Linear Inequations

linear inequations class 8 selina

Back Exercise: 15 Test Yourself Back Exercise: 15 Test Yourself Test Yourself Q1: Multiple Choice Type i. If x is […]

Test Yourself

Q1: Multiple Choice Type

i. If x is a whole number and 12 – x > 3x – 5; find the solution set

Step 1: Write the inequality \[ 12 – x > 3x – 5 \]Step 2: Add x to both sides \[ 12 > 4x – 5 \]Step 3: Add 5 to both sides \[ 17 > 4x \]Step 4: Divide both sides by 4 \[ x < 4.25 \]Step 5: Since x ∈ W (whole numbers: 0,1,2,…), solution set: \[ x = 0, 1, 2, 3, 4 \]Answer: d. {0, 1, 2, 3, 4}

ii. If 5x – 3 ≤ 12 and x ∈ {-5, -3, -1, 1, 3, 5, 7} the solution set is:

Step 1: Solve the inequality \[ 5x – 3 <= 12\\ 5x <= 15\\ x <= 3 \]Step 2: Select x from the replacement set ≤ 3 \[ x = -5, -3, -1, 1, 3 \]Answer: a. {-5, -3, -1, 1, 3}

iii. If x ∈ {-7, -4, -1, 2, 5} and 25 – 3(2x – 5) < 19; the solution set is:

Step 1: Expand the inequality \[ 25 – 6x + 15 < 19\\ 40 – 6x < 19 \]Step 2: Subtract 40 from both sides \[ -6x < -21 \]Step 3: Divide both sides by -6 (reverse inequality) \[ x > 3.5 \]Step 4: Select values from the set {-7, -4, -1, 2, 5} that satisfy x > 3.5 \[ x = 5 \]Answer: d. {5}

iv. If x is a real number and -4 < x ≤ 0, its solution set on the number line is:

Step 1: Represent -4 < x ≤ 0 on number line
Answer: b.

←──○──────────●──→
  -4          0
(x ∈ (-4, 0])

v. If x is an integer and 7 – 4x < 15, the solution set on the number line is:

Step 1: Solve the inequality \[ 7 – 4x < 15\\ -4x < 8\\ x > -2 \]Step 2: Since x ∈ Z, solution set: -1, 0, 1, 2, 3, 4, 5
Answer: c. {-1, 0, 1, 2, 3, 4, 5}

Number Line:

←──|──|──●──●──●──●──●──●──●──→
  -3 -2 -1  0  1  2  3  4  5

vi. Statement 1: A set from which the values of the variables involved in the inequation are chosen is called the solution set.
Statement 2: A linear inequation in one variable (or unknown) has exactly one solution.
Which of the following options is correct?

Step 1: Analyze
– Statement 1: A set from which values of variables are chosen is called the replacement set, not the solution set → Statement 1 is false
– Statement 2: A linear inequation in one variable can have infinite solutions, not exactly one → Statement 2 is false
Answer: b. Both statements are false

vii. Assertion (A) : The solution set for: x + 5 ≤ 10, if the replacement set is {x|x ≤ 5, x ∈ W} is {0, 1, 2, 3, 4, 5}.
Reason (R): The set of elements of the replacement set which satisfy the given inequation is called the solution set.

Step 1: Solve the inequality: \[ x + 5 \le 10\\ x \le 5 \]Step 2: Replacement set W = {0,1,2,3,4,5,…} → elements ≤ 5: \[ x = 0, 1, 2, 3, 4, 5 \]Step 3: Explanation:
R correctly defines the solution set.
Answer: a. Both A and R are correct, and R is the correct explanation for A

viii. Assertion (A): The solution set for x + 3 ≥ 15 is Φ if the replacement set is {x | x < 10, x ∈ N}.
Reason (R): If we change over the sides of an inequality, we must change the sign from < to > or > to < or ≥ to ≤ or ≤ to ≥.

Step 1: Solve the inequality: \[ x + 3 \ge 15\\ x \ge 12 \]Step 2: Replacement set {x ∈ N | x < 10} → no x satisfies x ≥ 12 → Solution set Φ
Step 3: Reason R talks about changing sides, which is irrelevant here → R is false.
Answer: c. A is true, but R is false

ix. Assertion (A): x < -2 and x ≥ 1 ⇒ Solution set S = {x | -2 < x ≤ 1, x ∈ R}.

Reason (R): Two inequations can be written in a combined expression.

Step 1: Consider x < -2 and x >= 1 → no x satisfies both simultaneously → Solution set Φ
Step 2: R is true in general (we can combine inequations using ∩ or ∪), but it does not justify A here.
Answer: d. A is false, but R is true

x. Assertion (A): The solution set in the system of negative integers for 0 > -4 – p is {-3, -2, -1}.

Reason (R): If the same quantity is subtracted from both sides of an inequation, the symbol of inequality is reversed.

Step 1: Solve the inequality: \[ 0 > -4 – p\\ p > -4 \]Step 2: System of negative integers → p = -3, -2, -1 ✅
Step 3: Reason R is false (subtracting the same quantity does NOT reverse inequality; multiplying/dividing by negative reverses it).
Answer: c. A is true, but R is false


Q2: If replacement set is the set of natural numbers, solve:

i. x – 5 < 0

Step 1: Solve the inequality: \[ x – 5 < 0\\ x < 5 \]Step 2: Since x ∈ N = {1, 2, 3, …}, select numbers < 5 \[ x = 1, 2, 3, 4 \]Answer: {1, 2, 3, 4}

ii. x + 1 ≤ 7

Step 1: Solve the inequality: \[ x + 1 \le 7\\ x \le 6 \]Step 2: Natural numbers ≤ 6 → \[ x = 1, 2, 3, 4, 5, 6 \]Answer: {1, 2, 3, 4, 5, 6}

iii. 3x – 4 > 6

Step 1: Solve the inequality: \[ 3x – 4 > 6\\ 3x > 10\\ x > \frac{10}{3} \approx 3.33 \]Step 2: Natural numbers > 3.33 → \[ x = 4, 5, 6, … \]Answer: {4, 5, 6, …}

iv. 4x + 1 ≥ 17

Step 1: Solve the inequality: \[ 4x + 1 \ge 17\\ 4x \ge 16\\ x \ge 4 \]Step 2: Natural numbers ≥ 4 → \[ x = 4, 5, 6, … \]Answer: {4, 5, 6, …}


Q3: If the replacement set = {-6, -3, 0, 3, 6, 9}, find the truth set of the following:

i. 2x – 1 > 9

Step 1: Solve the inequality: \[ 2x – 1 > 9\\ 2x > 10\\ x > 5 \]Step 2: Select elements from replacement set > 5: \[ x = 6, 9 \]Answer: {6, 9}

ii. 3x + 7 ≤ 1

Step 1: Solve the inequality: \[ 3x + 7 \le 1\\ 3x \le -6\\ x \le -2 \]Step 2: Select elements from replacement set ≤ -2: \[ x = -6, -3 \]Answer: {-6, -3}


Q4: Solve 9x – 7 ≤ 28 + 4x; x ∈ W

Step 1: Bring like terms together: \[ 9x – 7 \le 28 + 4x\\ 9x – 4x \le 28 + 7\\ 5x \le 35 \]Step 2: Divide both sides by 5: \[ x \le \frac{35}{5}\\ x \le 7 \]Step 3: x ∈ W (whole numbers = {0, 1, 2, 3, …}) \[ x = 0, 1, 2, 3, 4, 5, 6, 7 \]Answer: {0, 1, 2, 3, 4, 5, 6, 7}


Q5: Solve -5(x + 4) > 30; x ∈ Z

Step 1: Divide both sides by -5.
Note: When dividing an inequality by a negative number, reverse the inequality sign. \[ x + 4 < \frac{30}{-5}\\ x + 4 < -6 \]Step 2: Subtract 4 from both sides: \[ x < -6 – 4\\ x < -10 \]Step 3: x ∈ Z (integers), so the solution set is: \[ x = …, -13, -12, -11 \]Answer: {x ∈ Z | x < -10}


Q6: Solve \(\frac{2x + 1}{3} + 15 \le 17; x \in W\)

Step 1: Subtract 15 from both sides: \[ \frac{2x + 1}{3} \le 17 – 15\\ \frac{2x + 1}{3} \le 2 \]Step 2: Multiply both sides by 3: \[ 2x + 1 \le 6 \]Step 3: Subtract 1 from both sides: \[ 2x \le 5 \]Step 4: Divide both sides by 2: \[ x \le \frac{5}{2}\\ x \le 2.5 \]Step 5: x ∈ W (whole numbers = {0, 1, 2, …}) \[ x = 0, 1, 2 \]Answer: {0, 1, 2}


Q7: Solve -3 + x < 2; x ∈ N

Step 1: Solve the inequality: \[ -3 + x < 2\\ x < 2 + 3\\ x < 5 \]Step 2: x ∈ N (natural numbers = {1, 2, 3, …}) \[ x = 1, 2, 3, 4 \]Answer: {1, 2, 3, 4}


Q8: Solve and graph the solution set on a number line

i. x – 5 < -2; x ∈ N

Step 1: Solve the inequality: \[ x – 5 < -2\\ x < 3 \]Step 2: x ∈ N = {1, 2, 3, …} \[ x = 1, 2 \]Answer: {1, 2}

Number line:

←───|───|───●───●───|───|───→
   -1   0   1   2   3   4   

ii. 3x – 1 > 5; x ∈ W

Step 1: Solve the inequality: \[ 3x – 1 > 5\\ 3x > 6\\ x > 2 \]Step 2: x ∈ W = {0, 1, 2, …} \[ x = 3, 4, 5, … \]Answer: {3, 4, 5, …}

Number line:
                             °°°
←───|───|───|───|───●───●───●───→
   -1   0   1   2   3   4   5   

iii. -3x + 12 < -15; x ∈ R

Step 1: Solve the inequality: \[ -3x + 12 < -15\\ -3x < -27\\ x > 9 \quad \text{(divide by negative, reverse inequality)} \]Answer: {x ∈ R | x > 9}

Number line:
                          °°°
←───|───|───|───○───●───●───→
    6   7   8   9   10  11   

iv. 7 ≥ 3x – 8; x ∈ W

Step 1: Solve the inequality: \[ 3x – 8 \le 7\\ 3x \le 15\\ x \le 5 \]Step 2: x ∈ W = {0, 1, 2, 3, 4, 5}
Answer: {0, 1, 2, 3, 4, 5}

Number line:
                    
←───|───●───●───●───●───●───●───|───→
   -1   0   1   2   3   4   5   6

v. 8x – 8 ≤ -24; x ∈ Z

Step 1: Solve the inequality: \[ 8x – 8 \le -24\\ 8x \le -16\\ x \le -2 \]Step 2: x ∈ Z \[ x = …, -4, -3, -2 \]Answer: {x ∈ Z | x ≤ -2}

Number line:
 °°°                  
←───●───●───●───|───|───|───|───→
   -4  -3  -2  -1   0   1   2

vi. 8x – 9 ≥ 35 – 3x; x ∈ N

Step 1: Bring like terms together: \[ 8x + 3x – 9 \ge 35\\ 11x – 9 \ge 35\\ 11x \ge 44\\ x \ge 4 \]Step 2: x ∈ N = {1, 2, 3, …} \[ x = 4, 5, 6, … \]Answer: {4, 5, 6, …}

Number line:
                                     °°°                  
←───|───|───|───|───|───●───●───●───●───→
   -1   0   1   2   3   4   5   6   7

Q9: For each inequation, given below, represent the solution a number line:

i. \(\frac{5}{2} – 2x \ge \frac{1}{2}, x \in W\)

Step 1: Subtract 1/2 from both sides: \[ \frac{5}{2} – \frac{1}{2} – 2x \ge 0\\ 2 – 2x \ge 0 \]Step 2: Divide by -2 (reverse inequality): \[ x \le 1 \]Step 3: x ∈ W = {0, 1}
Answer: {0, 1}

Number line:
       
←───|───●───●───|───|───→
   -1   0   1   2   3

ii. 3(2x – 1) ≥ 2(2x + 3), x ∈ Z

Step 1: Expand both sides: \[ 6x – 3 \ge 4x + 6 \]Step 2: Subtract 4x from both sides: \[ 2x – 3 \ge 6\\ 2x \ge 9\\ x \ge \frac{9}{2} \]Step 3: x ∈ Z → x ≥ 5
Answer: {5, 6, 7, …}

Number line:
                                     °°°                  
←───|───|───|───|───|───|───|───●───●───→
   -2  -1   0   1   2   3   4   5   6

iii. 2(4 – 3x) ≤ 4(x – 5), x ∈ W

Step 1: Expand both sides: \[ 8 – 6x \le 4x – 20 \]Step 2: Add 6x to both sides: \[ 8 \le 10x – 20\\ 28 \le 10x\\ x \ge \frac{28}{10} = 2.8 \]Step 3: x ∈ W → x = 3, 4, 5, …
Answer: {3, 4, 5, …}

Number line:
                                  °°°                  
←───|───|───|───|───|───●───●───●───→
   -2  -1   0   1   2   3   4   5

iv. 4(3x + 1) > 2(4x – 1), x is a negative integer

Step 1: Expand both sides: \[ 12x + 4 > 8x – 2 \]Step 2: Subtract 8x from both sides: \[ 4x + 4 > -2\\ 4x > -6\\ x > -\frac{3}{2} \]Step 3: x ∈ negative integers → x = -1
Answer: {-1}

Number line:
         
←───|───|───●───|───|───|───→
   -3  -2  -1   0   1   2

v. \( \frac{4 – x}{2} < 3, x ∈ R\)

Step 1: Multiply both sides by 2: \[ 4 – x < 6\\ -x < 2 \]Step 2: Multiply by -1 (reverse inequality): \[ x > -2 \]Answer: {x ∈ R | x > -2}

Number line:
                         °°°                  
←───|───○───●───●───●───●───→
   -3  -2  -1   0   1   2

vi. -2(x + 8) ≤ 8, x ∈ R

Step 1: Expand the bracket: \[ -2(x + 8) ≤ 8,   x ∈ ℝ\\ -2(x + 8) ≤ 8 \\ ⇒ -2x – 16 ≤ 8 \]Step 2: Divide both sides by -2 (Inequality sign reverses) \[ x \ge -12 \]Answer: x ≥ -12,   x ∈ ℝ

Number line:
                                         °°°                  
←───|───●───●───●───●───●───●───●───●───●───→
   -13 -12 -11 -10 -9  -8  -7  -6  -5  -4

Q10: If x is a real 7(x – 2) + 2 > 2(5x + 9). Draw the solution set on the number line.

Step 1: Expand both sides: \[ 7x – 14 + 2 > 10x + 18\\ 7x – 12 > 10x + 18 \]Step 2: Subtract 10x from both sides: \[ 7x – 10x – 12 > 18\\ -3x – 12 > 18 \]Step 3: Add 12 to both sides: \[ -3x > 30 \]Step 4: Divide by -3 (reverse inequality): \[ x < -10 \]Answer: {x ∈ R | x < -10}

Number line:
 °°°                  
←───●───○───|───|───|───→
   -11 -10 -9  -8  -7


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