥ 4
2x - 1 ≥ 12
2x ≥ 13
x ≥ 6.5
Since x ∈ W, the solution set is {7, 8, 9, 10, ...}
Ans. (c) {7, 8, 9, 10, .......}
(v) If x is an integer and -2(x + 3) > 5; the solution set on the number line is:
(a) ...
(b) ...
(c) ...
(d) ...
Solution:
Given: -2(x + 3) > 5
Divide by -2 (sign reverses):
x + 3 < -2.5
x < -2.5 - 3
x < -5.5
Since x is an integer, the values are -6, -7, -8, etc. The number line should show dots on integers to the left of -5.
Ans. (b) (Graph showing integers less than -5.5, i.e., -6, -7...)
2. Solve: 7 > 3x - 8; x ∈ N.
Solution:
7 + 8 > 3x
15 > 3x
5 > x or x < 5
Since x ∈ N (Natural numbers = {1, 2, 3...}),
Solution set = {1, 2, 3, 4}
3. Solve: -17 < 9y - 8; y ∈ Z.
Solution:
-17 + 8 < 9y
-9 < 9y
-1 < y or y > -1
Since y ∈ Z (Integers),
Solution set = {0, 1, 2, 3, ...}
4. Solve: (2/3)x + 8 < 12 ; x ∈ W.
Solution:
(2/3)x < 12 - 8
(2/3)x < 4
2x < 12
x < 6
Since x ∈ W (Whole numbers),
Solution set = {0, 1, 2, 3, 4, 5}
5. Solve the inequation 8 - 2x ≥ x - 5; x ∈ N
Solution:
8 + 5 ≥ x + 2x
13 ≥ 3x
13/3 ≥ x
4.33 ≥ x or x ≤ 4.33
Since x ∈ N,
Solution set = {1, 2, 3, 4}
6. Solve the inequality 18 - 3(2x - 5) > 12; x ∈ W
Solution:
18 - 6x + 15 > 12
33 - 6x > 12
33 - 12 > 6x
21 > 6x
21/6 > x
3.5 > x or x < 3.5
Since x ∈ W,
Solution set = {0, 1, 2, 3}
7. Solve: 4x - 5 > 10 - x, x ∈ {0, 1, 2, 3, 4, 5, 6, 7}.
Solution:
4x + x > 10 + 5
5x > 15
x > 3
Given the replacement set {0, 1, 2, 3, 4, 5, 6, 7}, we select values greater than 3.
Solution set = {4, 5, 6, 7}
8. Solve: 15 - 2(2x - 1) < 15, x ∈ Z
Solution:
15 - 4x + 2 < 15
17 - 4x < 15
17 - 15 < 4x
2 < 4x
2/4 < x
0.5 < x or x > 0.5
Since x ∈ Z,
Solution set = {1, 2, 3, 4, ...}
9. Solve: (2x + 3)/5 > (4x - 1)/2, x ∈ W
Solution:
Cross-multiply (since denominators are positive):
2(2x + 3) > 5(4x - 1)
4x + 6 > 20x - 5
6 + 5 > 20x - 4x
11 > 16x
11/16 > x
0.6875 > x or x < 0.6875
Since x ∈ W,
Solution set = {0}
10. 5x + 4 > 8x - 11, x ∈ Z
Solution:
4 + 11 > 8x - 5x
15 > 3x
5 > x or x < 5
Since x ∈ Z,
Solution set = {..., 2, 3, 4}
11. (2x/5) + 1 < -3; x ∈ R
Solution:
2x/5 < -3 - 1
2x/5 < -4
2x < -20
x < -10
Solution set = {x : x ∈ R, x < -10}
12. x/2 > -1 + 3x/4; x ∈ N
Solution:
Multiply by 4 to clear denominators:
2x > -4 + 3x
4 > 3x - 2x
4 > x or x < 4
Since x ∈ N,
Solution set = {1, 2, 3}
13. (2/3)x + 5 ≤ (1/2)x + 6; x ∈ W
Solution:
(2/3)x - (1/2)x ≤ 6 - 5
(4x - 3x)/6 ≤ 1
x/6 ≤ 1
x ≤ 6
Since x ∈ W,
Solution set = {0, 1, 2, 3, 4, 5, 6}
14. Solve the inequation 5(x - 2) > 4(x + 3) - 24 and represent its solution on a number line. Given the replacement set is {-4, -3, -2, -1, 0, 1, 2, 3, 4}.
Solution:
5x - 10 > 4x + 12 - 24
5x - 10 > 4x - 12
5x - 4x > -12 + 10
x > -2
From the replacement set, values greater than -2 are {-1, 0, 1, 2, 3, 4}.
Number Line Representation: Dots on -1, 0, 1, 2, 3, 4.
15. Solve (2/3)(x - 1) + 4 < 10 and represent its solution on a number line. Given replacement set is {-8, -6, -4, 3, 6, 8, 12}
Solution:
(2/3)(x - 1) < 10 - 4
(2/3)(x - 1) < 6
2(x - 1) < 18
x - 1 < 9
x < 10
From the replacement set, values less than 10 are {-8, -6, -4, 3, 6, 8}.
Number Line Representation: Dots on -8, -6, -4, 3, 6, 8.
Test yourself
1. Multiple Choice Type:
Choose the correct answer from the options given below.
(i) If x is a whole number and 12 - x > 3x - 5; the solution set is:
(a) {1, 2, 3, 4}
(b) {1, 2, 3, 4, 5}
(c) {0, 1, 2, 3, 4, 5}
(d) {0, 1, 2, 3, 4}
Solution:
12 + 5 > 3x + x
17 > 4x
4.25 > x or x < 4.25
Since x is a whole number, solution set is {0, 1, 2, 3, 4}.
Ans. (d) {0, 1, 2, 3, 4,}
(ii) If 5x - 3 ≤ 12 and x ∈ {-5, -3, -1, 1, 3, 5, 7} the solution set is:
(a) {-5, -3, -1, 1, 3}
(b) {-5, -3, -1}
(c) {-5, -3, -1, 1}
(d) {-3, -1, 1, 3, 6}
Solution:
5x ≤ 15
x ≤ 3
From the set {-5, -3, -1, 1, 3, 5, 7}, values ≤ 3 are {-5, -3, -1, 1, 3}.
Ans. (a) {-5, -3, -1, 1, 3}
(iii) If x ∈ {-7, -4, -1, 2, 5} and 25 - 3(2x - 5) < 19 the solution set is:
(a) {-7, -4, -1, 2}
(b) {2, 5}
(c) {0, 1, 2, 3}
(d) {5}
Solution:
-3(2x - 5) < 19 - 25
-3(2x - 5) < -6
Divide by -3 (sign reverses):
2x - 5 > 2
2x > 7
x > 3.5
From the set {-7, -4, -1, 2, 5}, the value greater than 3.5 is {5}.
Ans. (d) {5}
(iv) If x is real number and -4 < x ≤ 0, its solution set on the number line is :
(a) ...
(b) ...
(c) ...
(d) ...
Solution:
The inequality -4 < x ≤ 0 represents real numbers between -4 and 0, excluding -4 (hollow circle) and including 0 (solid dot).
Ans. (b) (Graph showing hollow circle at -4, solid line to 0, solid dot at 0)
(v) If x is an integer and 7 - 4x < 15 the solution set on the number line is:
(a) ...
(b) ...
(c) ...
(d) ...
Solution:
-4x < 15 - 7
-4x < 8
Divide by -4 (sign reverses):
x > -2
Integers greater than -2 are -1, 0, 1, 2, 3...
Ans. (b) (Graph showing dots at -1, 0, 1, 2, 3...)
(vi) Statement 1: A set from which the values of the variables involved in the inequation are chosen is called the solution set.
Statement 2: A linear inequation in one variable (or unknown) has exactly one solution.
Which of the following options is correct?
(a) Both the statements are true.
(b) Both the statements are false.
(c) Statement 1 is true, and statement 2 is false.
(d) Statement 1 is false, and statement 2 is true.
Solution:
Statement 1 is False. The set is called the "Replacement Set". The "Solution Set" is the subset that satisfies the inequation.
Statement 2 is False. A linear inequation usually has a set of solutions (a range), not exactly one.
Ans. (b) Both the statements are false.
The following questions are Assertion-Reason based questions. Choose your answer based on the codes given below.
(1) Both A and R are correct, and R is the correct explanation for A.
(2) Both A and R are correct, and R is not the correct explanation for A.
(3) A is true, but R is false.
(4) A is false, but R is true.
(vii) Assertion (A): The solution set for: x + 5 ≤ 10, if the replacement set is {x | x ≤ 5, x ∈ W} is {0, 1, 2, 3, 4, 5}.
Reason (R): The set of elements of the replacement set which satisfy the given inequation is called the solution set.
Solution:
Inequation: x ≤ 5.
Replacement set elements (Whole numbers ≤ 5): {0, 1, 2, 3, 4, 5}.
All these elements satisfy x ≤ 5. So Solution set is {0, 1, 2, 3, 4, 5}. Assertion is True.
Reason is the correct definition and explains why we selected those elements. Reason is True and explains A.
Ans. (a) (1)
(viii) Assertion (A): The solution set for: x + 3 ≥ 15 is φ, if the replacement set is {x | x < 10, x ∈ N}
Reason (R): If we change over the sides of an inequality, we must change the sign from < to > or > to < or ≥ to ≤ or ≤ to ≥.
Solution:
Assertion: x ≥ 12. Replacement set is {1, 2, ..., 9}. No value satisfies x ≥ 12. Solution set is empty (φ). A is True.
Reason: Swapping sides (e.g., 3 < x becomes x > 3) requires changing the sign. R is True.
However, R is not the explanation for why the solution set is empty in A. The emptiness is due to the values in the replacement set, not algebraic manipulation rules.
Ans. (b) (2)
(ix) Assertion (A): x < -2 and x ≥ 1 Solution set S = {x | -2 < x ≤ 1, x ∈ R}.
Reason (R): Two inequations can be written in a combined expression.
Solution:
Assertion: "and" implies intersection. There are no numbers that are less than -2 AND greater than or equal to 1. The solution set is Empty. The expression given in A represents numbers between -2 and 1, which contradicts the condition. A is False.
Reason: True.
Ans. (d) (4)
(x) Assertion (A): The solution set in the system of negative integers for: 0 > -4 - p is {-3, -2, -1}.
Reason (R): If the same quantity is subtracted from both sides of an inequation, the symbol of inequality is reversed.
Solution:
Assertion: 4 > -p => -4 < p => p > -4. Negative integers greater than -4 are -3, -2, -1. A is True.
Reason: Subtracting quantities does NOT reverse the symbol. R is False.
Ans. (c) (3)
2. If the replacement set is the set of natural numbers, solve :
(i) x - 5 < 0
Solution:
x < 5
x ∈ N, so Solution set = {1, 2, 3, 4}
(ii) x + 1 ≤ 7
Solution:
x ≤ 6
x ∈ N, so Solution set = {1, 2, 3, 4, 5, 6}
(iii) 3x - 4 > 6
Solution:
3x > 10
x > 3.33
x ∈ N, so Solution set = {4, 5, 6, ...}
(iv) 4x + 1 ≥ 17
Solution:
4x ≥ 16
x ≥ 4
x ∈ N, so Solution set = {4, 5, 6, ...}
3. If the replacement set = {-6, -3, 0, 3, 6, 9} find the truth set of the following:
(i) 2x - 1 > 9
Solution:
2x > 10
x > 5
From replacement set, values > 5 are {6, 9}.
(ii) 3x + 7 ≤ 1
Solution:
3x ≤ -6
x ≤ -2
From replacement set, values ≤ -2 are {-6, -3}.
4. Solve: 9x - 7 ≤ 28 + 4x ; x ∈ W.
Solution:
9x - 4x ≤ 28 + 7
5x ≤ 35
x ≤ 7
x ∈ W, so Solution set = {0, 1, 2, 3, 4, 5, 6, 7}
5. Solve: -5(x + 4) > 30; x ∈ Z
Solution:
Divide by -5 (reverse sign):
x + 4 < -6
x < -10
x ∈ Z, so Solution set = {..., -13, -12, -11}
6. Solve: (2x + 1)/3 + 15 ≤ 17; x ∈ W.
Solution:
(2x + 1)/3 ≤ 2
2x + 1 ≤ 6
2x ≤ 5
x ≤ 2.5
x ∈ W, so Solution set = {0, 1, 2}
7. Solve: -3 + x < 2, x ∈ N.
Solution:
x < 5
x ∈ N, so Solution set = {1, 2, 3, 4}
8. Solve and graph the solution set on a number line.
(i) x - 5 < -2; x ∈ N
Solution:
x < 3
x ∈ N, Solution set = {1, 2}
Graph: Dots at 1 and 2.
(ii) 3x - 1 > 5; x ∈ W
Solution:
3x > 6
x > 2
x ∈ W, Solution set = {3, 4, 5, ...}
Graph: Dots at 3, 4, 5...
(iii) -3x + 12 < -15; x ∈ R
Solution:
-3x < -27
Divide by -3 (reverse sign):
x > 9
Graph: Hollow circle at 9, line extending to the right.
(iv) 7 ≥ 3x - 8; x ∈ W
Solution:
15 ≥ 3x
5 ≥ x or x ≤ 5
x ∈ W, Solution set = {0, 1, 2, 3, 4, 5}
Graph: Dots at 0, 1, 2, 3, 4, 5.
(v) 8x - 8 ≤ -24; x ∈ Z
Solution:
8x ≤ -16
x ≤ -2
x ∈ Z, Solution set = {..., -4, -3, -2}
Graph: Dots at -2, -3, -4...
(vi) 8x - 9 ≥ 35 - 3x; x ∈ N
Solution:
8x + 3x ≥ 35 + 9
11x ≥ 44
x ≥ 4
x ∈ N, Solution set = {4, 5, 6, ...}
Graph: Dots at 4, 5, 6...
9. For each inequation, given below, represent the solution on a number line:
(i) (5/2) - 2x ≥ 1/2, x ∈ W
Solution:
2.5 - 2x ≥ 0.5
2.5 - 0.5 ≥ 2x
2 ≥ 2x
1 ≥ x or x ≤ 1
x ∈ W, Solution set = {0, 1}
Graph: Dots at 0, 1.
(ii) 3(2x - 1) ≥ 2(2x + 3), x ∈ Z
Solution:
6x - 3 ≥ 4x + 6
2x ≥ 9
x ≥ 4.5
x ∈ Z, Solution set = {5, 6, 7, ...}
Graph: Dots at 5, 6, 7...
(iii) 2(4 - 3x) ≤ 4(x - 5), x ∈ W
Solution:
8 - 6x ≤ 4x - 20
8 + 20 ≤ 10x
28 ≤ 10x
2.8 ≤ x or x ≥ 2.8
x ∈ W, Solution set = {3, 4, 5, ...}
Graph: Dots at 3, 4, 5...
(iv) 4(3x + 1) > 2(4x - 1), x is a negative integer
Solution:
12x + 4 > 8x - 2
4x > -6
x > -1.5
x ∈ Negative Integers, Solution set = {-1}
Graph: Dot at -1.
(v) (4 - x)/2 < 3, x ∈ R
Solution:
4 - x < 6
-x < 2
x > -2
Graph: Hollow circle at -2, line extending right.
(vi) -2(x + 8) ≤ 8, x ∈ R
Solution:
x + 8 ≥ -4 (Divided by -2, sign reversed)
x ≥ -12
Graph: Solid dot at -12, line extending right.
10. If x is a real number and 7(x - 2) + 2 > 2(5x + 9). Draw the solution set on the number line.
Solution:
7x - 14 + 2 > 10x + 18
7x - 12 > 10x + 18
-12 - 18 > 10x - 7x
-30 > 3x
-10 > x or x < -10
Graph: Hollow circle at -10, line extending left.