Frank Brothers Solutions for ICSE Class 10 Mathematics Chapter 5 Linear Inequations

Exercise 5.1

Question 1. (i) x + 5 > 11
Answer:
x + 5 > 11
\( \implies \) x > 11 - 5
\( \implies \) x > 6
But {x : x ∈ N, N < 10}
Therefore, solution set x = {7,8,9}
In simple words: We subtract 5 from both sides. x must be greater than 6. From natural numbers less than 10, we get 7, 8, and 9.

📝 Teacher's Note: Show students that we subtract the same number from both sides. The inequality sign stays the same. Only natural numbers 1, 2, 3... are allowed here.

🎯 Exam Tip: Always write the constraint clearly. Here x must be a natural number and less than 10. Don't forget to list the final answer as a set.

 

Question 1. (ii) 2x + 1 < 17
Answer:
2x + 1 < 17
\( \implies \) 2x < 17 - 1
\( \implies \) x < \( \frac{16}{2} \)
\( \implies \) x < 8
But {x : x ∈ N, N < 10}
Therefore, solution set x = {1,2,3,4,5,6,7}
In simple words: First subtract 1, then divide by 2. x must be less than 8. From natural numbers less than 10, we get 1 to 7.

📝 Teacher's Note: When dividing by a positive number, the inequality sign does not change. Make this clear to students with examples.

🎯 Exam Tip: Show each step clearly. Write "subtract 1" and "divide by 2" as separate steps. List all values in the solution set.

 

Question 1. (iii) 3x - 5 ≤ 7
Answer:
3x - 5 ≤ 7
\( \implies \) 3x ≤ 7 + 5
\( \implies \) 3x ≤ 12
\( \implies \) x ≤ 4
But {x : x ∈ N, N < 10}
Therefore, solution set x = {1,2,3,4}
In simple words: Add 5 to both sides, then divide by 3. x must be 4 or less. From natural numbers, we get 1, 2, 3, and 4.

📝 Teacher's Note: The ≤ symbol means "less than or equal to". So 4 is included in the answer. Students often forget to include the equal part.

🎯 Exam Tip: When you see ≤ or ≥, include the boundary value in your answer. Here 4 is included because x ≤ 4.

 

Question 1. (iv) 8 - 3x ≥ 2
Answer:
8 - 3x ≥ 2
\( \implies \) -3x ≥ -6
\( \implies \) x ≤ 2
But {x : x ∈ N, N < 10}
Therefore, solution set x = {1,2}
In simple words: Subtract 8 from both sides. When we divide by -3, the inequality sign flips. x must be 2 or less.

📝 Teacher's Note: This is the most important rule - when you multiply or divide by a negative number, flip the inequality sign. Practice this many times.

🎯 Exam Tip: Always write "multiplying by -1 changes the sign" when you flip the inequality. This shows the examiner you know the rule.

 

Question 1. (v) 5 - 2x < 11
Answer:
5 - 2x < 11
\( \implies \) -2x < 11 - 5
\( \implies \) -2x < 6
\( \implies \) x > -3
But {x : x ∈ N, N < 10}
Therefore, solution set x = {1,2,3,4,5,6,7,8,9}
In simple words: After moving terms, we divide by -2 and flip the sign. x must be greater than -3. All natural numbers less than 10 work.

📝 Teacher's Note: Since all natural numbers are positive, they are all greater than -3. This is a good example to show the concept of natural numbers.

🎯 Exam Tip: Remember natural numbers are 1, 2, 3, 4... Since x > -3 and x is natural, all natural numbers less than 10 are in the answer.

 

Question 2. (i) 3x > 12
Answer:
3x > 12
x > \( \frac{12}{3} \)
x > 4
Since, replacement set is R
Solution set = {x: x ∈ R and x > 4}
In simple words: Divide both sides by 3. x must be greater than 4. Since we can use any real number, the answer is all real numbers greater than 4.

📝 Teacher's Note: R means real numbers - all numbers including decimals, fractions, positive and negative. This is different from natural numbers.

🎯 Exam Tip: When the replacement set is R, write your answer in set notation with the condition. Don't try to list all values.

 

Question 2. (ii) 2x - 3 < 7
Answer:
2x - 3 < 7
2x < 7 + 3
2x < 10
x < \( \frac{10}{2} \)
Since, replacement set is R
Solution set = {x : x ∈ R and x < 5}
In simple words: Add 3 to both sides, then divide by 2. x must be less than 5. Any real number less than 5 works.

📝 Teacher's Note: Examples of numbers less than 5: 4.9, 3, 0, -10. All these are in the solution set because the replacement set is R.

🎯 Exam Tip: Always mention the replacement set in your final answer. Write clearly whether it's natural numbers, integers, or real numbers.

 

Question 2. (iii) 3x + 2 ≤ 11
Answer:
3x + 2 ≤ 11
3x ≤ 11 - 2
3x ≤ 9
x ≤ \( \frac{9}{3} \)
x ≤ 3
Since, replacement set is R
Solution set = {x : x ∈ R and x ≤ 3}
In simple words: Subtract 2, then divide by 3. x must be 3 or less. Any real number up to and including 3 works.

📝 Teacher's Note: The ≤ symbol includes 3 in the answer. Numbers like 3, 2.99, 1, 0, -100 are all in the solution set.

🎯 Exam Tip: Don't forget the "equal to" part when you see ≤. The value 3 is included in the solution set.

 

Question 2. (iv) 14 - 3x ≥ 5
Answer:
14 - 3x ≥ 5
-3x ≥ 5 - 14
3x ≤ 9 (multiplying by -1 changes the sign)
x ≤ \( \frac{9}{3} \)
x ≤ 3
Since, replacement set is R
Solution set = {x : x ∈ R and x ≤ 3}
In simple words: Move terms, then divide by -3 and flip the inequality sign. x must be 3 or less.

📝 Teacher's Note: This problem has two key steps - moving terms and flipping the inequality sign. Make sure students write both steps clearly.

🎯 Exam Tip: Always write the reason when you change the inequality sign. Write "multiplying by -1 changes the sign" to show you know the rule.

 

Question 2. (v) 7x + 11 > 16 - 3x
Answer:
7x + 11 > 16 - 3x
7x + 3x > 16 - 11
10x > 5
x > \( \frac{5}{10} \) = 0.5
Since, replacement set is R
Solution set = {x : x ∈ R and x > 0.5}
In simple words: Move all x terms to one side and numbers to the other side. Then divide by 10. x must be greater than 0.5.

📝 Teacher's Note: When moving terms across the inequality sign, change their signs. +3x becomes -3x, +11 becomes -11, etc.

🎯 Exam Tip: Keep all x terms on one side and all number terms on the other side. This makes the problem easier to solve.

 

Question 2. (vi) 3x + 25 < 8x - 10
Answer:
3x + 25 < 8x - 10
25 + 10 < 8x - 3x
35 < 5x
x > 7
Since, replacement set is R
Solution set = {x : x ∈ R and x > 7}
In simple words: Move x terms to one side and numbers to the other. Then divide by 5. x must be greater than 7.

📝 Teacher's Note: Notice that 35 < 5x is the same as 5x > 35. Students can write it either way, but x > 7 is the clearest final answer.

🎯 Exam Tip: You can flip an inequality to make it clearer. 35 < 5x becomes 5x > 35. Both mean the same thing.

 

Question 2. (vii) 2(3x - 5) ≤ 8
Answer:
2(3x - 5) ≤ 8
6x - 10 ≤ 8
6x ≤ 8 + 10
6x ≤ 18
x ≤ 3
Since, replacement set is R
Solution set = {x : x ∈ R and x ≤ 3}
In simple words: First multiply out the bracket: 2 × 3x = 6x and 2 × (-5) = -10. Then solve normally.

📝 Teacher's Note: Always expand brackets first before solving. Remind students that 2 × (-5) = -10, not +10.

🎯 Exam Tip: Show the bracket expansion step clearly. Write "2(3x - 5) = 6x - 10" as a separate line before continuing.

 

Question 2. (viii) x + 7 ≥ 15 + 3x
Answer:
x + 7 ≥ 15 + 3x
x - 3x ≥ 15 - 7
-2x ≥ 8
2x ≤ -8
x ≤ -4
Since, replacement set is R
Solution set = {x : x ∈ R and x ≤ -4}
In simple words: Move terms to get -2x ≥ 8. When we divide by -2, we flip the inequality sign to get x ≤ -4.

📝 Teacher's Note: This is another example where we divide by a negative number and flip the sign. Practice this type of problem often.

🎯 Exam Tip: When you get negative coefficients for x, be extra careful about flipping the inequality sign when you solve for x.

 

Question 2. (ix) 2x - 7 ≥ 5x + 8
Answer:
2x - 7 ≥ 5x + 8
2x - 5x ≥ 8 + 7
-3x ≥ 15
3x ≤ -15
x ≤ -5
Since, replacement set is R
Solution set = {x : x ∈ R and x ≤ -5}
In simple words: Collect x terms and number terms. Then divide by -3 and flip the sign. x must be -5 or less.

📝 Teacher's Note: Examples of numbers ≤ -5 are -5, -6, -10, -100. All negative numbers smaller than -5 work.

🎯 Exam Tip: Remember that -6 is smaller than -5. On the number line, we go left for smaller numbers.

 

Question 2. (x) 9 - 4x ≤ 15 - 7x
Answer:
9 - 4x ≤ 15 - 7x
9 - 15 ≤ 4x - 7x
-6 ≤ -3x
6 ≥ 3x (multiplying by -1 changes the sign)
x ≤ 2
Since, replacement set is R
Solution set = {x : x ∈ R and x ≤ 2}
In simple words: Move terms to get -6 ≤ -3x. Divide by -3 and flip the sign. x must be 2 or less.

📝 Teacher's Note: You can also write 6 ≥ 3x as 3x ≤ 6. Both give the same answer x ≤ 2.

🎯 Exam Tip: When you flip an inequality, make sure you also flip the entire statement. -6 ≤ -3x becomes 6 ≥ 3x.

 

Question 3. 6 - 10x < 36
Answer:
-10x < 36 - 6
-10x < 30
10x > -30
x > -3
Solution set = {-2, -1, 0, 1, 2}
In simple words: Subtract 6, then divide by -10 and flip the sign. From the given options, numbers greater than -3 are -2, -1, 0, 1, 2.

📝 Teacher's Note: This question has specific integer options. Show students that -2 > -3, -1 > -3, etc. Use a number line if needed.

🎯 Exam Tip: When given specific options, check each one against your inequality. Here x > -3 means only -2, -1, 0, 1, 2 work.

 

Question 4. 3 - 2x ≥ x - 12
Answer:
3 + 12 ≥ x + 2x
15 ≥ 3x
x ≤ 5
Solution set = {1,2,3,4,5}
In simple words: Move terms to get 15 ≥ 3x. Divide by 3 to get x ≤ 5. From natural numbers, we get 1, 2, 3, 4, 5.

📝 Teacher's Note: The constraint here is natural numbers. So we only take positive integers 1, 2, 3, 4, 5. Zero is not a natural number.

🎯 Exam Tip: Read the question carefully for the replacement set. Natural numbers are 1, 2, 3, 4... (not including zero).

 

Question 5. 5x - 9 ≤ 15 - 7x
Answer:
5x + 7x ≤ 15 + 9
12x ≤ 24
x ≤ 2
Solution set = {0,1,2}
In simple words: Move terms to get 12x ≤ 24. Divide by 12 to get x ≤ 2. From whole numbers, we get 0, 1, 2.

📝 Teacher's Note: Whole numbers include 0, unlike natural numbers. Make sure students know the difference between these number sets.

🎯 Exam Tip: Whole numbers are 0, 1, 2, 3... Natural numbers are 1, 2, 3, 4... The difference is whether zero is included.

 

Question 6. 7 + 5x > x - 13
Answer:
-x + 5x > -13 - 7
4x > -20
x > -5
Solution set = {-4, -3, -2, -1}
In simple words: Collect like terms to get 4x > -20. Divide by 4 to get x > -5. From integers, we get -4, -3, -2, -1.

📝 Teacher's Note: This uses integers from a specific range. Students need to pick integers greater than -5 from the given set.

🎯 Exam Tip: When working with integers in a range, list them carefully. Here we want integers > -5, so -4, -3, -2, -1 work.

 

Question 7. 5x -14 < 18 - 3x
Answer:
5x + 3x < 18 + 14
8x < 32
x < 4
Solution set = {0,1,2,3}
In simple words: Move terms to get 8x < 32. Divide by 8 to get x < 4. From whole numbers, we get 0, 1, 2, 3.

📝 Teacher's Note: Since x < 4 (not ≤), the value 4 is not included in the answer. Only 0, 1, 2, 3 are less than 4.

🎯 Exam Tip: Pay attention to < versus ≤. When it's <, don't include the boundary value. When it's ≤, include the boundary value.

Answer 8.
Answer:
Step 1: Write the inequality
\( 2x + 7 \geq 5x - 14 \)

Step 2: Move all terms with x to left side and constants to right side
\( 2x - 5x \geq -14 - 7 \)

Step 3: Simplify
\( -3x \geq -21 \)

Step 4: Divide by -3 (flip the inequality sign)
\( 3x \leq 21 \)
\( x \leq 7 \)

Solution set = {2,3,5,7}
In simple words: We solve the inequality step by step. When we divide by a negative number, we flip the inequality sign. The answer includes all whole numbers less than or equal to 7.

📝 Teacher's Note: Students often forget to flip the inequality sign when dividing by negative numbers. Practice this rule with many examples so students remember it.

🎯 Exam Tip: Always write "flip the sign" when dividing by negative numbers. This shows the examiner you know the rule and gets you marks.

 

Answer 9.
Answer:
Step 1: Write the compound inequality
\( \frac{x}{4} + 3 \leq \frac{x}{3} + 4 \)

Step 2: Multiply everything by 12 to clear fractions
\( \frac{x + 12}{4} \leq \frac{x + 12}{3} \)
\( 3x + 36 \leq 4x + 48 \)

Step 3: Rearrange
\( 3x - 4x \leq 48 - 36 \)
\( -x \leq 12 \)
\( x \geq -12 \)

Solution set = {-11, -9, -7, -5, -3, -1}
In simple words: We clear the fractions by multiplying by 12. Then solve like a normal inequality. The solution includes all integers greater than or equal to -12.

📝 Teacher's Note: When dealing with fractions in inequalities, find the LCM of denominators and multiply the whole inequality by it. This makes the problem much easier.

🎯 Exam Tip: Always multiply the entire inequality by the LCM. Don't forget to flip the sign if you divide by a negative number.

 

Answer 10.
Answer:
Step 1: Write the inequality
\( \frac{x + 3}{3} \leq \frac{x + 8}{4} \)

Step 2: Multiply by 12 to clear fractions
\( 4x + 12 \leq 3x + 24 \)

Step 3: Rearrange
\( 4x - 3x \leq 24 - 12 \)
\( x \leq 12 \)

Solution set = {2,4,6,8,10,12}
In simple words: We multiply by 12 to remove fractions. Then solve the simple inequality. All even numbers up to 12 are the solution.

📝 Teacher's Note: Show students how multiplying by the LCM removes all fractions at once. This makes the inequality much simpler to solve.

🎯 Exam Tip: Write the LCM step clearly. Examiners give marks for showing this working step even if the final answer is wrong.

 

Answer 11.
Answer:
(i) x ∈ Z
\( x + 17 \leq 4x + 9 \)
\( x - 4x \leq 9 - 17 \)
\( -3x \leq -8 \)
\( 3x \geq 8 \)
\( x \geq \frac{8}{3} \)

Since \( x \in Z \)
Smallest value of x = [3]

(ii) x ∈ R
\( x + 17 \leq 4x + 9 \)
\( x - 4x \leq 9 - 17 \)
\( -3x \leq -8 \)
\( 3x \geq 8 \)
\( x \geq \frac{8}{3} \)

Since \( x \in R \)
Smallest value of x = \( \left[\frac{8}{3}\right] \)
In simple words: When x must be an integer (Z), we take the next whole number. When x can be any real number (R), we take the exact fraction value.

📝 Teacher's Note: Explain the difference between Z (integers) and R (real numbers). Students often confuse these and give wrong answers.

🎯 Exam Tip: Always check if the question asks for integers (Z) or real numbers (R). Write the correct symbol in your answer.

 

Answer 12.
Answer:
Step 1: Write the equation
\( \frac{2}{x^2} - \frac{5}{x} + 2 = 0 \)

Step 2: Multiply by \( x^2 \)
\( 2 - 5x + 2x^2 = 0 \)
\( 2x^2 - 5x + 2 = 0 \)

Step 3: Rearrange in standard form
\( x^2 - \frac{5}{2}x + 1 = 0 \)

Step 4: Multiply by 2
\( x^2 - 2x - \frac{1}{2}x + 1 = 0 \)

Step 5: Factor
\( x(x - 2) - \frac{1}{2}(x - 2) = 0 \)
\( (x - 2)(x - \frac{1}{2}) = 0 \)

Step 6: Solve
\( (x - 2) = 0, (x - \frac{1}{2}) = 0 \)
\( x = 2, x = \frac{1}{2} \)
In simple words: We multiply by x² to clear fractions. Then solve the quadratic equation by factoring. We get two solutions.

📝 Teacher's Note: When clearing fractions with variables in denominator, multiply by the highest power present. This removes all fractions at once.

🎯 Exam Tip: Always check your answers by substituting back into the original equation. This catches silly mistakes and gets you marks.

 

Answer 13.
Answer:
(i) \( 2x - 11 \leq 7 - 3x, x \in N \)
\( 2x - 11 \leq 7 - 3x \)
\( 2x + 3x \leq 7 + 11 \)
\( 5x \leq 18 \)
\( x \leq \frac{18}{5} \)
\( x \leq 3.6 \)

Since \( x \in N \)
Solution set = {1,2,3}
 

[Diagram: Number line showing points at 1, 2, and 3 marked with solid dots]

 

(ii) \( 3(5x + 3) \geq 2(9x - 17), x \in W \)
\( 3(5x + 3) \geq 2(9x - 17) \)
\( 15x + 9 \geq 18x - 34 \)
\( 15x - 18x \geq -34 - 9 \)
\( -3x \geq -43 \)
\( 3x \leq 43 \)
\( x \leq \frac{43}{3} \)

Solution set = \( \left[x \leq \frac{43}{3}\right] \)

 

 

 

[Diagram: Number line showing all values from negative infinity up to 43/3 approximately 14.33]

 

(iii) \( 2(3x - 5) > 5(13 - 2x), x \in W \)
\( 2(3x - 5) > 5(13 - 2x) \)
\( 6x - 10 > 65 - 10x \)
\( 6x + 10x > 65 + 10 \)
\( 16x > 75 \)
\( x > \frac{75}{16} \)
\( x > 4\frac{11}{16} \)

Solution set = \( \left[x > 4\frac{11}{16}\right] \)

 

 

 

 

[Diagram: Number line showing all values greater than 4 11/16 with an open circle at that point]

 

(iv) \( 3x - 9 \leq 4x - 7 < 2x + 5, x \in R \)
\( 3x - 9 \leq 4x - 7 \) and \( 4x - 7 < 2x + 5 \)
\( 3x - 4x \leq -7 + 9 \) and \( 4x - 2x < 5 + 7 \)
\( -x \leq 2 \) and \( 2x < 12 \)
\( x \geq -2 \) and \( x < 6 \)

Solution set = [-2≤x<6]

 

 

 

 

[Diagram: Number line showing closed circle at -2 and open circle at 6 with line segment between them]

 

(v) \( 2x - 7 < 5x + 2 \leq 3x + 14, x \in R \)
\( 2x - 7 < 5x + 2 \) and \( 5x + 2 \leq 3x + 14 \)
\( 2x - 5x < 2 + 7 \) and \( 5x - 3x \leq 14 - 2 \)
\( -3x < 9 \) and \( 2x \leq 12 \)
\( x > -3 \) and \( x \leq 6 \)

Solution set = [-3

[Diagram: Number line showing open circle at -3 and closed circle at 6 with line segment between them]

(vi) \( -3 \leq \frac{1}{2} - \frac{2x}{3} \leq 2 \frac{2}{3}, x \in N \)
\( -3 \leq \frac{1}{2} - \frac{2x}{3} \)
\( -3 \leq \frac{3 - 4x}{6} \)
\( -18 \leq 3 - 4x \)
\( -18 - 3 \leq -4x \)
\( -21 \leq -4x \)
\( x \leq \frac{21}{4} \)
\( x \leq 5\frac{1}{4} \)

and

\( \frac{1}{2} - \frac{2x}{3} \leq 2\frac{2}{3} \)
\( \frac{3 - 4x}{6} \leq \frac{8}{3} \)
\( 9 - 12x \leq 48 \)
\( -12x \leq 39 \)
\( 12x \geq -39 \)
\( x \geq -3\frac{1}{4} \)

Solution set = \( \left[-3\frac{1}{4} \leq x \leq 5\frac{1}{4}\right] \)


 

[Diagram: Number line showing closed circles at -3.25 and 5.25 with line segment between them]

(vii) \( 4\frac{3}{4} \geq x + \frac{5}{6} > \frac{1}{3}, x \in R \)
\( 4\frac{3}{4} \geq x + \frac{5}{6} \) and \( x + \frac{5}{6} > \frac{1}{3} \)
\( \frac{19}{4} \geq \frac{6x + 5}{6} \) and \( \frac{6x + 5}{6} > \frac{1}{3} \)
\( 114 \geq 24x + 20 \) and \( 18x + 15 > 6 \)
\( 114 - 20 \geq 24x \) and \( 18x > 6 - 15 \)
\( 94 \geq 24x \) and \( 18x > -9 \)
\( x \leq 3\frac{11}{12} \) and \( x > -\frac{1}{2} \)

Solution set = \( \left[-\frac{1}{2} < x \leq 3\frac{11}{12}\right] \)


 

[Diagram: Number line showing open circle at -0.5 and closed circle at 3.92 with line segment between them]

(viii) \( \frac{1}{3}(2x - 1) < \frac{1}{4}(x + 5) < \frac{1}{6}(3x + 4), x \in R \)
\( \frac{1}{3}(2x - 1) < \frac{1}{4}(x + 5) \) and \( \frac{1}{4}(x + 5) < \frac{1}{6}(3x + 4) \)
\( 4(2x - 1) < 3(x + 5) \) and \( 6(x + 5) < 4(3x + 4) \)
\( 8x - 4 < 3x + 15 \) and \( 6x + 30 < 12x + 16 \)
\( 8x - 3x < 15 + 4 \) and \( 6x - 12x < 16 - 30 \)
\( 5x < 19 \) and \( -6x < -14 \)
\( x < 3\frac{4}{5} \) and \( x > 2\frac{1}{3} \)

Solution set = \( \left[2\frac{1}{3} < x < 3\frac{4}{5}\right] \)


 

[Diagram: Number line showing open circles at 2.33 and 3.8 with line segment between them]

(ix) \( \frac{1}{3}(5x - 8) \geq \frac{1}{2}(4x - 7), x \in R \)
In simple words: These are compound inequalities. We solve each part separately and find where both conditions are true at the same time. The solution is the overlap region.

 

📝 Teacher's Note: For compound inequalities, solve each part separately first. Then find the intersection (overlap) of the solutions. Use number lines to show this visually.

🎯 Exam Tip: Always draw number lines for compound inequalities. This helps you find the correct intersection and shows your working to the examiner.

 

Answer 14.

Answer:
P = {x : -3 < x ≤ 7, x ∈ R} and Q = {x : -7 ≤ x < 3, x ∈ R}
P = {-2, -1, 0, 1, 2, 3, 4, 5, 6, 7} and Q = {-7, -6, -5, -4, -3, -2, -1, 0, 1, 2}

(i) P ∩ Q
P ∩ Q = {-2, -1, 0, 1, 2}
[Diagram: Number line showing points marked at -2, -1, 0, 1, 2 between -3 and 3]

(ii) Q' ∩ P
Q' ∩ P = {3, 4, 5, 6, 7}
[Diagram: Number line showing points marked at 3, 4, 5, 6, 7 between -3 and 9]

(iii) P-Q
P-Q = {3, 4, 5, 6, 7}
[Diagram: Number line showing points marked at 3, 4, 5, 6, 7 between 0 and 10]

In simple words: We found common elements (intersection), elements in P but not in Q (complement), and elements that are only in P. The number lines show these sets clearly.

📝 Teacher's Note: Draw two overlapping circles (Venn diagram) to show P and Q. Students can see intersection and difference easily. Use different colors for each set.

🎯 Exam Tip: Always write the final answer set clearly with curly braces. Draw number lines to show your answer. This gets full marks.

 

Answer 15.

Answer:
P = {x : 7x - 2 > 4x + 1, x ∈ R} and Q = {x : 9x - 45 ≥ 5(x - 5), x ∈ R}
7x - 2 > 4x + 1
7x - 4x > 1 + 2
3x > 3
x > 1
P = {2, 3, 4, 5, ...........}

9x - 45 ≥ 5x - 25
9x - 5x ≥ -25 + 45
4x ≥ 20
x ≥ 5
Q = {5, 6, 7, 8, 9, ...........}

(i) P ∩ Q
P ∩ Q = {2, 3, 4, 5}
[Diagram: Number line showing points marked at 2, 3, 4, 5 between 0 and 6]

(ii) P-Q
P-Q = {6, 7, 8, 9, ...........}
[Diagram: Number line showing points starting from 6 onwards between -3 and 13]

(iii) P ∩ Q'
P ∩ Q' = {6, 7, 8, 9, ...........}
[Diagram: Number line showing points starting from 6 onwards between -1 and 13]

In simple words: We solved two inequalities to find sets P and Q. Then we found their intersection (common elements) and difference (elements in P but not in Q).

📝 Teacher's Note: Show students how to solve inequalities step by step. Remind them that x > 1 means all numbers greater than 1, not equal to 1.

🎯 Exam Tip: Solve each inequality separately first. Then find the required operations. Always show your working clearly for full marks.

 

Answer 16.

Answer:
P = {x : 7x - 4 > 5x + 2, x ∈ R} and Q = {x : x - 19 ≥ 1 - 3x, x ∈ R}
7x - 4 > 5x + 2
7x - 5x > 2 + 4
2x > 6
x > 3
P = {4, 5, 6, 7, ...........}

x - 19 ≥ 1 - 3x
x + 3x ≥ 1 + 19
4x ≥ 20
x ≥ 5
Q = {5, 6, 7, 8, 9, .......}

(i) P ∩ Q
P ∩ Q = {5, 6, 7, 8, .........}
[Diagram: Number line showing points starting from 5 onwards between 0 and 8]

(ii) P' ∩ Q
P' ∩ Q = {∅}
In simple words: P' means all numbers not in P (numbers ≤ 3). Q has numbers ≥ 5. So there are no common numbers between them.

In simple words: We solved two inequalities to find P and Q. P has numbers greater than 3, Q has numbers greater than or equal to 5. Their intersection is numbers ≥ 5.

📝 Teacher's Note: Explain that P' means complement of P (all numbers not in P). When two sets have no common elements, we write empty set {∅}.

🎯 Exam Tip: For P' ∩ Q, first find P' (complement of P), then find common elements with Q. If no common elements exist, write empty set {∅}.