Selina Concise Solutions for ICSE Class 10 Mathematics Chapter 4 Linear Inequations In One Variable

Exercise 4A

Question 1. State, true or false:
(i) \( x < -y \Rightarrow -x > y \)
(ii) \( -5x \geq 15 \Rightarrow x \geq -3 \)
(iii) \( 2x \leq -7 \Rightarrow \frac{2x}{-4} \geq \frac{-7}{-4} \)
(iv) \( 7 > 5 \Rightarrow \frac{1}{7} < \frac{1}{5} \)
Answer:
(i) \( x < -y \Rightarrow -x > y \)
The given statement is true.

(ii) \( -5x \geq 15 \Rightarrow \frac{-5x}{-5} \geq \frac{15}{-5} \Rightarrow x \leq -3 \)
The given statement is false.

(iii) \( 2x \leq -7 \Rightarrow \frac{2x}{-4} \geq \frac{-7}{-4} \)
The given statement is true.

(iv) \( 7 > 5 \Rightarrow \frac{1}{7} < \frac{1}{5} \)
The given statement is true.
In simple words: When we multiply or divide by a negative number, the inequality sign flips. When we take reciprocals of positive numbers, bigger becomes smaller.

📝 Teacher's Note: Show students with examples: if -2x ≥ 10, then x ≤ -5. The sign flips because we divide by -2. Use a number line to show this clearly.

🎯 Exam Tip: Always remember - when you multiply or divide both sides by a negative number, flip the inequality sign. Write this rule clearly in your answer.

Question 2. State, whether the following statements are true or false:
(i) a < b, then a – c < b – c
(ii) If a > b, then a + c > b + c
(iii) If a < b, then ac > bc
(iv) If a > b, then \( \frac{a}{c} < \frac{b}{c} \)
(v) If a – c > b – d, then a + d > b + c
(vi) If a < b, and c > 0, then a – c > b – c
Where a, b, c and d are real numbers and c ≠ 0.
Answer:
(i) a < b
\( \Rightarrow \) a – c < b – c
The given statement is true.

(ii) If a > b
\( \Rightarrow \) a + c > b + c
The given statement is true.

(iii) If a < b
\( \Rightarrow \) ac < bc
The given statement is false.

(iv) If a > b
\( \Rightarrow \) \( \frac{a}{c} > \frac{b}{c} \)
The given statement is false.

(v) If a – c > b – d
\( \Rightarrow \) a + d > b + c
The given statement is true.

(vi) If a < b
\( \Rightarrow \) a – c < b – c (Since, c > 0)
The given statement is false.
In simple words: Adding or subtracting the same number keeps the inequality same. But multiplying or dividing can change things if the number is negative.

📝 Teacher's Note: Make students practice with actual numbers. Take a = 2, b = 5, c = -1. Show them how ac > bc becomes false because 2(-1) = -2 and 5(-1) = -5, so -2 > -5.

🎯 Exam Tip: Check if c is positive or negative. If c is negative, the inequality flips when you multiply or divide. Always mention this in your working.

Question 3. If x ∈ N, find the solution set of inequations.
(i) 5x + 3 ≤ 2x + 18
(ii) 3x – 2 < 19 – 4x
Answer:
(i) 5x + 3 ≤ 2x + 18
5x – 2x ≤ 18 – 3
3x ≤ 15
x ≤ 5
Since, x ∈ N, therefore solution set is {1, 2, 3, 4, 5}.

(ii) 3x – 2 < 19 – 4x
3x + 4x < 19 + 2
7x < 21
x < 3
Since, x ∈ N, therefore solution set is {1, 2}.
In simple words: We solve like normal equations but keep the inequality sign. Then we pick only natural numbers (1, 2, 3...) that satisfy our answer.

📝 Teacher's Note: Remind students that N means natural numbers = {1, 2, 3, 4, 5...}. Zero is not included. Make them list the numbers that work.

🎯 Exam Tip: Always write "Since x ∈ N" and then list the solution set in curly brackets. Don't forget to include all natural numbers that satisfy the condition.

Question 4. If the replacement set is the set of whole numbers, solve:
(i) x + 7 ≤ 11
(ii) 3x – 1 > 8
(iii) 8 – x > 5
(iv) 7 – 3x ≥ – \( \frac{1}{2} \)
(v) x – \( \frac{3}{2} \) < \( \frac{3}{2} \) – x
(vi) 18 ≤ 3x – 2
Answer:
(i) x + 7 ≤ 11
x ≤ 11 – 7
x ≤ 4
Since, the replacement set = W (set of whole numbers)
\( \Rightarrow \) Solution set = {0, 1, 2, 3, 4}

(ii) 3x – 1 > 8
3x > 8 + 1
x > 3
Since, the replacement set = W (set of whole numbers)
\( \Rightarrow \) Solution set = {4, 5, 6, …}

(iii) 8 – x > 5
– x > 5 – 8
– x > -3
x < 3
Since, the replacement set = W (set of whole numbers)
\( \Rightarrow \) Solution set = {0, 1, 2}

(iv) 7 – 3x ≥ – \( \frac{1}{2} \)
-3x ≥ – \( \frac{1}{2} \) – 7
-3x ≥ – \( \frac{15}{2} \)
x ≤ \( \frac{5}{2} \)
Since, the replacement set = W (set of whole numbers)
\( \therefore \) Solution set = {0, 1, 2}

(v) x – \( \frac{3}{2} \) < \( \frac{3}{2} \) – x
x + x < \( \frac{3}{2} \) + \( \frac{3}{2} \)
2x < 3
x < \( \frac{3}{2} \)
Since, the replacement set = W (set of whole numbers)
\( \therefore \) Solution set = {0, 1}

(vi) 18 ≤ 3x – 2
18 + 2 ≤ 3x
20 ≤ 3x
x ≥ \( \frac{20}{3} \)
Since, the replacement set = W (set of whole numbers)
\( \therefore \) Solution set = {7, 8, 9, …}
In simple words: Whole numbers include 0, 1, 2, 3... We solve the inequality and then pick only whole numbers that work.

📝 Teacher's Note: Make sure students know W = {0, 1, 2, 3, 4...} includes zero. When x ≥ 6.67, the smallest whole number is 7. Practice this concept well.

🎯 Exam Tip: Convert fractions to decimals to find the boundary. If x ≥ 6.67, then start from 7 for whole numbers. Always write the solution set clearly.

Question 5. Solve the inequation: 3 – 2x ≥ x – 12 given that x ∈ N.
Answer:
3 – 2x ≥ x – 12
-2x – x ≥ -12 – 3
-3x ≥ -15
x ≤ 5
Since, x ∈ N, therefore,
Solution set = {1, 2, 3, 4, 5}
In simple words: We rearrange the inequality by moving all x terms to one side. When we divide by -3, the inequality sign flips.

📝 Teacher's Note: Show students step by step how -3x ≥ -15 becomes x ≤ 5. The sign flips because we divide by -3. This is very important.

🎯 Exam Tip: When dividing by a negative number, always flip the inequality sign. Write this step clearly to show you understand the rule.

Question 6. If 25 – 4x ≤ 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:
25 – 4x ≤ 16
-4x ≤ 16 – 25
-4x ≤ -9
x ≥ \( \frac{9}{4} \)
x ≥ 2.25

(i) The smallest value of x, when x is a real number, is 2.25.
(ii) The smallest value of x, when x is an integer, is 3.
In simple words: When x can be any real number, the smallest value is exactly 2.25. When x must be a whole number, we take the next integer which is 3.

📝 Teacher's Note: Explain the difference between real numbers and integers. Real numbers include decimals, but integers are only whole numbers (...-2, -1, 0, 1, 2, 3...).

🎯 Exam Tip: Read carefully whether they want real numbers or integers. For integers, always round up to the next whole number when x ≥ (fraction).

Question 7. If the replacement set is the set of real numbers, solve:
(i) – 4x ≥ –16
(ii) 8 – 3x ≤ 20
(iii) 5 + \( \frac{x}{4} \) > \( \frac{x}{5} \) + 9
(iv) \( \frac{x + 3}{8} \) < \( \frac{x – 3}{5} \)
Answer:
(i) – 4x ≥ –16
x ≤ 4
Since, the replacement set of real numbers.
\( \therefore \) Solution set = {x: x ∈ R and x ≤ 4}

(ii) 8 – 3x ≤ 20
–3x ≤ 20 – 8
–3x ≤ 12
x ≥ –4
Since, the replacement set of real numbers.
\( \therefore \) Solution set = {x: x ∈ R and x ≥ –4}

(iii) 5 + \( \frac{x}{4} \) > \( \frac{x}{5} \) + 9
\( \frac{x}{4} \) – \( \frac{x}{5} \) > 9 – 5
\( \frac{x}{20} \) > 4
x > 80
Since, the replacement set of real numbers.
\( \therefore \) Solution set = {x: x ∈ R and x > 80}

(iv) \( \frac{x + 3}{8} \) < \( \frac{x – 3}{5} \)
5x + 15 < 8x – 24
5x – 8x < –24 – 15
–3x < –39
x > 13
Since, the replacement set of real numbers.
\( \therefore \) Solution set = {x: x ∈ R and x > 13}
In simple words: For real numbers, we write the solution as a set showing all numbers that work. We use R to mean all real numbers.

📝 Teacher's Note: For fractions, find common denominators. In part (iii), \( \frac{1}{4} \) – \( \frac{1}{5} \) = \( \frac{5-4}{20} \) = \( \frac{1}{20} \). Show this step clearly.

🎯 Exam Tip: For real number solutions, always write {x: x ∈ R and condition}. This is the standard notation that examiners expect to see.

Question 8. Find the smallest value of x for which \( 5 - 2x < 5\frac{1}{2} - \frac{5}{3}x \), where x is an integer.
Answer:
Step 1: Solve the inequality.
\( 5 - 2x < 5\frac{1}{2} - \frac{5}{3}x \)

Step 2: Convert mixed number to improper fraction.
\( 5 - 2x < \frac{11}{2} - \frac{5}{3}x \)

Step 3: Move x terms to one side and constants to other.
\( -2x + \frac{5}{3}x < \frac{11}{2} - 5 \)
\( -\frac{x}{3} < \frac{1}{2} \)
\( -x < \frac{3}{2} \)
\( x > -\frac{3}{2} \)
\( x > -1.5 \)

Step 4: Find the smallest integer value.
Since x is an integer and x > -1.5, the smallest value is x = -1.
In simple words: We solve the inequality step by step. The answer is x > -1.5. Since x must be an integer (whole number), the smallest value is -1.

📝 Teacher's Note: When working with mixed numbers, convert them to improper fractions first. This makes calculations easier. Also remind students that "smallest value greater than -1.5" means the first integer after -1.5.

🎯 Exam Tip: Always check if x must be integer, whole number, or real number. Write your final answer clearly: "The smallest value of x is -1."

Question 9. Find the largest value of x for which 2(x – 1) ≤ 9 – x and x ∈ W.
Answer:
Step 1: Expand the left side.
\( 2(x - 1) \leq 9 - x \)
\( 2x - 2 \leq 9 - x \)

Step 2: Move x terms to one side and constants to other.
\( 2x + x \leq 9 + 2 \)
\( 3x \leq 11 \)

Step 3: Divide both sides by 3.
\( x \leq \frac{11}{3} \)
\( x \leq 3.67 \)

Step 4: Find the largest whole number.
Since x ∈ W (whole numbers), the largest value is x = 3.
In simple words: We solve the inequality and get x ≤ 3.67. Since x must be a whole number (0, 1, 2, 3, 4...), the largest possible value is 3.

📝 Teacher's Note: W means whole numbers: {0, 1, 2, 3, 4...}. Students often forget that 0 is included in whole numbers. Show them that 3.67 means "3 and a bit more", so 3 is the largest whole number that fits.

🎯 Exam Tip: Write "x ∈ W" means whole numbers. Always find the exact decimal value first (3.67), then pick the correct whole number. Write your final answer clearly.

Question 10. Solve the inequation: \( 12 + 1\frac{5}{6}x \leq 5 + 3x \) and x ∈ R.
Answer:
Step 1: Convert mixed number to improper fraction.
\( 12 + \frac{11}{6}x \leq 5 + 3x \)

Step 2: Move x terms to one side and constants to other.
\( \frac{11}{6}x - 3x \leq 5 - 12 \)
\( -\frac{7}{6}x \leq -7 \)

Step 3: Divide by -7/6 and flip the inequality sign.
\( x \geq 6 \)

Solution set: {x: x ∈ R and x ≥ 6}
In simple words: We solve the inequality step by step. When we divide by a negative number, we flip the inequality sign. The answer is all real numbers greater than or equal to 6.

📝 Teacher's Note: This is the most important rule in inequalities - when you multiply or divide by a negative number, flip the inequality sign. Use a simple example like -2x > 4 becomes x < -2.

🎯 Exam Tip: Always write the solution set in proper notation: {x: x ∈ R and x ≥ 6}. Don't forget to flip the inequality when dividing by negative numbers.

Question 11. Given x ∈ {integers}, find the solution set of: -5 ≤ 2x – 3 < x + 2
Answer:
Step 1: Split into two inequalities.
\( -5 \leq 2x - 3 \) and \( 2x - 3 < x + 2 \)

Step 2: Solve first inequality.
\( -5 + 3 \leq 2x \)
\( -2 \leq 2x \)
\( x \geq -1 \)

Step 3: Solve second inequality.
\( 2x - x < 2 + 3 \)
\( x < 5 \)

Step 4: Combine conditions.
\( x \geq -1 \) and \( x < 5 \)
Since x ∈ {integers}

Solution set: {-1, 0, 1, 2, 3, 4}
In simple words: We split the compound inequality into two parts. Then we find x ≥ -1 and x < 5. The integers in this range are -1, 0, 1, 2, 3, 4.

📝 Teacher's Note: Compound inequalities have two conditions. Solve each part separately, then find numbers that satisfy both conditions. Use a number line to show students which integers fit.

🎯 Exam Tip: Always write the solution set as a list when dealing with integers: {-1, 0, 1, 2, 3, 4}. Don't forget to check both boundary values carefully.

Question 12. Given x ∈ {whole numbers}, find the solution set of: -1 ≤ 3 + 4x < 23
Answer:
Step 1: Split into two inequalities.
\( -1 \leq 3 + 4x \) and \( 3 + 4x < 23 \)

Step 2: Solve first inequality.
\( -4 \leq 4x \)
\( x \geq -1 \)

Step 3: Solve second inequality.
\( 4x < 20 \)
\( x < 5 \)

Step 4: Combine conditions.
\( x \geq -1 \) and \( x < 5 \)
Since x ∈ {whole numbers}

Solution set: {0, 1, 2, 3, 4}
In simple words: We get x ≥ -1 and x < 5. Since x must be a whole number (0, 1, 2, 3...), we pick 0, 1, 2, 3, 4. Note that -1 is not a whole number.

📝 Teacher's Note: Whole numbers start from 0, not -1. Even though x ≥ -1, we cannot include -1 because it's not a whole number. Remind students: whole numbers = {0, 1, 2, 3, 4...}

🎯 Exam Tip: Remember the difference between integers {...-2, -1, 0, 1, 2...} and whole numbers {0, 1, 2, 3...}. Write the final set clearly: {0, 1, 2, 3, 4}.

Exercise 4B

Question 1. Represent the following inequalities on real number lines:
(i) 2x – 1 < 5
(ii) 3x + 1 ≥ –5
(iii) 2(2x – 3) ≤ 6
(iv) – 4 < x < 4
(v) – 2 ≤ x < 5
(vi) 8 ≥ x > –3
(vii) – 5 < x ≤ –1
Answer:

(i) 2x – 1 < 5
2x < 6
x < 3
Solution on number line: Open circle at 3, arrow pointing left.

[Diagram: Number line showing x < 3 with open circle at 3 and arrow extending left]

(ii) 3x + 1 ≥ –5
3x ≥ –6
x ≥ –2
Solution on number line: Closed circle at –2, arrow pointing right.

[Diagram: Number line showing x ≥ -2 with closed circle at -2 and arrow extending right]

(iii) 2(2x – 3) ≤ 6
2x – 3 ≤ 3
2x ≤ 6
x ≤ 3
Solution on number line: Closed circle at 3, arrow pointing left.

[Diagram: Number line showing x ≤ 3 with closed circle at 3 and arrow extending left]

(iv) – 4 < x < 4
Solution on number line: Open circles at –4 and 4, line connecting them.

[Diagram: Number line showing -4 < x < 4 with open circles at both -4 and 4, with line segment between them]

(v) – 2 ≤ x < 5
Solution on number line: Closed circle at –2, open circle at 5, line connecting them.

[Diagram: Number line showing -2 ≤ x < 5 with closed circle at -2, open circle at 5, and line segment between them]

(vi) 8 ≥ x > –3
This is the same as –3 < x ≤ 8
Solution on number line: Open circle at –3, closed circle at 8, line connecting them.

[Diagram: Number line showing -3 < x ≤ 8 with open circle at -3, closed circle at 8, and line segment between them]

(vii) – 5 < x ≤ –1
Solution on number line: Open circle at –5, closed circle at –1, line connecting them.

[Diagram: Number line showing -5 < x ≤ -1 with open circle at -5, closed circle at -1, and line segment between them]

In simple words: We solve each inequality to find the range of x values. Then we draw number lines with open circles for < or > (not equal) and closed circles for ≤ or ≥ (equal allowed).

📝 Teacher's Note: Use different colored pens for open and closed circles. Show students that < and > use open circles (hollow), while ≤ and ≥ use closed circles (filled). This is very important for graphing.

🎯 Exam Tip: Always solve the inequality first, then draw the number line. Open circle = not included, closed circle = included. Draw arrows for one-sided inequalities, line segments for two-sided inequalities.

Question 2. For each graph given, write an inequation taking x as the variable:
Answer:

(i) From the graph: Closed circle at –1, arrow pointing right
Inequation: x ≥ –1, x ∈ R

(ii) From the graph: Closed circle at 2, arrow pointing right
Inequation: x ≥ 2, x ∈ R

(iii) From the graph: Closed circle at –4, open circle at 3, line segment between them
Inequation: –4 ≤ x < 3, x ∈ R

(iv) From the graph: Open circle at –1, closed circle at 5, line segment between them
Inequation: –1 < x ≤ 5, x ∈ R

In simple words: We look at each number line and write the inequality. Closed circles mean we include that number (≤ or ≥). Open circles mean we don't include that number (< or >).

📝 Teacher's Note: Teach students to read number lines from left to right. Closed circle = filled dot = include the number. Open circle = hollow dot = don't include the number. Practice with many examples.

🎯 Exam Tip: Look carefully at whether circles are open or closed. Write the inequation exactly as shown on the graph. Don't forget to mention x ∈ R (real numbers) when asked.

Question 3. For the following inequations, graph the solution set on the real number line:
(i) – 4 ≤ 3x – 1 < 8
(ii) x – 1 < 3 – x ≤ 5
Answer:

(i) – 4 ≤ 3x – 1 < 8
–4 ≤ 3x – 1 and 3x – 1 < 8
–1 ≤ x and x < 3
The solution set on the real number line: Closed circle at –1, open circle at 3, line segment between them.

[Diagram: Number line showing -1 ≤ x < 3 with closed circle at -1, open circle at 3, and line segment between them]

(ii) x – 1 < 3 – x ≤ 5
x – 1 < 3 – x and 3 – x ≤ 5
2x < 4 and –x ≤ 2
x < 2 and x ≥ –2
The solution set on the real number line: Closed circle at –2, open circle at 2, line segment between them.

[Diagram: Number line showing -2 ≤ x < 2 with closed circle at -2, open circle at 2, and line segment between them]

In simple words: We split compound inequalities into two parts, solve each separately, then combine the conditions. The final answer is where both conditions are true at the same time.

📝 Teacher's Note: Show students how to split compound inequalities into two separate inequalities. Solve each part, then find the overlap. Use colored pencils to show the overlapping region on the number line.

🎯 Exam Tip: Always split compound inequalities into two parts first. Solve each part separately, then find where both conditions are satisfied. Draw the number line carefully with correct circles.

Question 4. Represent the solution of each of the following inequalities on the real number line:
(i) 4x - 1 > x + 11
(ii) 7 - x ≤ 2 - 6x
(iii) x + 3 ≤ 2x + 9
(iv) 2 - 3x > 7 - 5x
(v) 1 + x ≥ 5x - 11
(vi) \( \frac{2x + 5}{3} > 3x - 3 \)

Answer:

(i) 4x - 1 > x + 11
3x > 12
x > 4

The solution on number line is: x > 4

[Diagram: Number line showing open circle at 4 with arrow pointing right, indicating x > 4]

(ii) 7 - x ≤ 2 - 6x
5x ≤ -5
x ≤ -1

The solution on number line is: x ≤ -1

[Diagram: Number line showing filled circle at -1 with arrow pointing left, indicating x ≤ -1]

(iii) x + 3 ≤ 2x + 9
-6 ≤ x

The solution on number line is: x ≥ -6

[Diagram: Number line showing filled circle at -6 with arrow pointing right, indicating x ≥ -6]

(iv) 2 - 3x > 7 - 5x
2x > 5
x > \( \frac{5}{2} \)
x > 2.5

The solution on number line is: x > 2.5

[Diagram: Number line showing open circle at 2.5 with arrow pointing right, indicating x > 2.5]

(v) 1 + x ≥ 5x - 11
12 ≥ 4x
3 ≥ x

The solution on number line is: x ≤ 3

[Diagram: Number line showing filled circle at 3 with arrow pointing left, indicating x ≤ 3]

(vi) \( \frac{2x + 5}{3} > 3x - 3 \)
2x + 5 > 9x - 9
-7x > -14
x < 2

The solution on number line is: x < 2

[Diagram: Number line showing open circle at 2 with arrow pointing left, indicating x < 2]

In simple words: We solve each inequality by moving terms to one side. Then we show the answer on a number line with circles and arrows.

📝 Teacher's Note: Show students that when we multiply or divide by a negative number, the inequality sign flips. Use a simple example like -2x > 4 becomes x < -2. Draw the number line clearly with open and closed circles.

🎯 Exam Tip: Always check your answer by putting a test value back into the original inequality. Draw number lines neatly with proper circles and arrows. Open circle means "not equal", filled circle means "can be equal".

Question 5. x ∈ {real numbers} and -1 < 3 - 2x ≤ 7, evaluate x and represent it on a number line.

Answer:
-1 < 3 - 2x ≤ 7
-1 < 3 - 2x and 3 - 2x ≤ 7
2x < 4 and -2x ≤ 4
x < 2 and x ≥ -2

Solution set = {-2 ≤ x < 2, x ∈ R}

Thus, the solution can be represented on a number line as: -2 ≤ x < 2

[Diagram: Number line showing filled circle at -2 and open circle at 2 with line segment between them, indicating -2 ≤ x < 2]

In simple words: This is a compound inequality. We break it into two parts and solve both. The answer is all numbers between -2 and 2, including -2 but not 2.

📝 Teacher's Note: Explain that compound inequalities have two conditions that must both be true. Show students how to split -1 < 3-2x ≤ 7 into two separate inequalities. Use a thermometer example - temperature between two values.

🎯 Exam Tip: Always split compound inequalities into two separate inequalities. Solve each one, then combine the results. Remember: filled circle means "included", open circle means "not included".

Question 6. List the elements of the solution set of the inequation -3 < x - 2 ≤ 9 - 2x; x ∈ N.

Answer:
-3 < x - 2 ≤ 9 - 2x
-3 < x - 2 and x - 2 ≤ 9 - 2x
-1 < x and 3x ≤ 11
-1 < x ≤ \( \frac{11}{3} \)

Since x ∈ N (natural numbers)
∴ Solution set = {1, 2, 3}

In simple words: We solve the compound inequality and get -1 < x ≤ 3.67 (approximately). Since x must be a natural number (1, 2, 3, 4...), we only take 1, 2, and 3.

📝 Teacher's Note: Remind students that N means natural numbers = {1, 2, 3, 4, 5...}. After solving, we only pick the natural numbers that satisfy our condition. Use counting numbers as an example.

🎯 Exam Tip: First solve the inequality completely. Then apply the given condition (like x ∈ N). List only the numbers that satisfy both the inequality and the given set condition.

Question 7. Find the range of values of x which satisfies -2\( \frac{2}{3} \) ≤ x + \( \frac{1}{3} \) < 3\( \frac{1}{3} \); x ∈ R. Graph these values of x on the number line.

Answer:
-2\( \frac{2}{3} \) ≤ x + \( \frac{1}{3} \) and x + \( \frac{1}{3} \) < 3\( \frac{1}{3} \)

\( \implies \) -\( \frac{8}{3} \) ≤ x + \( \frac{1}{3} \) and x + \( \frac{1}{3} \) < \( \frac{10}{3} \)

\( \implies \) -\( \frac{8}{3} \) - \( \frac{1}{3} \) ≤ x and x < \( \frac{10}{3} \) - \( \frac{1}{3} \)

\( \implies \) -\( \frac{9}{3} \) ≤ x and x < \( \frac{9}{3} \)

\( \implies \) -3 ≤ x and x < 3

∴ -3 ≤ x < 3

The required graph of the solution set is: -3 ≤ x < 3

[Diagram: Number line showing filled circle at -3 and open circle at 3 with line segment between them]

In simple words: We convert mixed numbers to improper fractions, then solve the compound inequality. The answer is all numbers from -3 to 3, including -3 but not 3.

📝 Teacher's Note: Show students how to convert mixed numbers to improper fractions first. Then solve step by step. Use pizza slices to explain fractions - makes it easier to understand.

🎯 Exam Tip: Convert all mixed numbers to improper fractions before solving. Work with fractions carefully. Show all steps clearly for full marks.

Question 8. Find the values of x, which satisfy the inequation: -2 ≤ \( \frac{1}{2} \) - \( \frac{2x}{3} \) < 1\( \frac{5}{6} \); x ∈ N. Graph the solution on the number line.

Answer:
-2 ≤ \( \frac{1}{2} \) - \( \frac{2x}{3} \) < 1\( \frac{5}{6} \)

-2 ≤ \( \frac{1}{2} \) - \( \frac{2x}{3} \) and \( \frac{1}{2} \) - \( \frac{2x}{3} \) < 1\( \frac{5}{6} \)

-\( \frac{5}{2} \) ≤ -\( \frac{2x}{3} \) and -\( \frac{2x}{3} \) < \( \frac{8}{6} \)

\( \frac{15}{4} \) ≥ x and x > -2

3.75 ≥ x and x > -2

Since x ∈ N
∴ Solution set = {1, 2, 3}

The required graph of the solution set is: {1, 2, 3}

[Diagram: Number line showing filled circles at points 1, 2, and 3]

In simple words: We solve the compound inequality and get -2 < x ≤ 3.75. Since x must be a natural number, we only take 1, 2, and 3.

📝 Teacher's Note: When dividing by a negative fraction, remember to flip the inequality sign. Convert 1⅚ to 11/6 first. Show each step on the board clearly.

🎯 Exam Tip: Be very careful with signs when multiplying or dividing by negative numbers. Convert mixed numbers to improper fractions. Apply the set condition (N) at the end.

Question 9. Given x ∈ {real numbers}, find the range of values of x for which -5 ≤ 2x - 3 < x + 2 and represent it on a number line.

Answer:
-5 ≤ 2x - 3 < x + 2
-5 ≤ 2x - 3 and 2x - 3 < x + 2
-2 ≤ 2x and x < 5
-1 ≤ x and x < 5

Required range is -1 ≤ x < 5.

The required graph is: -1 ≤ x < 5

[Diagram: Number line showing filled circle at -1 and open circle at 5 with line segment between them]

In simple words: We solve the compound inequality step by step. The answer includes all numbers from -1 to 5, including -1 but not 5.

📝 Teacher's Note: Break compound inequalities into two parts. Solve each part separately, then combine the results. Use a ruler to show the range of values on a number line.

🎯 Exam Tip: Write both inequalities clearly. Solve each one step by step. Combine the results correctly. Draw the number line with proper circles and shading.

Question 10. If 5x - 3 ≤ 5 + 3x ≤ 4x + 2, express it as a ≤ x ≤ b and then state the values of a and b.

Answer:
5x - 3 ≤ 5 + 3x ≤ 4x + 2
5x - 3 ≤ 5 + 3x and 5 + 3x ≤ 4x + 2
2x ≤ 8 and -x ≤ -3
x ≤ 4 and x ≥ 3

Thus, 3 ≤ x ≤ 4.
Hence, a = 3 and b = 4.

In simple words: We solve the compound inequality and get x between 3 and 4 (including both). So a = 3 and b = 4.

📝 Teacher's Note: Show students that the final answer must be in the form a ≤ x ≤ b. When we get x ≤ 4 and x ≥ 3, we combine them as 3 ≤ x ≤ 4.

🎯 Exam Tip: Always write the final answer in the required format a ≤ x ≤ b. Clearly state the values of a and b. Check your answer by substituting back.

Question 11. Solve the following inequation and graph the solution set on the number line: 2x - 3 < x + 2 ≤ 3x + 5, x ∈ R.

Answer:
2x - 3 < x + 2 ≤ 3x + 5
2x - 3 < x + 2 and x + 2 ≤ 3x + 5
x < 5 and -3 ≤ 2x
x < 5 and -1.5 ≤ x

Solution set = {-1.5 ≤ x < 5}

The solution set can be graphed on the number line as: -1.5 ≤ x < 5

[Diagram: Number line showing filled circle at -1.5 and open circle at 5 with line segment between them]

In simple words: We solve both parts of the compound inequality. The answer is all numbers from -1.5 to 5, including -1.5 but not 5.

📝 Teacher's Note: Explain that -1.5 is the same as -3/2. Students often get confused with decimal and fraction forms. Show both forms on the number line.

🎯 Exam Tip: Write the solution set in proper mathematical notation with curly brackets. Draw the number line clearly with correct circles and shading between them.

Question 12. Solve and graph the solution set of:
(i) 2x - 9 < 7 and 3x + 9 ≤ 25, x ∈ R
(ii) 2x - 9 ≤ 7 and 3x + 9 > 25, x ∈ I
(iii) x + 5 ≥ 4(x - 1) and 3 - 2x < -7, x ∈ R

Answer:
(i) 2x - 9 < 7 and 3x + 9 ≤ 25, x ∈ R
2x - 9 < 7 gives 2x < 16, so x < 8
3x + 9 ≤ 25 gives 3x ≤ 16, so x ≤ \( \frac{16}{3} \) ≈ 5.33

Solution: x ≤ \( \frac{16}{3} \)

[Diagram: Number line showing filled circle at 16/3 with arrow pointing left]

(ii) 2x - 9 ≤ 7 and 3x + 9 > 25, x ∈ I
2x - 9 ≤ 7 gives x ≤ 8
3x + 9 > 25 gives x > \( \frac{16}{3} \) ≈ 5.33

Since x ∈ I (integers): \( \frac{16}{3} \) < x ≤ 8
Solution set = {6, 7, 8}

[Diagram: Number line showing filled circles at points 6, 7, and 8]

(iii) x + 5 ≥ 4(x - 1) and 3 - 2x < -7, x ∈ R
x + 5 ≥ 4x - 4 gives -3x ≥ -9, so x ≤ 3
3 - 2x < -7 gives -2x < -10, so x > 5

Since we need x ≤ 3 AND x > 5, there is no solution.
Solution set = ∅ (empty set)

In simple words: For part (i), we take the smaller range. For part (ii), we find integers in the range. For part (iii), there are no numbers that satisfy both conditions.

📝 Teacher's Note: Show students how "AND" means both conditions must be true. In part (iii), no number can be both ≤3 and >5 at the same time, so the answer is empty set.

🎯 Exam Tip: For "AND" problems, find the intersection (overlap) of solutions. For integer problems, list all integers in the range. If there's no overlap, write "no solution" or ∅.

Question 13. Solve and graph the solution set of:
(i) \( 3x - 2 > 19 \) or \( 3 - 2x \geq -7 \), \( x \in R \)
(ii) \( 5 > p - 1 > 2 \) or \( 7 \leq 2p - 1 \leq 17 \), \( p \in R \)
Answer:
(i) \( 3x - 2 > 19 \) or \( 3 - 2x \geq -7 \)

Step 1: Solve first inequality
\( 3x - 2 > 19 \)
\( 3x > 21 \)
\( x > 7 \)

Step 2: Solve second inequality
\( 3 - 2x \geq -7 \)
\( -2x \geq -10 \)
\( x \leq 5 \)

Step 3: Find solution set
Graph of solution set of \( x > 7 \) or \( x \leq 5 \) = Graph of points which belong to \( x > 7 \) or \( x \leq 5 \) or both.

[Diagram: Number line showing \( x \leq 5 \) as a solid line from negative infinity to 5 (closed circle), and \( x > 7 \) as a solid line from 7 (open circle) to positive infinity]

(ii) \( 5 > p - 1 > 2 \) or \( 7 \leq 2p - 1 \leq 17 \)

Step 1: Solve first compound inequality
\( 5 > p - 1 > 2 \)
\( 6 > p > 3 \) or \( 3 < p < 6 \)

Step 2: Solve second compound inequality
\( 7 \leq 2p - 1 \leq 17 \)
\( 8 \leq 2p \leq 18 \)
\( 4 \leq p \leq 9 \)

Step 3: Find solution set
Graph of solution set of \( 6 > p > 3 \) or \( 4 \leq p \leq 9 \) = Graph of points which belong to \( 3 < p < 6 \) or \( 4 \leq p \leq 9 \) or both = Graph of points which belong to \( 3 < p \leq 9 \)

[Diagram: Number line showing solution from 3 (open circle) to 9 (closed circle)]

In simple words: For "or" problems, we solve each part separately. Then we take all values that work in any part. The final answer includes all these values together.

📝 Teacher's Note: Show students that "or" means "either this OR that OR both". Draw separate number lines first, then combine them. This makes it clear why we get the union of ranges.

🎯 Exam Tip: Always solve each inequality separately first. Then combine using "or" logic. Write the final answer clearly as a single range when possible.

Question 14. The diagram represents two inequations A and B on real number lines:
[Diagram: Two number lines showing A from -2 to 5 (closed interval) and B from -4 to 3 (closed at -4, open at 3)]
(i) Write down A and B in set builder notation.
(ii) Represent \( A \cup B \) and \( A \cap B' \) on two different number lines.
Answer:
(i) From the diagrams:
\( A = \{x \in R: -2 \leq x < 5\} \)
\( B = \{x \in R: -4 \leq x < 3\} \)

(ii) \( A \cap B = \{x \in R: -2 \leq x < 3\} \)
It can be represented on number line as:
[Diagram: Number line from -2 (closed) to 3 (open)]

\( B' = \{x \in R: x < -4 \text{ or } x \geq 3\} \)
\( A \cap B' = \{x \in R: 3 \leq x < 5\} \)
It can be represented on number line as:
[Diagram: Number line from 3 (closed) to 5 (open)]

In simple words: A ∩ B means values that are in both A and B. B' means all values NOT in B. A ∩ B' means values in A but not in B.

📝 Teacher's Note: Use Venn diagrams to show intersection and complement. Students understand better when they see the overlapping regions visually.

🎯 Exam Tip: For B', remember to write "x < -4 OR x ≥ 3". Don't forget the complement includes values on both sides of the original interval.

Question 15. Use real number line to find the range of values of x for which:
(i) \( x > 3 \) and \( 0 < x < 6 \)
(ii) \( x < 0 \) and \( -3 \leq x < 1 \)
(iii) \( -1 < x \leq 6 \) and \( -2 \leq x \leq 3 \)
Answer:
(i) \( x > 3 \) and \( 0 < x < 6 \)
Both the given inequations are true in the range where their graphs on the real number lines overlap.

The graphs of the given inequations can be drawn as:
[Diagram: Two number lines - first showing x > 3, second showing 0 < x < 6]

From both graphs, it is clear that their common range is \( 3 < x < 6 \)

(ii) \( x < 0 \) and \( -3 \leq x < 1 \)
Both the given inequations are true in the range where their graphs on the real number lines overlap.

The graphs of the given inequations can be drawn as:
[Diagram: Two number lines - first showing x < 0, second showing -3 ≤ x < 1]

From both graphs, it is clear that their common range is \( -3 \leq x < 0 \)

(iii) \( -1 < x \leq 6 \) and \( -2 \leq x \leq 3 \)
Both the given inequations are true in the range where their graphs on the real number lines overlap.

The graphs of the given inequations can be drawn as:
[Diagram: Two number lines - first showing -1 < x ≤ 6, second showing -2 ≤ x ≤ 3]

From both graphs, it is clear that their common range is \( -1 < x \leq 3 \)

In simple words: For "and" problems, we find where both conditions are true at the same time. This is the overlap region on the number line.

📝 Teacher's Note: Draw both inequalities on separate number lines first. Then show students how to find the common region by looking at where the thick lines overlap.

🎯 Exam Tip: For "and" problems, take the intersection. For "or" problems, take the union. Draw the number lines clearly to avoid mistakes.

Question 16. Illustrate the set \( \{x: -3 \leq x < 0 \text{ or } x > 2, x \in R\} \) on the real number line.
Answer:
Graph of solution set of \( -3 \leq x < 0 \) or \( x > 2 \) = Graph of points which belong to \( -3 \leq x < 0 \) or \( x > 2 \) or both.

Thus, the required graph is:
[Diagram: Number line showing two separate regions - from -3 (closed) to 0 (open), and from 2 (open) to positive infinity]

In simple words: This set includes two separate pieces - values from -3 to 0 (including -3 but not 0), and all values greater than 2.

📝 Teacher's Note: Show students that "or" creates separate regions on the number line. Unlike "and" which gives one connected region, "or" can give disconnected pieces.

🎯 Exam Tip: Use different symbols for open circles (not included) and closed circles (included). Be very careful with the inequality symbols.

Question 17. Given \( A = \{x: -1 < x \leq 5, x \in R\} \) and \( B = \{x: -4 \leq x < 3, x \in R\} \)
Represent on different number lines:
(i) \( A \cap B \)
(ii) \( A' \cap B \)
(iii) \( A - B \)
Answer:
(i) \( A \cap B = \{x: -1 < x < 3, x \in R\} \)
It can be represented on a number line as:
[Diagram: Number line from -1 (open) to 3 (open)]

(ii) Numbers which belong to B but do not belong to A = \( B - A \)
\( A' \cap B = \{x: -4 \leq x \leq -1, x \in R\} \)
It can be represented on a number line as:
[Diagram: Number line from -4 (closed) to -1 (closed)]

(iii) \( A - B = \{x: 3 \leq x \leq 5, x \in R\} \)
It can be represented on a number line as:
[Diagram: Number line from 3 (closed) to 5 (closed)]

In simple words: A ∩ B is the common part. A' ∩ B is the part of B that is not in A. A - B is the part of A that is not in B.

📝 Teacher's Note: Use two different colored pens to shade A and B on the same line. Students can easily see the intersections and differences when colors overlap.

🎯 Exam Tip: Always check your endpoints carefully. Open circles and closed circles make a big difference in set operations.

Question 18. P is the solution set of \( 7x - 2 > 4x + 1 \) and Q is the solution set of \( 9x - 45 \geq 5(x - 5) \);
where \( x \in R \). Represent:
(i) \( P \cap Q \)
(ii) \( P - Q \)
(iii) \( P \cap Q' \)
on different number lines.
Answer:
Step 1: Find P
\( P = \{x : 7x - 2 > 4x + 1, x \in R\} \)
\( 7x - 2 > 4x + 1 \)
\( 7x - 4x > 1 + 2 \)
\( 3x > 3 \)
\( x > 1 \)

Step 2: Find Q
\( Q = \{x : 9x - 45 \geq 5(x - 5), x \in R\} \)
\( 9x - 45 \geq 5x - 25 \)
\( 9x - 5x \geq -25 + 45 \)
\( 4x \geq 20 \)
\( x \geq 5 \)

Step 3: Find required sets
(i) \( P \cap Q = \{x : x \geq 5, x \in R\} \)
[Diagram: Number line from 5 (closed) to positive infinity]

(ii) \( P - Q = \{x : 1 < x < 5, x \in R\} \)
[Diagram: Number line from 1 (open) to 5 (open)]

(iii) \( P \cap Q' = \{x : 1 < x < 5, x \in R\} \)
[Diagram: Number line from 1 (open) to 5 (open)]

In simple words: P is all values greater than 1. Q is all values greater than or equal to 5. P ∩ Q is where both are true (x ≥ 5). P - Q is P but not Q (between 1 and 5).

📝 Teacher's Note: Show that P - Q and P ∩ Q' give the same result. This helps students understand the relationship between complement and difference operations.

🎯 Exam Tip: Solve each inequality completely before finding intersections or differences. Write P and Q clearly first, then work on the combinations.

Question 19. Find the range of values of x, which satisfy:
\( -\frac{1}{3} < \frac{x}{2} + 1 \frac{2}{3} < 5 \frac{1}{6} \)
Graph, in each of the following cases, the values of x on the different real number lines:
(i) x ∈ W (ii) x ∈ Z (iii) x ∈ R
Answer:
\( -\frac{1}{3} < \frac{x}{2} + 1 \frac{2}{3} < 5 \frac{1}{6} \)
\( -\frac{1}{3} - \frac{5}{3} < \frac{x}{2} < \frac{31}{6} - \frac{5}{3} \)
\( -\frac{6}{3} < \frac{x}{2} < \frac{21}{6} \)
\( -4 ≤ x < 7 \)

(i)If x ∈ W, range of values of x is {0, 1, 2, 3, 4, 5, 6}.

[Diagram: Number line showing filled circles at points 0, 1, 2, 3, 4, 5, 6]

(ii) If x ∈ Z, range of values of x is {-4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6}.

[Diagram: Number line showing filled circles at integer points from -4 to 6]

(iii)If x ∈ R, range of values of x is -4 ≤ x < 7.

[Diagram: Number line with filled circle at -4 and open circle at 7, with a line connecting them]

In simple words: We solve the inequality step by step. Then we show the answer on number lines for different types of numbers - whole numbers W, integers Z, and real numbers R.

📝 Teacher's Note: Show students that W means whole numbers (0, 1, 2...), Z means integers (...-2, -1, 0, 1, 2...), and R means all real numbers. Use different colored dots on number lines to make it clear.

🎯 Exam Tip: Always write the final inequality clearly. For number lines, use filled circles for ≤ or ≥, and open circles for < or >. Don't forget to show all three cases separately.

Question 20. Given: A = {x: -8 < 5x + 2 ≤ 17, x ∈ I}, B = {x: -2 ≤ 7 + 3x < 17, x ∈ R}
Where R = {real numbers} and I = {integers}. Represent A and B on two different number lines. Write down the elements of A ∩ B.
Answer:
A = {x: -8 < 5x + 2 ≤ 17, x ∈ I}
= {x: -10 < 5x ≤ 15, x ∈ I}
= {x: -2 < x ≤ 3, x ∈ I}
It can be represented on number line as: A = {-1, 0, 1, 2, 3}

[Diagram: Number line showing filled circles at -1, 0, 1, 2, 3]

B = {x: -2 ≤ 7 + 3x < 17, x ∈ R}
= {x: -9 ≤ 3x < 10, x ∈ R}
= {x: -3 ≤ x < 3.33, x ∈ R}
It can be represented on number line as:

[Diagram: Number line with filled circle at -3 and open circle at 3.33, with a line connecting them]

A ∩ B = {-1, 0, 1, 2, 3}
In simple words: We solve each inequality separately. Set A has integers from -1 to 3. Set B has all real numbers from -3 to 3.33. The intersection A ∩ B has the common elements.

📝 Teacher's Note: Explain that I means integers (whole numbers and their negatives). Show that when we find A ∩ B, we take only the values that are in both sets.

🎯 Exam Tip: Solve each inequality step by step. Draw number lines clearly. For A ∩ B, list only the integers that fall in the range of B. Write the final answer as a set.

Question 21. Solve the following inequation and represent the solution set on the number line 2x - 5 ≤ 5x + 4 < 11, where x ∈ I
Answer:
2x - 5 ≤ 5x + 4 and 5x + 4 < 11
2x - 5x ≤ 4 + 5 and 5x < 11 - 4
-3x ≤ 9 and 5x < 7
x ≥ -3 and x < \( \frac{7}{5} \)
x ≥ -3 and x < 1.4
Since x ∈ I, the solution set is {-3, -2, -1, 0, 1}
And the number line representation is:

[Diagram: Number line showing filled circles at -3, -2, -1, 0, 1]

In simple words: We solve the compound inequality by breaking it into two parts. Then we find integers that satisfy both conditions. The answer is integers from -3 to 1.

📝 Teacher's Note: Show students how to split a compound inequality into two separate inequalities. Remember to flip the inequality sign when dividing by a negative number.

🎯 Exam Tip: Always write both inequalities separately first. When x ∈ I, list only the integer values. Mark all integer solutions on the number line with filled circles.

Question 22. Given that x ∈ I, solve the inequation and graph the solution on the number line:
\( 3 ≥ \frac{x-4}{2} + \frac{x}{3} ≥ 2 \)
Answer:
\( 3 ≥ \frac{x-4}{2} + \frac{x}{3} ≥ 2 \)
\( 3 ≥ \frac{3x-12+2x}{6} ≥ 2 \)
\( 18 ≥ 5x - 12 ≥ 12 \)
\( 30 ≥ 5x ≥ 24 \)
\( 6 ≥ x ≥ 4.8 \)
Solution set = {5, 6}
It can be graphed on number line as:

[Diagram: Number line showing filled circles at 5 and 6]

In simple words: We solve the compound inequality by finding a common denominator. Then we solve for x and find the integers in that range.

📝 Teacher's Note: When combining fractions, find the LCD (lowest common denominator). Show students how to multiply through by the LCD to clear fractions.

🎯 Exam Tip: Write each step clearly when combining fractions. Since x ∈ I, only list integer values. Check your answer by substituting back into the original inequality.

Question 23. Given:
A = {x: 11x - 5 > 7x + 3, x ∈ R} and
B = {x: 18x - 9 ≥ 15 + 12x, x ∈ R}.
Find the range of set A ∩ B and represent it on number line.
Answer:
A = {x: 11x - 5 > 7x + 3, x ∈ R}
= {x: 4x > 8, x ∈ R}
= {x: x > 2, x ∈ R}

B = {x: 18x - 9 ≥ 15 + 12x, x ∈ R}
= {x: 6x ≥ 24, x ∈ R}
= {x: x ≥ 4, x ∈ R}

A ∩ B = {x: x ≥ 4, x ∈ R}
It can be represented on number line as:

[Diagram: Number line with filled circle at 4 and arrow extending to the right]

In simple words: Set A has all numbers greater than 2. Set B has all numbers greater than or equal to 4. The intersection is numbers greater than or equal to 4.

📝 Teacher's Note: Draw both sets A and B on separate number lines first, then show their intersection. Explain that A ∩ B means values that belong to both sets.

🎯 Exam Tip: Solve each inequality separately first. For intersection, take the more restrictive condition. Use filled circle for ≥ and arrow for "greater than."

Question 24. Find the set of values of x, satisfying:
\( 7x + 3 ≥ 3x - 5 \) and \( \frac{x}{4} - 5 ≤ \frac{5}{4} - x \), where x ∈ N.
Answer:
7x + 3 ≥ 3x - 5
4x ≥ -8
x ≥ -2

\( \frac{x}{4} - 5 ≤ \frac{5}{4} - x \)
\( \frac{x}{4} + x ≤ \frac{5}{4} + 5 \)
\( \frac{5x}{4} ≤ \frac{25}{4} \)
x ≤ 5

Since, x ∈ N
∴ Solution set = {1, 2, 3, 4, 5}
In simple words: We solve both inequalities. We need x ≥ -2 and x ≤ 5. Since x ∈ N (natural numbers), we take positive integers from 1 to 5.

📝 Teacher's Note: Remind students that N means natural numbers (1, 2, 3, ...). Natural numbers don't include 0 or negative numbers.

🎯 Exam Tip: Solve both inequalities separately. Remember N means natural numbers, so start from 1, not 0. List all natural numbers in the range.

Question 25. Solve:
(i) \( \frac{x}{2} + 5 ≤ \frac{x}{3} + 6 \), where x is a positive odd integer.
(ii) \( \frac{2x + 3}{3} ≥ \frac{3x - 1}{4} \), where x is a positive even integer.
Answer:
(i) \( \frac{x}{2} + 5 ≤ \frac{x}{3} + 6 \)
\( \frac{x}{2} - \frac{x}{3} ≤ 6 - 5 \)
\( \frac{x}{6} ≤ 1 \)
x ≤ 6

Since, x is a positive odd integer
∴ Solution set = {1, 3, 5}

(ii) \( \frac{2x + 3}{3} ≥ \frac{3x - 1}{4} \)
8x + 12 ≥ 9x - 3
-x ≥ -15
x ≤ 15

Since, x is a positive even integer
∴ Solution set = {2, 4, 6, 8, 10, 12, 14}
In simple words: For part (i), we find x ≤ 6 and take positive odd numbers. For part (ii), we find x ≤ 15 and take positive even numbers.

📝 Teacher's Note: Explain that positive odd integers are 1, 3, 5, 7, ... and positive even integers are 2, 4, 6, 8, ... Help students list them systematically.

🎯 Exam Tip: After solving the inequality, carefully list only the type of integers asked for. Check that all your answers satisfy both the inequality and the given condition (odd/even).

Question 26. Solve the inequation:
\( -2\frac{1}{2} + 2x ≤ \frac{4x}{5} ≤ \frac{4}{3} + 2x \), x ∈ W. Graph the solution set on the number line.
Answer:
\( -2\frac{1}{2} + 2x ≤ \frac{4x}{5} ≤ \frac{4}{3} + 2x \)
\( -2\frac{1}{2} ≤ \frac{4x}{5} - 2x ≤ \frac{4}{3} \)
\( -\frac{5}{2} ≤ -\frac{6x}{5} ≤ \frac{4}{3} \)
\( \frac{25}{12} ≥ x ≥ -\frac{10}{9} \)
In simple words: We solve the compound inequality and find the range for x. Since x ∈ W (whole numbers), we take whole numbers in that range.

📝 Teacher's Note: Show students how to work with mixed numbers and compound inequalities. Remember that W means whole numbers (0, 1, 2, 3, ...).

🎯 Exam Tip: Convert mixed numbers to improper fractions first. When dividing by negative numbers, flip the inequality signs. List only whole numbers in your final answer.

Question 27. Find three consecutive largest positive integers such that the sum of one-third of first, one-fourth of second and one-fifth of third is atmost 20.
Answer: Let the required integers be x, x + 1 and x + 2.
According to the given statement,
\( \frac{1}{3}x + \frac{1}{4}(x + 1) + \frac{1}{5}(x + 2) \leq 20 \)

Step 1: Multiply through by LCM of 3, 4, 5 which is 60.
\( \frac{20x + 15x + 15 + 12x + 24}{60} \leq 20 \)
\( 47x + 39 \leq 1200 \)

Step 2: Solve for x.
\( 47x \leq 1161 \)
\( x \leq 24.702 \)

Thus, the largest value of the positive integer x is 24.
Hence, the required integers are 24, 25 and 26.
In simple words: We write the three numbers as x, x+1, x+2. Then we make an inequality for their fractions. After solving, we get x ≤ 24. So the numbers are 24, 25, 26.

📝 Teacher's Note: Show students how to find LCM of 3, 4, 5. Many students forget to multiply the whole inequality by 60. Practice this step carefully.

🎯 Exam Tip: Always write "Let the integers be x, x+1, x+2" first. This gets you marks. Then show all algebraic steps clearly.

Question 28. Solve the given inequation and graph the solution on the number line. 2y – 3 < y + 1 ≤ 4y + 7, y ∈ R
Answer: 2y – 3 < y + 1 ≤ 4y + 7, y ∈ R

Step 1: Split the compound inequality.
\( 2y – 3 < y + 1 \) and \( y + 1 \leq 4y + 7 \)

Step 2: Solve first part.
\( 2y – y < 1 + 3 \)
\( y < 4 \)

Step 3: Solve second part.
\( y + 1 \leq 4y + 7 \)
\( 1 – 7 \leq 4y – y \)
\( –6 \leq 3y \)
\( y \geq –2 \)

Step 4: Combine the results.
\( –2 \leq y < 4 \)

In simple words: We break the double inequality into two parts. First part gives y < 4. Second part gives y ≥ –2. So y is between –2 and 4.

[Diagram: A number line showing a closed circle at –2 and an open circle at 4, with the region between them shaded.]

📝 Teacher's Note: Teach students to always split compound inequalities first. Use closed circle for ≤ or ≥, open circle for < or >.

🎯 Exam Tip: Write the final answer as –2 ≤ y < 4. Draw the number line graph correctly with proper circles. This gets full marks.

Question 29. Solve the inequation: 3z – 5 ≤ z + 3 < 5z – 9, z ∈ R. Graph the solution set on the number line.
Answer: 3z – 5 ≤ z + 3 < 5z – 9

Step 1: Split into two inequalities.
3z – 5 ≤ z + 3 and z + 3 < 5z – 9

Step 2: Solve first part.
3z – z ≤ 3 + 5
2z ≤ 8
z ≤ 4

Step 3: Solve second part.
z + 3 < 5z – 9
3 + 9 < 5z – z
12 < 4z
3 < z

Step 4: Combine results.
Since z ∈ R
∴ Solution set = {3 < z ≤ 4, z ∈ R}

In simple words: We split the problem into two parts. First gives z ≤ 4. Second gives z > 3. So z is between 3 and 4.

[Diagram: A number line showing an open circle at 3 and a closed circle at 4, with the region between them shaded.]

📝 Teacher's Note: Show students the difference between > and ≥ on the number line. Open circle means "not including", closed circle means "including".

🎯 Exam Tip: Write the solution as 3 < z ≤ 4. Draw open circle at 3, closed circle at 4. Check your inequality symbols carefully.

Question 30. Solve the following inequation and represent the solution set on the number line. –3 < –\(\frac{1}{2}\) – \(\frac{2x}{3}\) ≤ \(\frac{5}{6}\), x ∈ R
Answer: –3 < –\(\frac{1}{2}\) – \(\frac{2x}{3}\) ≤ \(\frac{5}{6}\)

Step 1: Multiply by 6 to clear fractions.
–18 < –3 – 4x ≤ 5

Step 2: Solve –18 < –3 – 4x.
–18 + 3 < –4x
–15 < –4x
Dividing by –4, we get \(\frac{–15}{–4}\) > x ≥ \(\frac{8}{–4}\)
So \(\frac{15}{4}\) > x, which means x < \(\frac{15}{4}\)

Step 3: Solve –3 – 4x ≤ 5.
–4x ≤ 5 + 3
–4x ≤ 8
x ≥ –2

Step 4: Combine results.
–2 ≤ x < \(\frac{15}{4}\)
Since \(\frac{15}{4}\) = 3.75
∴ x ∈ [–2, \(\frac{15}{4}\))

In simple words: We multiply by 6 to remove fractions. Then solve both parts. We get –2 ≤ x < 3.75.

[Diagram: A number line showing a closed circle at –2 and an open circle at 3.75, with the region between them shaded.]

📝 Teacher's Note: When dividing by negative numbers, remind students to flip the inequality sign. This is a very common mistake.

🎯 Exam Tip: Always multiply by LCM first to clear fractions. Show each step. Write final answer clearly as –2 ≤ x < 15/4.

Question 31. Solve the following inequation and represent the solution set on the number line: 4x – 19 < \(\frac{3x}{5}\) – 2 ≤ \(\frac{–2}{5}\) + x, x ∈ R
Answer: Consider the given inequation:
4x – 19 < \(\frac{3x}{5}\) – 2 ≤ \(\frac{–2}{5}\) + x, x ∈ R

Step 1: Add 2 to all parts.
4x – 19 + 2 < \(\frac{3x}{5}\) – 2 + 2 ≤ \(\frac{–2}{5}\) + x + 2, x ∈ R
4x – 17 < \(\frac{3x}{5}\) ≤ x + \(\frac{8}{5}\), x ∈ R

Step 2: Split into two inequalities.
4x – \(\frac{3x}{5}\) < 17 and \(\frac{–8}{5}\) ≤ x – \(\frac{3x}{5}\), x ∈ R

Step 3: Solve first part.
\(\frac{20x – 3x}{5}\) < 17 and \(\frac{–8}{5}\) ≤ \(\frac{5x – 3x}{5}\), x ∈ R
\(\frac{17x}{5}\) < 17 and \(\frac{–8}{5}\) ≤ \(\frac{2x}{5}\), x ∈ R
\(\frac{x}{5}\) < 1 and –4 ≤ x, x ∈ R
x < 5 and –4 ≤ x, x ∈ R

Step 4: Combine results.
–4 ≤ x < 5, where x ∈ R

In simple words: We solve this step by step by splitting into two parts. First part gives x < 5. Second part gives x ≥ –4. So x is between –4 and 5.

[Diagram: A number line showing a closed circle at –4 and an open circle at 5, with the region between them shaded.]

📝 Teacher's Note: Show students how to work with fractions systematically. Find common denominators first, then multiply through to clear fractions.

🎯 Exam Tip: Keep track of your inequality signs. Write each algebraic step clearly. Final answer should be –4 ≤ x < 5.

Question 32. Solve the following in equation, write the solution set and represent it on the number line: –\(\frac{x}{3}\) ≤ \(\frac{x}{2}\) – 1 < \(\frac{1}{6}\), x ∈ R
Answer: The given inequation is
–\(\frac{x}{3}\) ≤ \(\frac{x}{2}\) – 1 < \(\frac{1}{6}\), x ∈ R

Step 1: Rewrite as –\(\frac{x}{3}\) ≤ \(\frac{x}{2}\) – \(\frac{4}{3}\) < \(\frac{1}{6}\)

Step 2: Split into two parts.
–\(\frac{x}{3}\) ≤ \(\frac{x}{2}\) – \(\frac{4}{3}\) and \(\frac{x}{2}\) – \(\frac{4}{3}\) < \(\frac{1}{6}\)

Step 3: Solve first part.
–\(\frac{x}{3}\) – \(\frac{x}{2}\) ≤ –\(\frac{4}{3}\)
\(\frac{–2x – 3x}{6}\) ≤ –\(\frac{4}{3}\)
\(\frac{–5x}{6}\) ≤ –\(\frac{4}{3}\)
–5x ≤ –8
x ≥ \(\frac{8}{5}\) = 1.6

Step 4: Solve second part.
\(\frac{x}{2}\) – \(\frac{4}{3}\) < \(\frac{1}{6}\)
\(\frac{x}{2}\) < \(\frac{1}{6}\) + \(\frac{4}{3}\)
\(\frac{x}{2}\) < \(\frac{1 + 8}{6}\) = \(\frac{9}{6}\) = \(\frac{3}{2}\)
x < 3

Step 5: Combine results.
∴ Solution set = {x : 1.6 ≤ x < 3}

In simple words: We split the problem into two parts. First gives x ≥ 1.6. Second gives x < 3. So x is between 1.6 and 3.

[Diagram: A number line showing a closed circle at 1.6 and an open circle at 3, with the region between them shaded.]

📝 Teacher's Note: When working with fractions, find common denominators first. Students often make sign errors when dividing by negative numbers.

🎯 Exam Tip: Convert fractions to decimals for clarity. Write solution set clearly as 1.6 ≤ x < 3. Show all fraction work step by step.

Question 33. Find the values of x, which satisfy the inequation –2\(\frac{5}{6}\) < \(\frac{1}{2}\) – \(\frac{2x}{3}\) ≤ 2, x ∈ W,. Graph the solution set on the number line.
Answer: We need to find the values of x, such that
x satisfies the inequation –2\(\frac{5}{6}\) < \(\frac{1}{2}\) – \(\frac{2x}{3}\) ≤ 2, x ∈ W

Step 1: Convert mixed number to improper fraction.
–2\(\frac{5}{6}\) = –\(\frac{17}{6}\)
So we have –\(\frac{17}{6}\) < \(\frac{3 – 4x}{6}\) ≤ \(\frac{12}{6}\)

Step 2: Multiply by 6.
–17 > 4x – 3 ≥ –12
–12 ≤ 4x – 3 < 17
–12 + 3 ≤ 4x – 3 + 3 < 17 + 3
–9 ≤ 4x < 20

Step 3: Divide by 4.
–\(\frac{9}{4}\) ≤ x < 5
–2.25 ≤ x < 5

Step 4: Since x ∈ W, the values of x are 0, 1, 2, 3, 4.
And the required line is shown on the number line.

In simple words: We solve the inequality to get –2.25 ≤ x < 5. Since x must be a whole number, x can be 0, 1, 2, 3, or 4.

[Diagram: A number line showing solid dots at 0, 1, 2, 3, and 4.]

📝 Teacher's Note: Remind students that W means whole numbers (0, 1, 2, 3, ...). After solving, pick only whole number values in the range.

🎯 Exam Tip: Always check what set x belongs to (R, Z, W, N). List all valid values clearly: x = 0, 1, 2, 3, 4.

Question 34. Solve the following in equation and write the solution set: 13x – 5 < 15x + 4 < 7x + 12, x ∈ R
Answer: 13x – 5 < 15x + 4 < 7x + 12, x ∈ R

Step 1: Split into two parts.
13x – 5 < 15x + 4 and 15x + 4 < 7x + 12

Step 2: Solve first part.
13x < 15x + 9
0 < 2x + 9
–9 < 2x
–\(\frac{9}{2}\) < x

Step 3: Solve second part.
15x < 7x + 8
8x < 8
x < 1

Step 4: Combine results.
–\(\frac{9}{2}\) < x < 1
Since x ∈ R, the solution set is (–\(\frac{9}{2}\), 1)

In simple words: We split into two parts. First gives x > –4.5. Second gives x < 1. So x is between –4.5 and 1.

📝 Teacher's Note: Show students how to solve each inequality separately, then find the intersection. Use interval notation for the final answer.

🎯 Exam Tip: Write solution as –9/2 < x < 1 or in interval notation as (–9/2, 1). Both forms are correct. Show all algebraic steps.

Question 35. Solve the following inequation, write the solution set and represent it on the number line. -3(x - 7) ≥ 15 - 7x > x+1/3, x ∈ R.
Answer:
Step 1: Break the compound inequality into two parts.
\( -3(x - 7) \geq 15 - 7x \) and \( 15 - 7x > \frac{x - 1}{3} \)

Step 2: Solve the first inequality.
\( -3(x - 7) \geq 15 - 7x \)
\( -3x + 21 \geq 15 - 7x \)
\( -3x + 7x \geq 15 - 21 \)
\( 4x \geq -6 \)
\( x \geq -\frac{3}{2} \)
\( x \geq -1.5 \)

Step 3: Solve the second inequality.
\( 15 - 7x > \frac{x - 1}{3} \)
\( 45 - 21x > x - 1 \)
\( 45 - 1 > x + 21x \)
\( 44 > 22x \)
\( 2 > x \)

Step 4: Combine both conditions.
\( x \geq -1.5 \) and \( 2 > x \)
Therefore: \( -1.5 \leq x < 2 \)

Solution set: \( \{x : x \in R, -1.5 \leq x < 2\} \)

[Diagram: Number line showing interval from -1.5 to 2, with a closed circle at -1.5 and an open circle at 2, with the line shaded between these points.]

In simple words: We solved two inequalities together. The answer is all numbers from -1.5 to 2, including -1.5 but not including 2.

📝 Teacher's Note: Show students how compound inequalities work like "and" conditions. Both parts must be true at the same time. Draw the number line step by step.

🎯 Exam Tip: Always write the solution set in proper set notation. Show both inequalities solved separately, then combine them. Draw the number line clearly with open and closed circles.

Question 36. Solve the following inequation and represent the solution set on a number line. \( -8\frac{1}{2} < -\frac{1}{2} - 4x \leq 7\frac{1}{2} \), x ∈ I
Answer:
Step 1: Convert mixed fractions to improper fractions.
\( -\frac{17}{2} < -\frac{1}{2} - 4x \leq \frac{15}{2} \), x ∈ I

Step 2: Solve the left inequality.
\( -\frac{17}{2} < -\frac{1}{2} - 4x \)
\( -\frac{17}{2} + \frac{1}{2} < -4x \)
\( -\frac{16}{2} < -4x \)
\( -8 < -4x \)
\( 8 > 4x \)
\( x < 2 \)

Step 3: Solve the right inequality.
\( -\frac{1}{2} - 4x \leq \frac{15}{2} \)
\( -\frac{1}{2} - \frac{15}{2} \leq 4x \)
\( -\frac{16}{2} \leq 4x \)
\( -8 \leq 4x \)
\( x \geq -2 \)

Step 4: Combine both conditions.
\( x \geq -2 \) and \( x < 2 \)
Therefore: \( -2 \leq x < 2 \)

Step 5: Since x ∈ I (integers), find integer values.
\( x = \{-2, -1, 0, 1\} \)

[Diagram: Number line showing points at -2, -1, 0, and 1 marked with solid dots, representing the integer solutions.]

In simple words: We found that x must be between -2 and 2. Since x must be a whole number (integer), the answers are -2, -1, 0, and 1.

📝 Teacher's Note: Remind students that when x ∈ I, they must only pick integer values from the solution range. Mixed fractions should be converted to improper fractions first.

🎯 Exam Tip: Always check if the domain is integers (I), real numbers (R), or natural numbers (N). This changes your final answer. List all integer solutions clearly.