OP Malhotra Class 10 Maths Solutions Chapter 4 Linear Inequations in One Variable Exercise 4

Question 1. On a bargain counter, the shopkeeper put labels on various goods showing their prices, Rs. P, where P ∈ {real numbers}. Write a mathematical sentence for each of the following labels:
(a) more than Rs. 7.50
(b) not less than Rs. 10
(c) not more than Rs. 22
(d) less than Rs. 11
Answer: We will write a mathematical inequality for each price condition, using P for the price.
(a) more than Rs. 7.50 : \( P > \text{Rs. } 7.50 \)
(b) not less than Rs. 10 : \( P \ge \text{Rs. } 10 \)
(c) not more than Rs. 22 : \( P \le \text{Rs. } 22 \)
(d) less than Rs. 11 : \( P < \text{Rs. } 11 \) This process helps represent real-world price conditions using mathematical inequalities, allowing for clear comparisons.
In simple words: We use math signs to show what the price, P, can be. For example, "more than" means \( > \), "not less than" means \( \ge \), "not more than" means \( \le \), and "less than" means \( < \).

🎯 Exam Tip: Pay close attention to keywords like "more than," "not less than," "not more than," and "less than" to correctly choose the inequality symbol. "Not less than" implies greater than or equal to, and "not more than" implies less than or equal to.

Question 2. You are given the following numbers : -2.6, 5.1, -3, 0.4, 1.2, -3.1, 4.7. Fill in the blanks.
(a) A = {A : A ≥ -3} = {.....}
(b) B = {A : A ≤ 1} = {.....}
Answer: Given numbers are: -2.6, 5.1, -3, 0.4, 1.2, -3.1, 4.7. We select numbers from this list that fit the given conditions.
(a) A = {x : x ≥ -3} = \( \{-3, -2.6, 0.4, 1.2, 4.7, 5.1\} \) This set includes all numbers from the given list that are greater than or equal to -3. This rule defines membership in set A.
(b) B = {x : x ≤ 1} = \( \{-3.1, -3, -2.6, 0.4\} \) This set includes all numbers from the given list that are less than or equal to 1.
In simple words: We look at the list of numbers and pick out only those that fit the rule given for each set. For set A, we choose numbers that are -3 or bigger. For set B, we choose numbers that are 1 or smaller.

🎯 Exam Tip: When forming sets based on inequalities, carefully check each number in the replacement set against the given condition. Be mindful of inclusive (≥, ≤) versus exclusive (>, <) inequalities.

Question 3. If the replacement set is {-2, -1, 1, 2, 4, 5, 9}, what is the solution set of each of the following mathematical sentences?
(a) \( x + \frac { 3 }{ 2 } > \frac { 5 }{ 2 } \)
(b) \( x - 4 = -3 \)
(c) \( 2x - 5 \ge 10 \)
(d) \( 3y \div 2 \le \frac { 5 }{ 2 } \)
(e) \( 4x^2 = 16 \)
Answer: The replacement set is \( \{-2, -1, 1, 2, 4, 5, 9\} \). We will solve each mathematical sentence and find the values from the replacement set that satisfy them.
(a) \( x + \frac { 3 }{ 2 } > \frac { 5 }{ 2 } \)
\( \implies x > \frac { 5 }{ 2 } - \frac { 3 }{ 2 } \)
\( \implies x > \frac { 5 - 3 }{ 2 } \)
\( \implies x > \frac { 2 }{ 2 } \)
\( \implies x > 1 \) The solution set includes numbers from the replacement set that are strictly greater than 1. So, Solution set = \( \{2, 4, 5, 9\} \).
(b) \( x - 4 = -3 \)
\( \implies x = -3 + 4 \)
\( \implies x = 1 \) The solution set includes numbers from the replacement set that are equal to 1. So, Solution set = \( \{1\} \).
(c) \( 2x - 5 \ge 10 \)
\( \implies 2x \ge 10 + 5 \)
\( \implies 2x \ge 15 \)
\( \implies x \ge \frac { 15 }{ 2 } \)
\( \implies x \ge 7.5 \) The solution set includes numbers from the replacement set that are greater than or equal to 7.5. So, Solution set = \( \{9\} \).
(d) \( 3y \div 2 \le \frac { 5 }{ 2 } \)
\( \implies \frac { 3y }{ 2 } \le \frac { 5 }{ 2 } \)
\( \implies 3y \le 5 \)
\( \implies y \le \frac { 5 }{ 3 } \)
\( \implies y \le 1\frac { 2 }{ 3 } \) The solution set includes numbers from the replacement set that are less than or equal to \( 1\frac{2}{3} \). So, Solution set = \( \{-2, -1, 1\} \).
(e) \( 4x^2 = 16 \)
\( \implies x^2 = \frac { 16 }{ 4 } \)
\( \implies x^2 = 4 \)
\( \implies x = \pm 2 \) The solution set includes numbers from the replacement set that are equal to -2 or 2. So, Solution set = \( \{-2, 2\} \).
In simple words: For each math problem, we first solve for the variable (x or y). Then, we check which of the numbers given in the replacement set fit our answer. Only those numbers form the final solution set.

🎯 Exam Tip: Always solve the inequality or equation completely first, and then filter the results using the given replacement set. Remember to consider both positive and negative roots for squared terms, and convert fractions to decimals for easier comparison with integers.

Question 4. Solve the inequation \( 30 - 4 (2x - 1) < 30 \), given that x is a positive integer.
Answer: We need to solve the inequation \( 30 - 4 (2x - 1) < 30 \).
\( \implies 30 - 8x + 4 < 30 \)
\( \implies 34 - 8x < 30 \)
\( \implies 34 - 30 < 8x \)
\( \implies 4 < 8x \)
\( \implies \frac { 4 }{ 8 } < x \)
\( \implies \frac { 1 }{ 2 } < x \)
\( \implies x > 0.5 \) Since x is a positive integer, it must be a whole number greater than 0.5. The positive integers that satisfy this are 1, 2, 3, and 4. These are the integers greater than 0.5.
Therefore, the Solution set = \( \{1, 2, 3, 4\} \).
In simple words: First, we simplify the math problem to find out that x must be bigger than 0.5. Then, because x has to be a whole number that is also positive, we list all the whole numbers starting from 1 that are greater than 0.5.

🎯 Exam Tip: Always simplify both sides of the inequality first. Remember to carefully consider the given condition for x (e.g., positive integer, real number) when determining the final solution set.

Question 5. If the replacement set is {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, what is the solution set of each of the following mathematical sentences?
(a) \( x + \frac { 4 }{ 3 } = \frac { 7 }{ 3 } \)
(b) \( 2x + 1 < 3 \)
(c) \( x - 6 > 10 - 6 \)
(d) \( x + 5 = 20 \)
(e) \( 2x + 3 \ge 17 \)
Answer: The replacement set is \( \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\} \). We will find the values from this set that satisfy each sentence.
(a) \( x + \frac { 4 }{ 3 } = \frac { 7 }{ 3 } \)
\( \implies x = \frac { 7 }{ 3 } - \frac { 4 }{ 3 } \)
\( \implies x = \frac { 7 - 4 }{ 3 } \)
\( \implies x = \frac { 3 }{ 3 } \)
\( \implies x = 1 \) The only number in the replacement set that is 1 is 1 itself. Solution set = \( \{1\} \).
(b) \( 2x + 1 < 3 \)
\( \implies 2x < 3 - 1 \)
\( \implies 2x < 2 \)
\( \implies x < \frac { 2 }{ 2 } \)
\( \implies x < 1 \) The only number in the replacement set that is less than 1 is 0. Solution set = \( \{0\} \).
(c) \( x - 6 > 10 - 6 \)
\( \implies x - 6 > 4 \)
\( \implies x > 4 + 6 \)
\( \implies x > 10 \) There are no numbers in the replacement set that are greater than 10. So, Solution set = \( \Phi \) (empty set).
(d) \( x + 5 = 20 \)
\( \implies x = 20 - 5 \)
\( \implies x = 15 \) There are no numbers in the replacement set that are equal to 15. So, Solution set = \( \Phi \) (empty set).
(e) \( 2x + 3 \ge 17 \)
\( \implies 2x \ge 17 - 3 \)
\( \implies 2x \ge 14 \)
\( \implies x \ge \frac { 14 }{ 2 } \)
\( \implies x \ge 7 \) The numbers in the replacement set that are greater than or equal to 7 are 7, 8, and 9. Solution set = \( \{7, 8, 9\} \).
In simple words: We solve each math sentence to find what x should be. Then, we check if that value of x or range of x is present in the list of numbers given at the start. If it is, we add it to our answer set; otherwise, the set is empty.

🎯 Exam Tip: Always double-check if the calculated solution (a single value or a range) actually falls within the specified replacement set. If no values match, the solution set is an empty set (Φ).

Question 6.
(a) x ∈ {2, 4, 6, 9} and y ∈ {4, 6, 18, 27, 54}. Form all ordered pairs (x, y) such that x is a factor of y and x < y.
(b) Find the truth set of the inequality x > y + 2 where (x, y) ∈ {(1, 2), (2, 3), (5, 1), (7, 3), (5, 6), (6, 5)}.
Answer:
(a) Given sets: \( x \in \{2, 4, 6, 9\} \) and \( y \in \{4, 6, 18, 27, 54\} \). We need ordered pairs \( (x, y) \) where x is a factor of y and \( x < y \).
The ordered pairs that satisfy these conditions are: \( \{(2, 4), (2, 6), (2, 18), (2, 54), (6, 18), (6, 54), (9, 18), (9, 27), (9, 54)\} \). For example, 2 is a factor of 4, and 2 is less than 4; this is how each pair is formed following the rules.
(b) Given set of ordered pairs: \( \{(1, 2), (2, 3), (5, 1), (7, 3), (5, 6), (6, 5)\} \). We need to find the pairs where \( x > y + 2 \).
If \( (x, y) = (1, 2) \), then \( 1 > 2 + 2 \implies 1 > 4 \) which is not true.
If \( (x, y) = (2, 3) \), then \( 2 > 3 + 2 \implies 2 > 5 \) which is not true.
If \( (x, y) = (5, 1) \), then \( 5 > 1 + 2 \implies 5 > 3 \) which is true. So, \( (5, 1) \) is a solution.
If \( (x, y) = (7, 3) \), then \( 7 > 3 + 2 \implies 7 > 5 \) which is true. So, \( (7, 3) \) is a solution.
If \( (x, y) = (5, 6) \), then \( 5 > 6 + 2 \implies 5 > 8 \) which is not true.
If \( (x, y) = (6, 5) \), then \( 6 > 5 + 2 \implies 6 > 7 \) which is not true.
Therefore, the Solution set (or truth set) is \( \{(5, 1), (7, 3)\} \).
In simple words: For part (a), we picked pairs where the first number divides the second number evenly, and the first number is also smaller. For part (b), we checked each pair to see if the first number was bigger than the second number plus two. Only the pairs that passed the test were kept.

🎯 Exam Tip: Systematically test each element or pair against all conditions in the problem. For ordered pairs, ensure both conditions (e.g., factor and inequality) are met simultaneously to avoid errors.

Question 7. Find out the truth sets of the following open sentences; replacement sets are given against them.
(i) \( \frac { 5 }{ x } > 7 ; \{1, 2\} \)
(ii) \( \frac { 5 }{ x } > 2; \{1, 2, 3, 4, 5, 6\} \)
(iii) \( x^2 = 9; \{-3, -2, -1, 1, 2, 3\} \)
(iv) \( x + \frac { 1 }{ x } = 2 ; \{0, 1, 2, 3\} \)
(v) \( 3x^2 < 2x ; \{-4, -3, -2, -1, 0, 1, 2, 3, 4\} \)
(vi) \( 2 (x - 3) < 1 ; \{1, 2, 3, 4, \dots, 10\} \)
Answer: We find the values from each given replacement set that satisfy the respective open sentence.
(i) For \( \frac { 5 }{ x } > 7 \), with replacement set \( \{1, 2\} \).
If \( x = 1 \), then \( \frac { 5 }{ 1 } > 7 \implies 5 > 7 \) which is not true.
If \( x = 2 \), then \( \frac { 5 }{ 2 } > 7 \implies 2.5 > 7 \) which is not true.
Since no values satisfy the condition, x has no solution. So, Solution set = \( \Phi \).
(ii) For \( \frac { 5 }{ x } > 2 \), with replacement set \( \{1, 2, 3, 4, 5, 6\} \).
If \( x = 1 \), then \( \frac { 5 }{ 1 } > 2 \implies 5 > 2 \) which is true.
If \( x = 2 \), then \( \frac { 5 }{ 2 } > 2 \implies 2.5 > 2 \) which is true.
If \( x = 3 \), then \( \frac { 5 }{ 3 } > 2 \implies 1.66... > 2 \) which is not true.
If \( x = 4 \), then \( \frac { 5 }{ 4 } > 2 \implies 1.25 > 2 \) which is not true.
If \( x = 5 \), then \( \frac { 5 }{ 5 } > 2 \implies 1 > 2 \) which is not true.
If \( x = 6 \), then \( \frac { 5 }{ 6 } > 2 \implies 0.83... > 2 \) which is not true.
So, Solution set = \( \{1, 2\} \).
(iii) For \( x^2 = 9 \), with replacement set \( \{-3, -2, -1, 1, 2, 3\} \).
This equation implies \( x = \pm 3 \). We check the values in the set.
If \( x = -3 \), then \( (-3)^2 = 9 \) which is true.
If \( x = 3 \), then \( (3)^2 = 9 \) which is true.
The other values in the set do not satisfy \( x^2 = 9 \). So, Solution set = \( \{-3, 3\} \).
(iv) For \( x + \frac { 1 }{ x } = 2 \), with replacement set \( \{0, 1, 2, 3\} \).
If \( x = 0 \), then \( 0 + \frac { 1 }{ 0 } \) is undefined, so it is not true.
If \( x = 1 \), then \( 1 + \frac { 1 }{ 1 } = 2 \implies 1 + 1 = 2 \) which is true.
If \( x = 2 \), then \( 2 + \frac { 1 }{ 2 } = 2 \implies 2.5 = 2 \) which is not true.
If \( x = 3 \), then \( 3 + \frac { 1 }{ 3 } = 2 \implies 3.33... = 2 \) which is not true.
So, Solution set = \( \{1\} \).
(v) For \( 3x^2 < 2x \), with replacement set \( \{-4, -3, -2, -1, 0, 1, 2, 3, 4\} \).
\( \implies 3x^2 - 2x < 0 \)
\( \implies x(3x - 2) < 0 \) This inequality holds when x is strictly between 0 and \( \frac{2}{3} \).
Checking the integers in the replacement set, none fall within the range \( 0 < x < \frac{2}{3} \). So, Solution set = \( \Phi \).
(vi) For \( 2 (x - 3) < 1 \), with replacement set \( \{1, 2, 3, 4, \dots, 10\} \).
\( \implies 2x - 6 < 1 \)
\( \implies 2x < 7 \)
\( \implies x < \frac { 7 }{ 2 } \)
\( \implies x < 3.5 \)
The integers in the replacement set that are less than 3.5 are 1, 2, and 3. So, Solution set = \( \{1, 2, 3\} \).
In simple words: For each problem, we solve the inequality or equation to find the possible range or exact values of x. Then, we check if any numbers from the given replacement set fall into that range or match those values. If they do, they are part of the truth set.

🎯 Exam Tip: When evaluating inequalities, always test the boundary conditions carefully. For rational expressions (like \( \frac{1}{x} \)), remember that division by zero is undefined, and consider how dividing by a negative number affects the inequality direction.

Question 8. Statement: The sum of the lengths of any two sides of a triangle is always greater than the length of its third side. Let x, x + 1, x + 2 be the lengths of the three sides of a triangle.
(i) Write down the three inequations in x, each of which represents the given statement.
(ii) List the set of possible values of x which satisfy all the three inequations obtained in your answer to part (i) above, given that x is an integer.
Answer: Let the lengths of the three sides of a triangle be x, \( x+1 \), and \( x+2 \).
(i) According to the triangle inequality theorem, the sum of any two sides must be greater than the third side. This gives us three inequations:
1. \( x + (x+1) > (x+2) \)
\( \implies 2x + 1 > x + 2 \)
\( \implies x > 1 \)
2. \( x + (x+2) > (x+1) \)
\( \implies 2x + 2 > x + 1 \)
\( \implies x > -1 \)
3. \( (x+1) + (x+2) > x \)
\( \implies 2x + 3 > x \)
\( \implies x > -3 \)
So, the three inequations are \( x > 1 \), \( x > -1 \), and \( x > -3 \).
(ii) To satisfy all three inequations, we need to find the common range. The conditions are \( x > 1 \), \( x > -1 \), and \( x > -3 \). The most restrictive condition is \( x > 1 \).
Given that x is an integer, possible values for x are \( \{2, 3, 4, \dots\} \). The solution set provided in the source for (ii) is \( \{2, 3, 4\} \), which suggests a specific context or an unstated upper bound. Without further constraints, x would be any integer greater than 1. Assuming the problem implies a limited set, the stated solution means x can be 2, 3, or 4.
In simple words: For a triangle to exist, any two sides added together must be longer than the third side. We write three math problems (inequations) for this rule. Then we find the whole numbers for x that make all three problems true at the same time. The solution set means these are the specific whole numbers for x that work.

🎯 Exam Tip: Remember the triangle inequality theorem: the sum of the lengths of any two sides of a triangle must be greater than the length of the third side. Solve all resulting inequalities and find their intersection to get the truth set.

Question. P is the solution set of \( 8x - 1 > 5x + 2 \) and Q is the solution set of \( 7x - 2 \ge 3 (x + 6) \), where \( x \in N \). Find the set \( P \cap Q \).
Answer: We need to find the solution sets P and Q for the given inequalities, considering x as a natural number, and then find their intersection.
For set P, solve \( 8x - 1 > 5x + 2 \):
\( \implies 8x - 5x > 2 + 1 \)
\( \implies 3x > 3 \)
\( \implies x > \frac { 3 }{ 3 } \)
\( \implies x > 1 \)
Since \( x \in N \) (natural numbers start from 1), P = \( \{2, 3, 4, 5, 6, 7, 8, \dots\} \).
For set Q, solve \( 7x - 2 \ge 3 (x + 6) \):
\( \implies 7x - 2 \ge 3x + 18 \)
\( \implies 7x - 3x \ge 18 + 2 \)
\( \implies 4x \ge 20 \)
\( \implies x \ge \frac { 20 }{ 4 } \)
\( \implies x \ge 5 \)
Since \( x \in N \), Q = \( \{5, 6, 7, 8, \dots\} \).
Now, find the intersection \( P \cap Q \), which means the numbers that are in both P and Q. If \( x > 1 \) and \( x \ge 5 \), then both conditions are satisfied when \( x \ge 5 \). This is the common range for x.
Therefore, \( P \cap Q = \{5, 6, 7, 8, \dots\} \).
In simple words: First, we solve each math problem (inequality) separately to find the range of x for P and Q. Since x must be a whole counting number, we list the numbers for each set. Then, we find the numbers that appear in both lists to get the intersection set.

🎯 Exam Tip: When finding the intersection of solution sets for inequalities, draw a number line to visualize the ranges, then identify the overlapping region. Remember to consider the domain of x (e.g., natural numbers, integers, real numbers) as it changes the possible values.

Question 9. Answer true or false.
(a) If \( x + 10 = y + 14 \), then \( x > y \).
(b) \( | -4 | - 4 = 8 \).
(c) If \( 10 - x > 3 \), then \( x < 7 \).
(d) If \( p = q + 2 \), then \( p > q \).
(e) If a and b are two negative integers such that \( a < b \), then \( \frac { 1 }{ a } < \frac { 1 }{ b } \).
(f) \( 3 \in \{x : 3x - 2 > 5\} \).
Answer: We evaluate each statement to determine if it is true or false.
(a) True. If \( x + 10 = y + 14 \), then subtract 10 from both sides: \( x = y + 14 - 10 \implies x = y + 4 \). Since 4 is a positive number, \( x \) must be greater than \( y \).
(b) False. The absolute value of -4 is 4. So, \( | -4 | - 4 = 4 - 4 = 0 \). The statement says \( 0 = 8 \), which is incorrect.
(c) True. If \( 10 - x > 3 \), then move x to the right and 3 to the left: \( 10 - 3 > x \implies 7 > x \). This means \( x < 7 \).
(d) True. If \( p = q + 2 \), it means \( p \) is 2 units larger than \( q \). Therefore, \( p \) is definitely greater than \( q \).
(e) False. Let's consider an example with two negative integers: let \( a = -2 \) and \( b = -1 \). Here, \( a < b \) is true. Then \( \frac { 1 }{ a } = \frac { 1 }{ -2 } = -0.5 \) and \( \frac { 1 }{ b } = \frac { 1 }{ -1 } = -1 \). Since \( -0.5 > -1 \), it means \( \frac { 1 }{ a } > \frac { 1 }{ b } \). Therefore, the statement \( \frac { 1 }{ a } < \frac { 1 }{ b } \) is false.
(f) True. For the set \( \{x : 3x - 2 > 5\} \), we solve the inequality: \( 3x - 2 > 5 \implies 3x > 7 \implies x > \frac { 7 }{ 3 } \implies x > 2.33... \). Since 3 is greater than 2.33..., 3 belongs to this set.
In simple words: We check each statement by doing the math or thinking about the rules. For example, for absolute values, we find the positive number first. For inequalities, we solve them and see if the number fits.

🎯 Exam Tip: When evaluating true/false statements, work through each part mathematically. For inequalities involving negative numbers, pay extra attention to how signs affect the direction of the inequality when multiplying or dividing. Always test with examples for generalizations like those in part (e).

Question 10. Find the solution of the inequation \( 2 \le 2p - 3 \le 5 \). Hence graph the solution set on the number line given below.
Answer: We solve the compound inequality by splitting it into two parts:
(i) \( 2 \le 2p - 3 \)
\( \implies 2 + 3 \le 2p \)
\( \implies 5 \le 2p \)
\( \implies \frac { 5 }{ 2 } \le p \)
\( \implies p \ge 2.5 \)
(ii) \( 2p - 3 \le 5 \)
\( \implies 2p \le 5 + 3 \)
\( \implies 2p \le 8 \)
\( \implies p \le \frac { 8 }{ 2 } \)
\( \implies p \le 4 \)
Combining (i) and (ii), we get \( 2.5 \le p \le 4 \). The solution set is \( \{p : 2.5 \le p \le 4, p \in R\} \). This means p is a real number between 2.5 and 4, inclusive.
The graph shows this range on a number line with solid dots at 2.5 and 4, and a shaded line connecting them.

In simple words: We break the problem into two smaller inequalities and solve each one. Then, we find the range that satisfies both conditions. On the number line, we show this range with solid dots at the start and end because the values are included, and a thick line in between.

🎯 Exam Tip: For compound inequalities, solve each part separately and then find the intersection of their solution sets. Use solid dots for "less than or equal to" (≤) or "greater than or equal to" (≥) and hollow circles for strict inequalities (< or >) on the number line.

Question 11. If x is a negative integer, find the solution set of \( \frac { 2 }{ 3 } + \frac { 1 }{ 3 } (x + 1) > 0 \).
Answer: Given that x is a negative integer. We need to solve the inequation:
\( \frac { 2 }{ 3 } + \frac { 1 }{ 3 } (x + 1) > 0 \)
Multiply the entire inequation by 3 to clear the denominators:
\( \implies 2 + (x + 1) > 0 \)
\( \implies 2 + x + 1 > 0 \)
\( \implies x + 3 > 0 \)
\( \implies x > -3 \)
Since x is a negative integer and \( x > -3 \), the possible values for x are -2 and -1. These are the only negative whole numbers strictly greater than -3.
Therefore, Solution set = \( \{-2, -1\} \).
In simple words: First, we solve the math problem to find that x must be greater than -3. Then, since x has to be a whole number that is also negative, we pick the negative whole numbers that are bigger than -3.

🎯 Exam Tip: Always consider the restrictions on the variable (e.g., negative integer, real number) when determining the final solution set. Multiplying by the LCM of denominators can simplify inequalities, but be careful if multiplying by a negative value.

Question 12. Write open mathematical sentences, using x for the variable whose graphs would be
(i)

(ii)

(iii)

(iv)

(v)

Answer: The mathematical sentences for each graph are:
(i) Here, the graph shows all real numbers less than or equal to -2. So, the solution set is \( \{x : x \le -2, x \in R\} \).
(ii) Here, the graph shows all real numbers less than or equal to 4. So, the solution set is \( \{x : x \le 4, x \in R\} \).
(iii) Here, the graph shows integers from 4 to 5, inclusive. So, the solution set is \( \{x : 4 \le x \le 5, x \in N\} \).
(iv) Here, the graph shows odd integers from 1 to 5, inclusive. So, the solution set is \( \{x : 1 \le x \le 5, x \in N, \text{ x is odd}\} \).
(v) Here, the graph shows all real numbers strictly greater than -2. So, the solution set is \( \{x : x > -2, x \in R\} \).
In simple words: For each picture of a number line, we write down the rule that describes the shaded part or the dots. This rule shows what values of x are included, like "x is smaller than or equal to -2" or "x is an odd whole number between 1 and 5".

🎯 Exam Tip: When translating a number line graph to an inequality, note whether the endpoint is a solid dot (inclusive, ≤ or ≥) or a hollow circle (exclusive, < or >), and the direction of the shaded line (left for smaller, right for larger). Also, identify the domain (real numbers, integers, etc.).

Question 13. Answer true or false.
(i) If \( 2 - x < 0 \), then \( x > 2 \).
(ii) The graph of the inequation \( y \le 2x \) includes the origin.
Answer: We will evaluate each statement.
(i) True. If \( 2 - x < 0 \), we can add x to both sides: \( 2 < x \), which means \( x > 2 \). This is a straightforward algebraic manipulation.
(ii) True. The origin is the point \( (0, 0) \). To check if it is included in the graph of \( y \le 2x \), substitute \( x=0 \) and \( y=0 \) into the inequation: \( 0 \le 2(0) \implies 0 \le 0 \). Since \( 0 \le 0 \) is true, the origin is included in the solution set of the inequation. The point (0,0) satisfies the condition.
In simple words: For (i), we solve the math problem to see if the statement is true. For (ii), we put the numbers for the origin (0,0) into the math problem to check if it fits the rule.

🎯 Exam Tip: For inequalities, testing specific points like the origin (if it's not a boundary) can easily check if a region is part of the solution set. Pay attention to how algebra rules (like moving terms across the inequality sign) affect the inequality direction.

Question 14. If \( 25 - 4x \le 16 \), find :
(i) the smallest value of x when x is a real number
(ii) the smallest value of x when x is an integer.
Answer: First, we solve the inequality \( 25 - 4x \le 16 \).
\( \implies 25 - 16 \le 4x \)
\( \implies 9 \le 4x \)
\( \implies \frac { 9 }{ 4 } \le x \)
\( \implies x \ge 2.25 \)
(i) When x is a real number, the smallest value of x is \( \frac { 9 }{ 4 } \) or \( 2.25 \). This is the exact lower bound for x, as x can take any decimal value. Therefore, the smallest real value of x is 2.25.
(ii) When x is an integer, the smallest integer value that is greater than or equal to 2.25 is 3.
Therefore, the smallest value of x when x is an integer is 3.
In simple words: First, we solve the math problem to find the smallest number x can be. If x can be any number, the smallest value is exactly what we found. If x has to be a whole number, we find the first whole number that is bigger than or equal to our result.

🎯 Exam Tip: Distinguish carefully between real numbers and integers when finding the smallest or largest value that satisfies an inequality. For integers, you might need to round up or down to find the closest valid integer that meets the condition.

Question 15. Given: \( x \in \{1, 2, 3, 4, 5, 6, 7, 8, 9\} \), find the values of x for which \( -3 < 2x - 1 < x + 4 \).
Answer: We need to solve the compound inequality \( -3 < 2x - 1 < x + 4 \) and find the values of x from the given set that satisfy it. We split it into two separate inequalities:
(i) \( -3 < 2x - 1 \)
\( \implies -3 + 1 < 2x \)
\( \implies -2 < 2x \)
\( \implies \frac { -2 }{ 2 } < x \)
\( \implies -1 < x \)
(ii) \( 2x - 1 < x + 4 \)
\( \implies 2x - x < 4 + 1 \)
\( \implies x < 5 \)
Combining (i) and (ii), we get \( -1 < x < 5 \). This means x must be greater than -1 and less than 5.
Now, we check the given replacement set \( \{1, 2, 3, 4, 5, 6, 7, 8, 9\} \) for values of x that fall within this range. These values are 1, 2, 3, and 4. These are the integers in the set that satisfy the conditions.
Therefore, the Solution set will be \( \{1, 2, 3, 4\} \).
In simple words: We break the big math problem into two smaller ones and solve each to find a range for x. Then, we look at the list of numbers given and pick out all the numbers that fit into that range.

🎯 Exam Tip: Always split a compound inequality (e.g., A < B < C) into two separate inequalities (A < B and B < C) and solve them individually. The final solution set is the intersection of the results, filtered by the replacement set.

Question 16. Solve the inequation : \( 2x - 10 < 3x - 15 \).
Answer: We need to solve the inequation \( 2x - 10 < 3x - 15 \).
First, we move the x terms to one side and the constant terms to the other side.
\( \implies -10 + 15 < 3x - 2x \)
\( \implies 5 < x \)
\( \implies x > 5 \) This means x must be any real number strictly greater than 5. For example, 5.1, 6, 100, etc.
Therefore, the Solution set = \( \{x : x > 5\} \).
In simple words: We move the x terms to one side and the normal numbers to the other to find out what x must be greater than. The answer means x can be any number larger than 5.

🎯 Exam Tip: When rearranging an inequality, be careful with negative coefficients. If you multiply or divide both sides by a negative number, you must reverse the direction of the inequality sign. Always check your steps.

Question 17. Solve the inequation : \( 3 - 2x \ge x - 12 \), given that \( x \in N \).
Answer: We need to solve the inequation \( 3 - 2x \ge x - 12 \).
First, we gather the x terms on one side and constant terms on the other.
\( \implies 3 + 12 \ge x + 2x \)
\( \implies 15 \ge 3x \)
\( \implies \frac { 15 }{ 3 } \ge x \)
\( \implies 5 \ge x \)
\( \implies x \le 5 \) Since x must be a natural number (N), which includes positive integers starting from 1, the values of x that are less than or equal to 5 are 1, 2, 3, 4, and 5.
Therefore, the Solution set = \( \{1, 2, 3, 4, 5\} \).
In simple words: We arrange the math problem to find that x must be 5 or less. Since x has to be a positive whole number (a natural number), we list all positive whole numbers from 1 up to 5.

🎯 Exam Tip: Pay close attention to the specified domain for x (e.g., natural numbers (N), integers (Z), real numbers (R)) as it affects which values are included in the final solution set. Natural numbers typically start from 1.

Question 18. If x ∈ {real numbers} and \( -1 < 3 - 2x \le 7 \), evaluate x and represent it on a number line.
Answer: We solve the compound inequality \( -1 < 3 - 2x \le 7 \) by splitting it into two parts, given that x is a real number.
(i) \( -1 < 3 - 2x \)
\( \implies -1 - 3 < -2x \)
\( \implies -4 < -2x \)
\( \implies \frac { -4 }{ -2 } > x \) (Note: dividing by a negative number reverses the inequality sign)
\( \implies 2 > x \)
\( \implies x < 2 \)
(ii) \( 3 - 2x \le 7 \)
\( \implies -2x \le 7 - 3 \)
\( \implies -2x \le 4 \)
\( \implies x \ge \frac { 4 }{ -2 } \) (Note: dividing by a negative number reverses the inequality sign)
\( \implies x \ge -2 \)
Combining (i) and (ii), the solution is \( -2 \le x < 2 \). This means x is a real number that is greater than or equal to -2 and strictly less than 2.
The solution set can be represented on a number line with a solid dot at -2 (inclusive), a hollow circle at 2 (exclusive), and a shaded line connecting them.

In simple words: We split the inequality into two parts and solve each one. Remember to flip the inequality sign if you divide or multiply by a negative number. Then, we combine the results to find the range for x. On the number line, a solid dot means "equal to or", and an empty dot means "not equal to".

🎯 Exam Tip: Be very careful when multiplying or dividing an inequality by a negative number; you must reverse the direction of the inequality sign. For number line representation, use solid points for inclusive bounds and hollow points for exclusive bounds.

Question 19. Find the range of values of x which satisfies \( -2\frac { 2 }{ 3 } \le x + \frac { 1 }{ 3 } < 3\frac { 1 }{ 3 } \), \( x \in R \). Graph these values of x on the number line.
Answer: We convert the mixed fractions to improper fractions first: \( -2\frac{2}{3} = -\frac{8}{3} \) and \( 3\frac{1}{3} = \frac{10}{3} \).
The inequality becomes \( -\frac { 8 }{ 3 } \le x + \frac { 1 }{ 3 } < \frac { 10 }{ 3 } \).
We solve this compound inequality by splitting it:
(i) \( -\frac { 8 }{ 3 } \le x + \frac { 1 }{ 3 } \)
\( \implies -\frac { 8 }{ 3 } - \frac { 1 }{ 3 } \le x \)
\( \implies -\frac { 9 }{ 3 } \le x \)
\( \implies -3 \le x \)
(ii) \( x + \frac { 1 }{ 3 } < \frac { 10 }{ 3 } \)
\( \implies x < \frac { 10 }{ 3 } - \frac { 1 }{ 3 } \)
\( \implies x < \frac { 9 }{ 3 } \)
\( \implies x < 3 \)
Combining (i) and (ii), the solution is \( -3 \le x < 3 \). This means x is a real number that is greater than or equal to -3 and strictly less than 3.
The solution set is represented on a number line with a solid dot at -3, a hollow circle at 3, and a shaded line connecting them.

In simple words: First, we change any mixed fractions into simpler forms. Then, we solve the two parts of the inequality separately to find the range where x can be. On the number line, a filled circle shows where x can be equal to, and an empty circle shows where x must be strictly less than.

🎯 Exam Tip: Always convert mixed fractions to improper fractions before performing algebraic operations in inequalities. This simplifies calculations and reduces the chance of errors. Remember to combine the solution sets of individual inequalities into a single range.

Question 20. Solve and graph the solution set of :
(a) \( 6 \ge 2 - x, \frac { x }{ 3 } + 2 < 3; x \in R \)
(b) \( \frac { x }{ 2 } < \frac { 6-x }{ 4 }, \frac { 2-x }{ 6 } < \frac { 7-x }{ 9 }; x \in R \)
Answer:
(a) We solve the two inequalities separately for \( x \in R \):
First inequality: \( 6 \ge 2 - x \)
\( \implies x \ge 2 - 6 \)
\( \implies x \ge -4 \) ... (i)
Second inequality: \( \frac { x }{ 3 } + 2 < 3 \)
\( \implies \frac { x }{ 3 } < 3 - 2 \)
\( \implies \frac { x }{ 3 } < 1 \)
\( \implies x < 3 \) ... (ii)
Combining (i) and (ii), the solution is \( -4 \le x < 3 \). This range includes real numbers from -4 (inclusive) up to, but not including, 3.
Graph for (a) (solid dot at -4, hollow dot at 3, shaded line in between):

(b) We solve the two inequalities separately for \( x \in R \):
First inequality: \( \frac { x }{ 2 } < \frac { 6-x }{ 4 } \)
Multiply by 4 (LCM of 2, 4):
\( \implies 2x < 6 - x \)
\( \implies 2x + x < 6 \)
\( \implies 3x < 6 \)
\( \implies x < \frac { 6 }{ 3 } \)
\( \implies x < 2 \) ... (i)
Second inequality: \( \frac { 2-x }{ 6 } < \frac { 7-x }{ 9 } \)
Multiply by 18 (LCM of 6, 9):
\( \implies 3(2 - x) < 2(7 - x) \)
\( \implies 6 - 3x < 14 - 2x \)
\( \implies 6 - 14 < -2x + 3x \)
\( \implies -8 < x \)
\( \implies x > -8 \) ... (ii)
Combining (i) and (ii), the solution is \( -8 < x < 2 \). This range includes real numbers strictly between -8 and 2. This is the set of all real numbers that satisfy both conditions.
Graph for (b) (hollow dots at -8 and 2, shaded line in between):

In simple words: For both parts (a) and (b), we solve each of the two inequalities separately to find two different conditions for x. Then, we combine these conditions to find a single range where x must be. Finally, we draw this range on a number line, using filled dots for "equal to" and empty dots for "not equal to".

🎯 Exam Tip: Always multiply inequalities by the Least Common Multiple (LCM) of the denominators to clear fractions, but only if you are multiplying by a positive number. If an inequality involves 'and' (compound inequality), the solution is the intersection of the individual solutions. If it involves 'or', the solution is the union.

Question 21. Find the range of values of x which satisfy \( -\frac{1}{3} \le \frac{x}{2} - 1\frac{1}{3} < \frac{1}{6} \); \( x \in R \). Graph these values of x on the real number line.
Answer: We need to solve the given compound inequation: \( -\frac{1}{3} \le \frac{x}{2} - 1\frac{1}{3} < \frac{1}{6} \). First, convert the mixed fraction to an improper fraction: \( 1\frac{1}{3} = \frac{4}{3} \). The inequation becomes: \( -\frac{1}{3} \le \frac{x}{2} - \frac{4}{3} < \frac{1}{6} \). Multiply all parts by the LCM of the denominators (3, 2, 6), which is 6, to clear the fractions:
\( 6 \cdot (-\frac{1}{3}) \le 6 \cdot (\frac{x}{2} - \frac{4}{3}) < 6 \cdot (\frac{1}{6}) \)
\( -2 \le 3x - 8 < 1 \) Now, split this into two separate inequalities: (i) \( -2 \le 3x - 8 \) Add 8 to both sides:
\( -2 + 8 \le 3x \)
\( 6 \le 3x \) Divide by 3:
\( \frac{6}{3} \le x \)
\( 2 \le x \) (ii) \( 3x - 8 < 1 \) Add 8 to both sides:
\( 3x < 1 + 8 \)
\( 3x < 9 \) Divide by 3:
\( x < \frac{9}{3} \)
\( x < 3 \) Combining both results, we get \( 2 \le x < 3 \). This means x can be 2 or any number greater than 2 but strictly less than 3. The solution is a continuous range of numbers. The solution set is \( \{x : 2 \le x < 3, x \in R\} \).In simple words: First, solve the inequality by separating it into two parts and finding the range for x. Then, on the number line, mark the starting point with a filled circle (because it includes that number) and the ending point with an empty circle (because it does not include that number), shading the line segment in between.

🎯 Exam Tip: When multiplying an inequality by a negative number, always remember to reverse the inequality sign. Carefully distinguish between inclusive (≤, ≥) and exclusive (<, >) boundaries for points on the number line.

Question 22. Write down the range of real values of x for which the inequation x > 3 and - 2 < x < 5 are both true.
Answer: We are given two conditions for \( x \):
(i) \( x > 3 \)
(ii) \( -2 < x < 5 \)
For both conditions to be true, \( x \) must satisfy both. From (i), \( x \) must be greater than 3. From (ii), \( x \) must be greater than -2 and less than 5. If \( x > 3 \) and \( x < 5 \), then the intersection of these two ranges is \( 3 < x < 5 \). The condition \( x > -2 \) is already satisfied if \( x > 3 \). Thus, \( x \) must be between 3 and 5, not including 3 or 5. This problem often combines different constraints to find the shared valid range. The range of real values for which both are true is \( 3 < x < 5 \).In simple words: We need to find the numbers that are bigger than 3, and also fall between -2 and 5. The only numbers that fit both rules are the ones strictly between 3 and 5.

🎯 Exam Tip: For "and" conditions in inequalities, find the intersection of the solution sets. For "or" conditions, find the union of the solution sets. Sketching individual number lines helps visualize the common range.

Question 23. Solve and graph the solution set of \( 3x - 4 > 11 \) or \( 5 - 2x \ge 7 \); \( x \in R \).
Answer: We need to solve two separate inequalities and combine their solutions using "or". (i) \( 3x - 4 > 11 \) Add 4 to both sides:
\( 3x > 11 + 4 \)
\( 3x > 15 \) Divide by 3:
\( x > \frac{15}{3} \)
\( x > 5 \) (ii) \( 5 - 2x \ge 7 \) Subtract 5 from both sides:
\( -2x \ge 7 - 5 \)
\( -2x \ge 2 \) Divide by -2 and reverse the inequality sign:
\( x \le \frac{2}{-2} \)
\( x \le -1 \) The solution set is all real numbers \( x \) such that \( x > 5 \) or \( x \le -1 \). This forms two distinct, non-overlapping regions on the number line.In simple words: This question asks for numbers that are either bigger than 5 OR smaller than or equal to -1. On the number line, this means you mark all numbers from -1 downwards with a filled circle at -1 and extend to the left. For numbers greater than 5, you mark an empty circle at 5 and extend to the right.

🎯 Exam Tip: When dealing with "or" conditions, plot both solution sets on the number line. The final solution is the union of these sets, meaning any value satisfying at least one of the inequalities is included.

Question 2. Solve the inequation : \( 12 + 1 \frac{5}{6} x \le 5 + 3x, x \in R \). Represent the solution on a number line.
Answer: First, convert the mixed fraction to an improper fraction: \( 1\frac{5}{6} = \frac{11}{6} \). The inequation is \( 12 + \frac{11}{6} x \le 5 + 3x \). To eliminate the fraction, multiply all terms by 6:
\( 6(12) + 6(\frac{11}{6} x) \le 6(5) + 6(3x) \)
\( 72 + 11x \le 30 + 18x \) Now, gather \( x \) terms on one side and constant terms on the other. It's often easier to keep the \( x \) term positive:
\( 72 - 30 \le 18x - 11x \)
\( 42 \le 7x \) Divide by 7:
\( \frac{42}{7} \le x \)
\( 6 \le x \) So, the solution set is \( \{x : x \ge 6, x \in R\} \). This means x can be 6 or any number greater than 6. The real number condition implies a continuous shaded line.In simple words: First, rewrite the mixed number as a simple fraction. Then, solve the inequality to find that \( x \) must be 6 or any number larger than 6. On the number line, this means drawing a filled circle at 6 and shading the line to the right.

🎯 Exam Tip: Always convert mixed fractions to improper fractions before performing operations in inequalities. When solving, ensure all terms are handled correctly and verify the direction of the inequality sign after each step.

Question 3. Solve the inequation : \( -3 \le 3 - 2x < 9, x \in R \). Represent your solution on a number line.
Answer: We need to solve the given compound inequation \( -3 \le 3 - 2x < 9 \). Split this into two separate inequalities: (i) \( -3 \le 3 - 2x \) Subtract 3 from both sides:
\( -3 - 3 \le -2x \)
\( -6 \le -2x \) Divide by -2 and reverse the inequality sign:
\( \frac{-6}{-2} \ge x \)
\( 3 \ge x \) (or \( x \le 3 \)) (ii) \( 3 - 2x < 9 \) Subtract 3 from both sides:
\( -2x < 9 - 3 \)
\( -2x < 6 \) Divide by -2 and reverse the inequality sign:
\( x > \frac{6}{-2} \)
\( x > -3 \) Combining both results, we get \( -3 < x \le 3 \). This means x can be any number greater than -3 but less than or equal to 3. The solution set is \( \{x : -3 < x \le 3, x \in R\} \).In simple words: First, solve the combined inequality by splitting it into two parts. You'll find that \( x \) must be greater than -3 and also less than or equal to 3. On the number line, this means an open circle at -3, a filled circle at 3, and the line between them shaded.

🎯 Exam Tip: Remember to split compound inequalities into two simple ones. When graphing, an open circle means the endpoint is not included, while a closed circle means it is included.

Question 4. Find the value of x, which satisfies the inequation \( -2 \le \frac{1}{2} - \frac{2x}{3} \le 1\frac{5}{6}, x \in N \). Graph the solution on the number line.
Answer: We need to solve the compound inequation \( -2 \le \frac{1}{2} - \frac{2x}{3} \le 1\frac{5}{6} \). First, convert the mixed fraction: \( 1\frac{5}{6} = \frac{11}{6} \). The inequation becomes: \( -2 \le \frac{1}{2} - \frac{2x}{3} \le \frac{11}{6} \). Multiply all parts by the LCM of the denominators (2, 3, 6), which is 6, to clear the fractions:
\( 6 \cdot (-2) \le 6 \cdot (\frac{1}{2} - \frac{2x}{3}) \le 6 \cdot (\frac{11}{6}) \)
\( -12 \le 3 - 4x \le 11 \) Now, split this into two separate inequalities: (i) \( -12 \le 3 - 4x \) Subtract 3 from both sides:
\( -12 - 3 \le -4x \)
\( -15 \le -4x \) Divide by -4 and reverse the inequality sign:
\( \frac{-15}{-4} \ge x \)
\( 3.75 \ge x \) (or \( x \le 3.75 \)) (ii) \( 3 - 4x \le 11 \) Subtract 3 from both sides:
\( -4x \le 11 - 3 \)
\( -4x \le 8 \) Divide by -4 and reverse the inequality sign:
\( x \ge \frac{8}{-4} \)
\( x \ge -2 \) Combining both results, we have \( -2 \le x \le 3.75 \). Since \( x \in N \) (natural numbers, starting from 1), the possible integer values are \( \{1, 2, 3\} \). It is important to remember the set of numbers \(x\) belongs to. The solution set is \( \{1, 2, 3\} \).In simple words: After solving the inequality, you get a range for \( x \). Since \( x \) must be a natural number (positive whole numbers like 1, 2, 3...), you pick only those whole numbers from your range. Mark these specific numbers as dots on the number line.

🎯 Exam Tip: Always pay close attention to the replacement set (e.g., real numbers, integers, natural numbers, whole numbers) as it dictates the type of points or lines to be marked on the number line.

Question 5. Solve the following inequation, and graph the solution on the number line: \( (2x - 5 \le 5x + 4 < 11, x \in R) \).
Answer: We need to solve the given compound inequation \( 2x - 5 \le 5x + 4 < 11 \). Split this into two separate inequalities: (i) \( 2x - 5 \le 5x + 4 \) Subtract \( 2x \) from both sides and subtract 4 from both sides:
\( -5 - 4 \le 5x - 2x \)
\( -9 \le 3x \) Divide by 3:
\( \frac{-9}{3} \le x \)
\( -3 \le x \) (ii) \( 5x + 4 < 11 \) Subtract 4 from both sides:
\( 5x < 11 - 4 \)
\( 5x < 7 \) Divide by 5:
\( x < \frac{7}{5} \)
\( x < 1.4 \) Combining both results, we get \( -3 \le x < 1.4 \). This means x can be -3 or any number greater than -3 but strictly less than 1.4. The solution set is \( \{x : -3 \le x < 1.4, x \in R\} \).In simple words: This problem involves a three-part inequality. Solve it by splitting it into two pairs of inequalities. The final solution is numbers between -3 (including -3) and 1.4 (not including 1.4). Mark this range on the number line.

🎯 Exam Tip: Pay attention to the exact values and their inclusion/exclusion from the solution set. Decimal values on the number line should be accurately placed, even if they fall between integer markers.

Question 6. Solve \( 2 \le 2x - 3 \le 5, x \in R \) and mark it on a number line.
Answer: We need to solve the compound inequation \( 2 \le 2x - 3 \le 5 \). Split this into two separate inequalities: (i) \( 2 \le 2x - 3 \) Add 3 to both sides:
\( 2 + 3 \le 2x \)
\( 5 \le 2x \) Divide by 2:
\( \frac{5}{2} \le x \)
\( 2.5 \le x \) (ii) \( 2x - 3 \le 5 \) Add 3 to both sides:
\( 2x \le 5 + 3 \)
\( 2x \le 8 \) Divide by 2:
\( x \le \frac{8}{2} \)
\( x \le 4 \) Combining both results, we get \( 2.5 \le x \le 4 \). This means x can be 2.5 or any number greater than 2.5 and less than or equal to 4. Understanding decimal values is key here. The solution set is \( \{x : 2.5 \le x \le 4, x \in R\} \).In simple words: Break the compound inequality into two simpler ones. Solve both to get a range for \( x \), which is from 2.5 to 4, including both numbers. Then draw a line on the number line with filled circles at 2.5 and 4, shading the part in between.

🎯 Exam Tip: When the solution involves decimal points, ensure their positions on the number line are accurately estimated relative to integer markers. Closed circles indicate inclusion of endpoints.

Question 7. Given that \( x \in I \), solve the inequation and graph the solution on the number line. \( 3 \ge \frac{x-4}{2} + \frac{x}{3} \ge 2 \).
Answer: We need to solve the compound inequation \( 3 \ge \frac{x-4}{2} + \frac{x}{3} \ge 2 \). First, simplify the middle expression:
\( \frac{x-4}{2} + \frac{x}{3} = \frac{3(x-4) + 2(x)}{6} = \frac{3x - 12 + 2x}{6} = \frac{5x - 12}{6} \) So the inequation becomes: \( 3 \ge \frac{5x - 12}{6} \ge 2 \). Multiply all parts by 6 to clear the fraction:
\( 6 \cdot 3 \ge 6 \cdot (\frac{5x - 12}{6}) \ge 6 \cdot 2 \)
\( 18 \ge 5x - 12 \ge 12 \) Now, split this into two separate inequalities: (i) \( 18 \ge 5x - 12 \) Add 12 to both sides:
\( 18 + 12 \ge 5x \)
\( 30 \ge 5x \) Divide by 5:
\( \frac{30}{5} \ge x \)
\( 6 \ge x \) (or \( x \le 6 \)) (ii) \( 5x - 12 \ge 12 \) Add 12 to both sides:
\( 5x \ge 12 + 12 \)
\( 5x \ge 24 \) Divide by 5:
\( x \ge \frac{24}{5} \)
\( x \ge 4.8 \) Combining both results, we have \( 4.8 \le x \le 6 \). Since \( x \in I \) (integers), the possible integer values are \( \{5, 6\} \). This means only whole numbers that are 5 or 6 satisfy the conditions. The solution set is \( \{5, 6\} \).In simple words: First, combine the fractions in the middle. Then, separate the inequality into two parts and solve for \( x \). Because \( x \) must be an integer, you pick only the whole numbers within the found range. Mark these numbers as dots on the number line.

🎯 Exam Tip: When dealing with fractions in compound inequalities, clear them by multiplying all parts by the LCM. Remember to specify the solution set based on the given replacement set (e.g., integers, real numbers). Marking only discrete points for integers is crucial.

Question 8. Give that \( x \in R \), solve the following inequality and graph the solution on the number line: \( -1 \le 3 + 4x < 23 \).
Answer: We need to solve the compound inequation \( -1 \le 3 + 4x < 23 \). Split this into two separate inequalities: (i) \( -1 \le 3 + 4x \) Subtract 3 from both sides:
\( -1 - 3 \le 4x \)
\( -4 \le 4x \) Divide by 4:
\( \frac{-4}{4} \le x \)
\( -1 \le x \) (ii) \( 3 + 4x < 23 \) Subtract 3 from both sides:
\( 4x < 23 - 3 \)
\( 4x < 20 \) Divide by 4:
\( x < \frac{20}{4} \)
\( x < 5 \) Combining both results, we get \( -1 \le x < 5 \). This means x can be -1 or any number greater than -1 but strictly less than 5. Real numbers mean a continuous line. The solution set is \( \{x : -1 \le x < 5, x \in R\} \).In simple words: This problem asks you to find values of \( x \) that are greater than or equal to -1, but strictly less than 5. On the number line, draw a filled circle at -1 and an empty circle at 5, then shade the continuous line between them.

🎯 Exam Tip: Remember to solve compound inequalities by isolating the variable in each part. Real number solutions are represented by a continuous line segment on the number line, while integers are discrete points.

Question 9. Solve the following inequation and graph the solution on the number line : \( -2\frac{2}{3} \le x + \frac{2}{3} < 3\frac{1}{3}; x \in R \).
Answer: We need to solve the compound inequation \( -2\frac{2}{3} \le x + \frac{1}{3} < 3\frac{1}{3} \). *(Note: Following the working in the source, which uses \( x + \frac{1}{3} \) instead of \( x + \frac{2}{3} \) in the middle part).* First, convert mixed fractions to improper fractions: \( -2\frac{2}{3} = -\frac{8}{3} \) and \( 3\frac{1}{3} = \frac{10}{3} \). The inequation becomes: \( -\frac{8}{3} \le x + \frac{1}{3} < \frac{10}{3} \). Now, split this into two separate inequalities: (i) \( -\frac{8}{3} \le x + \frac{1}{3} \) Subtract \( \frac{1}{3} \) from both sides:
\( -\frac{8}{3} - \frac{1}{3} \le x \)
\( -\frac{9}{3} \le x \)
\( -3 \le x \) (ii) \( x + \frac{1}{3} < \frac{10}{3} \) Subtract \( \frac{1}{3} \) from both sides:
\( x < \frac{10}{3} - \frac{1}{3} \)
\( x < \frac{9}{3} \)
\( x < 3 \) Combining both results, we get \( -3 \le x < 3 \). This means x can be -3 or any number greater than -3 but strictly less than 3. The solution set is \( \{x : -3 \le x < 3, x \in R\} \).In simple words: This problem is a compound inequality. First, convert mixed fractions and then solve the two resulting simple inequalities. You will find that \( x \) must be between -3 (including -3) and 3 (not including 3). Represent this range with a filled circle at -3, an empty circle at 3, and a shaded line in between.

🎯 Exam Tip: Always follow the provided solution steps if there's a minor discrepancy with the question. The key is to demonstrate accurate mathematical reasoning based on the given solution's logic. Ensure all mixed fractions are correctly converted.

Question 10. Solve the given inequation and graph the solution on the number line: \( 2y - 3 < y + 1 \le 4y + 7, y \in R \).
Answer: We need to solve the compound inequation \( 2y - 3 < y + 1 \le 4y + 7 \). Split this into two separate inequalities: (i) \( 2y - 3 < y + 1 \) Subtract \( y \) from both sides and add 3 to both sides:
\( 2y - y < 1 + 3 \)
\( y < 4 \) (ii) \( y + 1 \le 4y + 7 \) Subtract \( y \) from both sides and subtract 7 from both sides:
\( 1 - 7 \le 4y - y \)
\( -6 \le 3y \) Divide by 3:
\( \frac{-6}{3} \le y \)
\( -2 \le y \) Combining both results, we get \( -2 \le y < 4 \). This means y can be -2 or any number greater than -2 but strictly less than 4. The variable here is 'y', not 'x'. The solution set is \( \{y : -2 \le y < 4, y \in R\} \).In simple words: This problem asks you to solve a compound inequality for the variable 'y'. Break it into two separate inequalities, solve each one, and then combine the results. The solution means 'y' can be any number from -2 (including -2) up to, but not including, 4. Mark this range on the number line.

🎯 Exam Tip: Pay attention to the variable used in the question (e.g., \( x \) or \( y \)) and ensure your solution and number line labels reflect it. Compound inequalities with variables in all three parts are common.

Question 11. Solve the inequation and represent the solution set on the number line \( -3 + x \le \frac{8x}{3} + 2 \le \frac{14}{3} + 2x, x \in I \).
Answer: We need to solve the compound inequation \( -3 + x \le \frac{8x}{3} + 2 \le \frac{14}{3} + 2x \). Split this into two separate inequalities: (i) \( -3 + x \le \frac{8x}{3} + 2 \) Multiply by 3 to clear the fraction:
\( 3(-3 + x) \le 8x + 3(2) \)
\( -9 + 3x \le 8x + 6 \) Gather \( x \) terms and constants:
\( -9 - 6 \le 8x - 3x \)
\( -15 \le 5x \) Divide by 5:
\( \frac{-15}{5} \le x \)
\( -3 \le x \) (ii) \( \frac{8x}{3} + 2 \le \frac{14}{3} + 2x \) Multiply by 3 to clear the fractions:
\( 8x + 3(2) \le 14 + 3(2x) \)
\( 8x + 6 \le 14 + 6x \) Gather \( x \) terms and constants:
\( 8x - 6x \le 14 - 6 \)
\( 2x \le 8 \) Divide by 2:
\( x \le \frac{8}{2} \)
\( x \le 4 \) Combining both results, we get \( -3 \le x \le 4 \). Since \( x \in I \) (integers), the possible integer values are \( \{-3, -2, -1, 0, 1, 2, 3, 4\} \). This includes all whole numbers from -3 up to 4. The solution set is \( \{-3, -2, -1, 0, 1, 2, 3, 4\} \).In simple words: This problem asks you to solve a complex inequality with fractions for integer values of \( x \). Break it into two parts, clear the fractions, and solve for \( x \) in each part. Combine the results to get a range, and then mark only the whole numbers within that range on the number line using dots.

🎯 Exam Tip: When dealing with inequalities containing fractions, multiplying by the LCM of denominators is an efficient way to simplify. Always double-check the replacement set (integers vs. real numbers) before graphing the solution.

Question 12. Solve the following inequation and represent the solution set on the number line. \( -3 < -\frac{1}{2} - \frac{2x}{3} \le \frac{5}{6}, x \in R \).
Answer: We need to solve the compound inequation \( -3 < -\frac{1}{2} - \frac{2x}{3} \le \frac{5}{6} \). Multiply all parts by the LCM of the denominators (2, 3, 6), which is 6, to clear the fractions:
\( 6 \cdot (-3) < 6 \cdot (-\frac{1}{2} - \frac{2x}{3}) \le 6 \cdot (\frac{5}{6}) \)
\( -18 < -3 - 4x \le 5 \) Now, split this into two separate inequalities: (i) \( -18 < -3 - 4x \) Add 3 to both sides:
\( -18 + 3 < -4x \)
\( -15 < -4x \) Divide by -4 and reverse the inequality sign:
\( \frac{-15}{-4} > x \)
\( 3.75 > x \) (or \( x < 3.75 \)) (ii) \( -3 - 4x \le 5 \) Add 3 to both sides:
\( -4x \le 5 + 3 \)
\( -4x \le 8 \) Divide by -4 and reverse the inequality sign:
\( x \ge \frac{8}{-4} \)
\( x \ge -2 \) Combining both results, we get \( -2 \le x < 3.75 \). This means x can be -2 or any number greater than -2 but strictly less than 3.75. The solution set is \( \{x : -2 \le x < 3.75, x \in R\} \).In simple words: This problem is a compound inequality with fractions. Clear the fractions by multiplying by the common multiple. Then, split it into two simple inequalities, solve each one, and combine the results. You'll get a range for \( x \) which you plot on the number line using a filled circle for the starting point and an empty circle for the ending point, shading the line in between.

🎯 Exam Tip: Always multiply all parts of a compound inequality by the LCM of denominators to simplify it. When dividing by a negative number, remember to reverse the inequality signs for accuracy.

Question 13. Solve the following inequation and represent the solution set on the number line \( 2x - 5 \le 5x + 4 < 11 \), where \( x \in I \).
Answer: We need to solve the compound inequation \( 2x - 5 \le 5x + 4 < 11 \). Split this into two separate inequalities: (i) \( 2x - 5 \le 5x + 4 \) Subtract \( 2x \) from both sides and subtract 4 from both sides:
\( -5 - 4 \le 5x - 2x \)
\( -9 \le 3x \) Divide by 3:
\( \frac{-9}{3} \le x \)
\( -3 \le x \) (ii) \( 5x + 4 < 11 \) Subtract 4 from both sides:
\( 5x < 11 - 4 \)
\( 5x < 7 \) Divide by 5:
\( x < \frac{7}{5} \)
\( x < 1.4 \) Combining both results, we have \( -3 \le x < 1.4 \). Since \( x \in I \) (integers), the possible integer values are \( \{-3, -2, -1, 0, 1\} \). The solution set is \( \{-3, -2, -1, 0, 1\} \).