RS Aggarwal Solutions for Class 11 Chapter 06 Linear Inequations (In one variable)

Question 1. Fill in the blanks with correct inequality sign (>, <, ≥, ≤).
(i) 5x < 20 ⇒ x ………. 4
(ii) –3x > 9 ⇒ x ………. –3
(iii) 4x > –16 ⇒ x ………. –4
(iv) –6x ≤ –18 ⇒ x ………. 3
(v) x > –3 ⇒ –2x ………. 6
(vi) a < b and c > 0 ⇒ ……….
(vii) p – q = –3 ⇒ p ………. q
(viii) u – v = 2 ⇒ u ………. v
Answer: (i) Start with 5x < 20. Divide both sides by 5 to get x < 4.

(ii) Start with -3x > 9. Divide both sides by 3 to get x > -3.

(iii) Start with 4x > -16. Divide both sides by 4 to get x > -4.

(iv) Start with -6x ≤ -18. Divide both sides by 6 to get -x ≤ -3. Multiply by -1 on both sides (reverse the inequality sign) to get x ≥ 3.

(v) Start with x > -3. Multiply both sides by 2 to get 2x > -6. Now multiply both sides by -1 (reverse the inequality sign) to get -2x < 6. This means -2x < 6.

(vi) Since a < b and c > 0, divide both sides of the first inequality by c to get a/c < b/c.

(vii) From p - q = -3, rearrange to get p = q - 3. This shows that p is always less than q, so p < q.

(viii) From u - v = 2, rearrange to get u = v + 2. This shows that u is always greater than v, so u > v.

Exam Tip: Remember the key rule: when you multiply or divide by a negative number, always flip the inequality sign. This is the most common mistake students make.

Question 2. Solve each of the following inequations and represent the solution set on the number line. 6x ≤ 25, where (i) x ∈ N, (ii) x ∈ Z.
Answer: (i) When x ∈ N (natural numbers): Divide both sides by 6 to get x ≤ 25/6, which equals x ≤ 4.166. Since x must be a natural number, the possible values are x = {1, 2, 3, 4}. On the number line, mark filled dots at 1, 2, 3, and 4.

(ii) When x ∈ Z (integers): Divide both sides by 6 to get x ≤ 25/6, which equals x ≤ 4.166. Since x must be an integer, the possible values are x = {..., -3, -2, -1, 0, 1, 2, 3, 4}. On the number line, mark filled dots at all these integer points.

Exam Tip: Always check which set (natural numbers, integers, or real numbers) the variable belongs to - this determines which values you include in the solution.

Question 3. Solve each of the following inequations and represent the solution set on the number line. –2x > 5, where (i) x ∈ Z, (ii) x ∈ R.
Answer: (i) When x ∈ Z (integers): Multiply both sides by -1 (flip the inequality sign) to get 2x < -5. Divide both sides by 2 to get x < -2.5. Since x must be an integer, the possible values are x = {..., -2, -1, 0, 1, 2}. On the number line, mark filled dots at all these points, and an empty circle or marking at -2.5 shows the boundary.

(ii) When x ∈ R (real numbers): Multiply both sides by -1 (flip the inequality sign) to get 2x < -5. Divide both sides by 2 to get x < -5/2. The solution set is all real numbers less than -5/2, written as x ∈ (-∞, -5/2). On the number line, draw an open circle at -2.5 and shade everything to the left.

Exam Tip: Use a filled dot for integers that satisfy the inequality and an open circle for boundary points in real number solutions.

Question 4. Solve each of the following inequations and represent the solution set on the number line. 3x + 8 > 2, where (i) x ∈ Z, (ii) x ∈ R.
Answer: (i) When x ∈ Z (integers): Subtract 8 from both sides to get 3x > -6. Divide both sides by 3 to get x > -2. Since x must be an integer, the possible values are x = {-1, 0, 1, 2, 3, ...}. On the number line, mark filled dots at all these integer points starting from -1.

(ii) When x ∈ R (real numbers): Subtract 8 from both sides to get 3x > -6. Divide both sides by 3 to get x > -2. The solution set includes all real numbers greater than -2, written as x ∈ (-2, ∞). On the number line, draw an open circle at -2 and shade everything to the right.

Exam Tip: Use an open circle for the boundary point when the inequality is strict (> or <), and a filled dot or closed circle when it includes the boundary (≥ or ≤).

Question 5. Solve each of the following inequations and represent the solution set on the number line. 5x + 2 < 17, where (i) x ∈ Z, (ii) x ∈ R.
Answer: (i) When x ∈ Z (integers): Subtract 2 from both sides to get 5x < 15. Divide both sides by 5 to get x < 3. Since x must be an integer, the possible values are x = {..., -2, -1, 0, 1, 2}. On the number line, mark filled dots at all these integer points up to 2.

(ii) When x ∈ R (real numbers): Subtract 2 from both sides to get 5x < 15. Divide both sides by 5 to get x < 3. The solution set includes all real numbers less than 3, written as x ∈ (-∞, 3). On the number line, draw an open circle at 3 and shade everything to the left.

Exam Tip: Always perform the same operation on both sides to maintain the inequality. When multiplying or dividing by a negative number, reverse the inequality sign.

Question 6. Solve each of the following inequations and represent the solution set on the number line. 3x - 4 > x + 6, where x ∈ R.
Answer: Given 3x - 4 > x + 6. Add 4 to both sides to get 3x > x + 10. Subtract x from both sides to get 2x > 10. Divide both sides by 2 to get x > 5. The solution set includes all real numbers greater than 5, written as x ∈ (5, ∞). On the number line, draw an open circle at 5 and shade everything to the right.

Exam Tip: Collect all variable terms on one side and constants on the other to simplify the inequality. Check your answer by substituting a test value.

Question 7. Solve each of the following inequations and represent the solution set on the number line. 3 - 2x ≥ 4x - 9, where x ∈ R.
Answer: Given 3 - 2x ≥ 4x - 9. Subtract 3 from both sides to get -2x ≥ 4x - 12. Subtract 4x from both sides to get -6x ≥ -12. Divide both sides by 6 to get -x ≥ -2. Multiply both sides by -1 (flip the inequality sign) to get x ≤ 2. The solution set includes all real numbers less than or equal to 2, written as x ∈ (-∞, 2]. On the number line, draw a filled circle at 2 and shade everything to the left.

Exam Tip: Always remember to flip the inequality sign when multiplying or dividing by a negative number - this is critical for getting the correct answer.

Question 8. Solve each of the following inequations and represent the solution set on the number line. \( \frac{5x - 8}{3} \geq \frac{4x - 7}{2} \), where x ∈ R.
Answer: Given \( \frac{5x - 8}{3} \geq \frac{4x - 7}{2} \). Multiply both sides by 6 (the LCD of 3 and 2) to get 2(5x - 8) ≥ 3(4x - 7), which simplifies to 10x - 16 ≥ 12x - 21. Add 16 to both sides to get 10x ≥ 12x - 5. Subtract 12x from both sides to get -2x ≥ -5. Multiply both sides by -1 (flip the inequality sign) to get 2x ≤ 5. Divide both sides by 2 to get x ≤ 5/2 or x ≤ 2.5. The solution set is x ∈ (-∞, 2.5]. On the number line, draw a filled circle at 2.5 and shade everything to the left.

Exam Tip: When dealing with fractions, multiply through by the least common denominator to clear all denominators at once before solving.

Question 9. Solve each of the following inequations and represent the solution set on the number line. \( \frac{5x}{4} - \frac{4x - 1}{3} > 1 \), where x ∈ R.
Answer: Given \( \frac{5x}{4} - \frac{4x - 1}{3} > 1 \). Multiply both sides by 12 (the LCD) to get 3(5x) - 4(4x - 1) > 12, which simplifies to 15x - 16x + 4 > 12. Combine like terms to get -x + 4 > 12. Subtract 4 from both sides to get -x > 8. Multiply both sides by -1 (flip the inequality sign) to get x < -8. The solution set is x ∈ (-∞, -8). On the number line, draw an open circle at -8 and shade everything to the left.

Exam Tip: Carefully distribute negative signs when clearing fractions - sign errors are the most common mistakes in these problems.

Question 10. Solve each of the following inequations and represent the solution set on the number line. \( \frac{1}{2} \left( \frac{2}{3}x + 1 \right) \geq \frac{1}{3}(x - 2) \), where x ∈ R.
Answer: Given \( \frac{1}{2} \left( \frac{2}{3}x + 1 \right) \geq \frac{1}{3}(x - 2) \). Distribute to get \( \frac{1}{3}x + \frac{1}{2} \geq \frac{1}{3}x - \frac{2}{3} \). Subtract \( \frac{1}{3}x \) from both sides to get \( \frac{1}{2} \geq -\frac{2}{3} \). This is always true since 0.5 is greater than -0.667. Therefore, the solution is true for all values of x, written as x ∈ ℝ or x ∈ (-∞, ∞). On the number line, shade the entire line to show all real numbers are solutions.

Exam Tip: If variables cancel out and you get a true statement, the solution is all real numbers. If you get a false statement, there is no solution.

Question 11. Solve each of the following inequations and represent the solution set on the number line. \( \frac{2x - 1}{12} - \frac{x - 1}{3} < \frac{3x + 1}{4} \), where x ∈ R.
Answer: Given \( \frac{2x - 1}{12} - \frac{x - 1}{3} < \frac{3x + 1}{4} \). Multiply both sides by 12 to get (2x - 1) - 4(x - 1) < 3(3x + 1), which simplifies to 2x - 1 - 4x + 4 < 9x + 3. Combine like terms to get 3 - 2x < 9x + 3. Subtract 3 from both sides to get -2x < 9x. Subtract 9x from both sides to get -11x < 0. Divide both sides by -11 (flip the inequality sign) to get x > 0. The solution set is x ∈ (0, ∞). On the number line, draw an open circle at 0 and shade everything to the right.

Exam Tip: Break down complex fractions step by step. Multiply by the LCD to clear all denominators before combining like terms.

Question 12. Solve each of the following in equations and represent the solution set on the number line.
\( \frac{x}{4} < \frac{(5x - 2)}{3} - \frac{(7x - 3)}{5} \), where x ∈ R.
Answer: We are given: \( \frac{x}{4} < \frac{(5x - 2)}{3} - \frac{(7x - 3)}{5} \), where x ∈ R.

To clear fractions, we multiply both sides by 60:
\( \frac{x}{4}(60) < \frac{(5x - 2)}{3}(60) - \frac{(7x - 3)}{5}(60) \)

\( 15x < 20(5x - 2) - 12(7x - 3) \)
\( 15x < 100x - 40 - 84x + 36 \)
\( 15x < 16x - 4 \)

Subtracting 16x from both sides:
\( 15x - 16x < 16x - 4 - 16x \)
\( -x < -4 \)

When we multiply both sides by -1, the inequality sign reverses:
\( (-x)(-1) > (-4)(-1) \)
\( x > 4 \)

In simple words: After clearing the fractions and simplifying, we find that x must be greater than 4. This means any number larger than 4 will satisfy the original inequality.

Exam Tip: Always remember to flip the inequality sign when multiplying or dividing both sides by a negative number - this is a critical rule that many students overlook.

Question 13. Solve each of the following in equations and represent the solution set on the number line.
\( \frac{(2x - 1)}{3} \geq \frac{(3x - 2)}{4} - \frac{(2 - x)}{5} \), where x ∈ R.
Answer: We are given: \( \frac{(2x - 1)}{3} \geq \frac{(3x - 2)}{4} - \frac{(2 - x)}{5} \), where x ∈ R.

To clear fractions, we multiply both sides by 60:
\( (60)\frac{(2x - 1)}{3} \geq (60)\frac{(3x - 2)}{4} - (60)\frac{(2 - x)}{5} \)

\( 20(2x - 1) \geq 15(3x - 2) - 12(2 - x) \)
\( 40x - 20 \geq 45x - 30 - 24 + 12x \)
\( 40x - 20 \geq 57x - 54 \)

Adding 20 to both sides:
\( 40x - 20 + 20 \geq 57x - 54 + 20 \)
\( 40x \geq 57x - 34 \)

Subtracting 57x from both sides:
\( 40x - 57x \geq 57x - 34 - 57x \)
\( -17x \geq -34 \)

When we multiply both sides by -1, the inequality sign reverses:
\( (-17x)(-1) \leq (-34)(-1) \)
\( 17x \leq 34 \)

Dividing both sides by 17:
\( \frac{17x}{17} \leq \frac{34}{17} \)
\( x \leq 2 \)

In simple words: When you solve this step by step, you get x is less than or equal to 2. This means any number from negative infinity up to and including 2 works.

Exam Tip: Track each operation carefully - especially when multiplying by negative numbers, as the inequality reverses. Double-check your final answer by testing a value that should work.

Question 14. Solve each of the following in equations and represent the solution set on the number line.
\( \frac{x - 3}{x + 1} < 0, x \in \mathbb{R} \)
Answer: We are given: \( \frac{x - 3}{x + 1} < 0, x \in \mathbb{R} \)

First, we examine the signs of the numerator and denominator:

Signs of (x - 3):
\( x - 3 = 0 \rightarrow x = 3 \) (Adding 3 to both sides)
\( x - 3 < 0 \rightarrow x < 3 \) (Adding 3 to both sides)
\( x - 3 > 0 \rightarrow x > 3 \) (Adding 3 to both sides)

Signs of (x + 1):
\( x + 1 = 0 \rightarrow x = -1 \) (Subtracting 1 from both sides)
\( x + 1 < 0 \rightarrow x < -1 \) (Subtracting 1 from both sides)
\( x + 1 > 0 \rightarrow x > -1 \) (Subtracting 1 from both sides)

The fraction \( \frac{x - 3}{x + 1} \) is not defined when x = -1.

For the fraction to be negative, we need either a negative numerator with a positive denominator, or a positive numerator with a negative denominator. This occurs when - 1 < x < 3.

Therefore, \( x \in (-1, 3) \)

In simple words: A fraction is negative when its top and bottom have different signs. The fraction equals zero at x = 3 and is undefined at x = -1, so we check what values between and outside these points make the fraction negative.

Exam Tip: Always mark critical points (where numerator and denominator equal zero) on a number line, then test intervals to determine which ones satisfy the inequality.

Question 15. Solve each of the following in equations and represent the solution set on the number line.
\( \frac{x - 3}{x + 4} < 0, x \in \mathbb{R} \)
Answer: We are given: \( \frac{x - 3}{x + 4} < 0, x \in \mathbb{R} \)

Signs of (x - 3):
\( x - 3 = 0 \rightarrow x = 3 \) (Adding 3 to both sides)
\( x - 3 < 0 \rightarrow x < 3 \) (Adding 3 to both sides)
\( x - 3 > 0 \rightarrow x > 3 \) (Adding 3 to both sides)

Signs of (x + 4):
\( x + 4 = 0 \rightarrow x = -4 \) (Subtracting 4 from both sides)
\( x + 4 < 0 \rightarrow x < -4 \) (Subtracting 4 from both sides)
\( x + 4 > 0 \rightarrow x > -4 \) (Subtracting 4 from both sides)

The fraction \( \frac{x - 3}{x + 4} \) is not defined when x = -4.

For the fraction to be negative, the numerator and denominator must have opposite signs. This happens when - 4 < x < 3.

Therefore, \( x \in (-4, 3) \)

In simple words: The fraction becomes negative in the region between its critical points (-4 and 3), where the top and bottom have different signs. Outside this region, they have the same sign, making the fraction positive.

Exam Tip: Use a sign chart or number line to organize your critical values - this visual approach makes it much easier to identify which intervals satisfy the inequality.

Question 16. Solve each of the following in equations and represent the solution set on the number line.
\( \frac{2x - 3}{3x - 7} < 0, x \in \mathbb{R} \)
Answer: We are given: \( \frac{2x - 3}{3x - 7} < 0, x \in \mathbb{R} \)

Signs of (2x - 3):
\( 2x - 3 = 0 \rightarrow x = \frac{3}{2} \)
(Adding 3 to both sides and then dividing both sides by 2)

\( 2x - 3 < 0 \rightarrow x < \frac{3}{2} \)
(Adding 3 to both sides and then dividing both sides by 2)

\( 2x - 3 > 0 \rightarrow x > \frac{3}{2} \)
(Adding 3 to both sides and then dividing both sides by 2)

Signs of (3x - 7):
\( 3x - 7 = 0 \rightarrow x = \frac{7}{3} \)
(Adding 7 to both sides and then dividing both sides by 3)

\( 3x - 7 < 0 \rightarrow x < \frac{7}{3} \)
(Adding 7 to both sides and then dividing both sides by 3)

\( 3x - 7 > 0 \rightarrow x > \frac{7}{3} \)
(Adding 7 to both sides and then dividing both sides by 3)

The denominator equals zero when \( x = \frac{7}{3} \), so the fraction is undefined there.

For the fraction to be negative, the numerator and denominator must have opposite signs. This happens when \( \frac{3}{2} < x < \frac{7}{3} \).

In simple words: Find where the numerator and denominator each equal zero. Then test the regions created by these critical points to see where they have opposite signs.

Exam Tip: When dealing with fractional coefficients in the numerator and denominator, convert zeros to fractions first, then carefully order them on your number line before testing intervals.

Question 17. Solve each of the following in equations and represent the solution set on the number line.
\( \frac{x - 7}{x - 2} \geq 0, x \in \mathbb{R} \)
Answer: We are given: \( \frac{x - 7}{x - 2} \geq 0, x \in \mathbb{R} \)

Signs of (x - 7):
\( x - 7 = 0 \rightarrow x = 7 \) (Adding 7 to both sides)
\( x - 7 > 0 \rightarrow x > 7 \) (Adding 7 to both sides)
\( x - 7 < 0 \rightarrow x < 7 \) (Adding 7 to both sides)

Signs of (x - 2):
\( x - 2 = 0 \rightarrow x = 2 \) (Adding 2 to both sides)
\( x - 2 > 0 \rightarrow x > 2 \) (Adding 2 to both sides)
\( x - 2 < 0 \rightarrow x < 2 \) (Adding 2 to both sides)

At x = 2, the denominator \( \frac{x - 7}{x - 2} \) is undefined.

For the fraction to be non-negative (either positive or zero), we need both numerator and denominator to have the same sign, or the numerator to equal zero. This happens when x < 2 or x = 7 or x > 7.

Therefore, \( x \in (-\infty, 2) \cup [7, \infty) \)

In simple words: The fraction is zero when x = 7 (making it non-negative), and it's positive when the numerator and denominator have the same sign. Include x = 7 since the condition allows equality, but exclude x = 2 where the fraction is undefined.

Exam Tip: Remember that ≥ includes equality, so if the numerator can equal zero, that value is included in your solution. Always exclude values that make the denominator zero.

Question 18. Solve each of the following in equations and represent the solution set on the number line.
\( \frac{3}{x - 2} < 2, x \in \mathbb{R} \)
Answer: We are given: \( \frac{3}{x - 2} < 2, x \in \mathbb{R} \)

Subtracting 2 from both sides:
\( \frac{3}{x - 2} - 2 < 2 - 2 \)

\( \frac{3 - 2(x - 2)}{x - 2} < 0 \)

\( \frac{3 - 2x + 4}{x - 2} < 0 \)

\( \frac{7 - 2x}{x - 2} < 0 \)

Signs of (7 - 2x):
\( 7 - 2x = 0 \rightarrow x = \frac{7}{2} \)
(Subtracting 7 from both sides, then multiplying by -1 and dividing by 2)

\( 7 - 2x < 0 \rightarrow x > \frac{7}{2} \)
(Subtracting 7 from both sides, then multiplying by -1 and dividing by 2)

\( 7 - 2x > 0 \rightarrow x < \frac{7}{2} \)
(Subtracting 7 from both sides, then multiplying by -1 and dividing by 2)

Signs of (x - 2):
\( x - 2 = 0 \rightarrow x = 2 \) (Adding 2 to both sides)
\( x - 2 < 0 \rightarrow x < 2 \) (Adding 2 to both sides)
\( x - 2 > 0 \rightarrow x > 2 \) (Adding 2 to both sides)

At x = 2, the fraction \( \frac{7 - 2x}{x - 2} \) is not defined.

For the fraction to be negative, the numerator and denominator must have opposite signs. This occurs when x < 2 and x > \frac{7}{2}, which is impossible, OR when x > 2 and x < \frac{7}{2}. Since \frac{7}{2} = 3.5 and 2 < 3.5, we get 2 < x < \frac{7}{2}.

Combining, the solution is: x < 2 or x > \frac{7}{2}

Therefore, \( x \in (-\infty, 2) \cup \left(\frac{7}{2}, \infty\right) \)

In simple words: After rearranging and simplifying, you need intervals where the numerator and denominator have opposite signs. This happens in two separate regions on either side of the critical points.

Exam Tip: Always move all terms to one side of the inequality and combine over a common denominator before analyzing signs - this approach is more systematic and less error-prone.

Question 19. Solve each of the following in equations and represent the solution set on the number line.
\( \frac{1}{x - 1} \leq 2, x \in \mathbb{R} \)
Answer: We are given: \( \frac{1}{x - 1} \leq 2, x \in \mathbb{R} \)

Subtracting 2 from both sides:
\( \frac{1}{x - 1} - 2 \leq 2 - 2 \)

\( \frac{1 - 2(x - 1)}{x - 1} \leq 0 \)

\( \frac{1 - 2x + 2}{x - 1} \leq 0 \)

\( \frac{3 - 2x}{x - 1} \leq 0 \)

Signs of (3 - 2x):
\( 3 - 2x = 0 \rightarrow x = \frac{3}{2} \)
(Subtracting 3 from both sides, then multiplying by -1 and dividing by 2)

\( 3 - 2x < 0 \rightarrow x > \frac{3}{2} \)
(Subtracting 3 from both sides, then multiplying by -1 and dividing by 2)

\( 3 - 2x > 0 \rightarrow x < \frac{3}{2} \)
(Subtracting 3 from both sides, then multiplying by -1 and dividing by 2)

Signs of (x - 1):
\( x - 1 = 0 \rightarrow x = 1 \) (Adding 1 to both sides)
\( x - 1 < 0 \rightarrow x < 1 \) (Adding 1 to both sides)
\( x - 1 > 0 \rightarrow x > 1 \) (Adding 1 to both sides)

At x = 1, the fraction \( \frac{3 - 2x}{x - 1} \) is not defined.

For the fraction to be non-positive (either negative or zero), we need the numerator and denominator to have opposite signs, or the numerator to equal zero. This happens when x < 1 and x \geq \frac{3}{2} (impossible) OR when x > 1 and x \leq \frac{3}{2}. Since 1 < \frac{3}{2}, we get 1 < x \leq \frac{3}{2}. We also include x = \frac{3}{2} from the inequality ≤.

Therefore, \( x \in (-\infty, 1) \cup \left[\frac{3}{2}, \infty\right) \)

In simple words: Rearrange to get a single fraction, then identify where that fraction is negative or zero by checking the signs of its parts. Remember the inequality includes equality.

Exam Tip: For ≤ and ≥ inequalities, carefully note when to use closed brackets [ ] versus open parentheses ( ) - include points where the numerator is zero, but always exclude points where the denominator is zero.

Question 20. Solve each of the following in equations and represent the solution set on the number line.
\( \frac{5x + 8}{4 - x} < 2, x \in \mathbb{R} \)
Answer: We are given: \( \frac{5x + 8}{4 - x} < 2, x \in \mathbb{R} \)

Subtracting 2 from both sides:
\( \frac{5x + 8}{4 - x} - 2 < 2 - 2 \)

\( \frac{5x + 8 - 2(4 - x)}{4 - x} < 0 \)

\( \frac{5x + 8 - 8 + 2x}{4 - x} < 0 \)

\( \frac{7x}{4 - x} < 0 \)

Dividing both sides by 7:
\( \frac{x}{4 - x} < 0 \)

Signs of x:
\( x = 0 \)
\( x < 0 \)
\( x > 0 \)

Signs of (4 - x):
\( 4 - x = 0 \rightarrow x = 4 \)
(Subtracting 4 from both sides, then dividing by -1)

\( 4 - x < 0 \rightarrow x > 4 \)
(Subtracting 4 from both sides, then multiplying by -1)

\( 4 - x > 0 \rightarrow x < 4 \)
(Subtracting 4 from both sides, then multiplying by -1)

At x = 4, the fraction \( \frac{x}{4 - x} \) is not defined.

For the fraction to be negative, the numerator and denominator must have opposite signs. This happens when x < 0 or x > 4.

Therefore, \( x \in (-\infty, 0) \cup (4, \infty) \)

In simple words: After simplifying the inequality, you get a fraction that is negative when its parts have opposite signs. The critical values are 0 and 4, and you check which intervals satisfy the condition.

Exam Tip: Always simplify rational inequalities by combining fractions first. Be systematic about finding critical points and testing the sign of the expression in each region.

Question 21. Solve each of the following in equations and represent the solution set on the number line.
\( |3x - 7| > 4, x \in \mathbb{R} \)
Answer: We are given: \( |3x - 7| > 4, x \in \mathbb{R} \)

When an absolute value is greater than a positive number, we have two cases:
\( 3x - 7 < -4 \) or \( 3x - 7 > 4 \)
(Because if \( |u| > a \) where a > 0, then \( u < -a \) or \( u > a \))

Case 1: \( 3x - 7 < -4 \)

Adding 7 to both sides:
\( 3x - 7 + 7 < -4 + 7 \)
\( 3x < 3 \)

Dividing by 3 on both sides:
\( x < 1 \)

Case 2: \( 3x - 7 > 4 \)

Adding 7 to both sides:
\( 3x - 7 + 7 > 4 + 7 \)
\( 3x > 11 \)

Dividing by 3 on both sides:
\( x > \frac{11}{3} \)

Combining both cases:

Therefore, \( x \in (-\infty, 1) \cup \left(\frac{11}{3}, \infty\right) \)

In simple words: When something's absolute value (its distance from zero) is greater than 4, it must be either far to the left (less than -4) or far to the right (greater than 4) on a number line.

Exam Tip: For absolute value inequalities, remember the key distinction: \( |x| > a \) gives an "or" statement (two separate regions), while \( |x| < a \) gives an "and" statement (one middle region).

Question 22. Solve each of the following in equations and represent the solution set on the number line.
\( |5 - 2x| \leq 3, x \in \mathbb{R} \)
Answer: We are given: \( |5 - 2x| \leq 3, x \in \mathbb{R} \)

When an absolute value is at most a positive number, we have:
\( 5 - 2x \geq -3 \) or \( 5 - 2x \leq 3 \)

Case 1: \( 5 - 2x \geq -3 \)

Subtracting 5 from both sides:
\( 5 - 2x - 5 \geq -3 - 5 \)
\( -2x \geq -8 \)

Multiplying both sides by -1 (inequality reverses):
\( -2x(-1) \leq -8(-1) \)
\( 2x \leq 8 \)

Dividing by 2 on both sides:
\( x \leq 4 \)

Case 2: \( 5 - 2x \leq 3 \)

Subtracting 5 from both sides:
\( 5 - 2x - 5 \leq 3 - 5 \)
\( -2x \leq -2 \)

Multiplying both sides by -1 (inequality reverses):
\( -2x(-1) \geq -2(-1) \)
\( 2x \geq 2 \)

Dividing by 2 on both sides:
\( x \geq 1 \)

Combining both cases, we need x ≤ 4 AND x ≥ 1:

Therefore, \( x \in [1, 4] \)

In simple words: The absolute value of something being at most 3 means that thing is trapped in a range - it can't be too far positive or too far negative. So we get a closed interval between two values.

Exam Tip: For \( |u| \leq a \), the solution is a single closed interval \( -a \leq u \leq a \). For \( |u| > a \), it's two separate open intervals: \( u < -a \) or \( u > a \).

Question 23. Solve each of the following in equations and represent the solution set on the number line.
\( |4x - 5| \leq \frac{1}{3}, x \in \mathbb{R} \)
Answer: We are given: \( |4x - 5| \leq \frac{1}{3}, x \in \mathbb{R} \)

When an absolute value is at most a positive number, we have two simultaneous conditions:
\( 4x - 5 \leq \frac{1}{3} \) or \( 4x - 5 \geq -\frac{1}{3} \)

Case 1: \( 4x - 5 \leq \frac{1}{3} \)

Adding 5 to both sides:
\( 4x - 5 + 5 \leq \frac{1}{3} + 5 \)
\( 4x \leq \frac{1 + 15}{3} \)
\( 4x \leq \frac{16}{3} \)

Dividing by 4 on both sides:
\( x \leq \frac{16}{12} = \frac{4}{3} \)

Case 2: \( 4x - 5 \geq -\frac{1}{3} \)

Adding 5 to both sides:
\( 4x - 5 + 5 \geq -\frac{1}{3} + 5 \)
\( 4x \geq \frac{-1 + 15}{3} \)
\( 4x \geq \frac{14}{3} \)

Dividing by 4 on both sides:
\( x \geq \frac{14}{12} = \frac{7}{6} \)

Combining both cases, we need x ≤ \frac{4}{3} AND x ≥ \frac{7}{6}:

Therefore, \( x \in \left[\frac{7}{6}, \frac{4}{3}\right] \)

In simple words: The absolute value inequality gives us a range - all values of x between two boundaries (inclusive) that keep the expression inside the absolute value close enough to zero.

Exam Tip: When combining fractions in the middle steps, find a common denominator carefully. Also, convert the final answer to the simplest form if needed.

Question 24. Solve each of the following in equations and represent the solution set on the number line.
\( \frac{1}{|x| - 3} \leq \frac{1}{2}, x \in \mathbb{R} \)
Answer: Given: \( \frac{1}{|x| - 3} \leq \frac{1}{2}, x \in \mathbb{R} \)

For absolute value, we have: \( |x| = -x \) when \( x < 0 \) and \( |x| = x \) when \( x \geq 0 \)

To find the domain of \( \frac{1}{|x| - 3} \leq \frac{1}{2} \), we need \( |x| + 3 \neq 0 \), which gives \( x \neq -3 \) or \( x \neq 3 \). Thus, the domain is \( -3 < x < 3 \).

We now split into intervals: \( x < -3, -3 < x < 0, 0 \leq x < 3, x > 3 \)

For \( x < -3 \): After simplifying, we get \( \frac{2 - (-x - 3)}{2(-x - 3)} \leq 0 \), which becomes \( \frac{x + 5}{-2x - 6} \leq 0 \)

Signs of \( x + 5 \): equals zero at \( x = -5 \); positive when \( x > -5 \); negative when \( x < -5 \)

Signs of \( -2x - 6 \): equals zero at \( x = -3 \); positive when \( x < -3 \); negative when \( x > -3 \)

From this interval, the solution is \( x \leq -5 \) or \( x > -3 \)

For \( -3 < x < 0 \): The condition simplifies similarly, yielding \( -3 < x < 0 \)

For \( 0 \leq x < 3 \): After simplification, we find \( 0 \leq x < 3 \)

For \( x \geq 3 \): The solution is \( x \leq -5 \) or \( x > -3 \), which combined with this interval gives no overlap in the intended range.

Merging all intervals: \( x \leq -5 \) or \( -3 < x < 0 \) or \( 0 \leq x < 3 \) or \( x \geq 5 \)

Therefore, \( x \in (-\infty, -5] \cup (-3, 3) \cup [5, \infty) \)
In simple words: We check different regions of the number line by considering when \( x \) is negative, zero, or positive. In each region, we find which values satisfy the inequality. The final answer combines all these regions together.

Exam Tip: Always determine the domain first to identify values where the expression is undefined. Split the problem into cases based on the absolute value definition.

Question 25. Solve each of the following in equations and represent the solution set on the number line.
\( \frac{1}{|x - 3| - x} < 2, x \in \mathbb{R} \)
Answer: Given: \( \frac{1}{|x - 3| - x} < 2, x \in \mathbb{R} \)

For absolute value, \( |x - 3| = x - 3 \) when \( x \geq 3 \) and \( |x - 3| = -(x - 3) \) when \( x < 3 \)

The expression \( \frac{1}{|x - 3| - x} \) is not defined when \( x = 0 \). The domain is \( x < 0 \) or \( 0 < x < 3 \) or \( x \geq 3 \)

For \( x < 0 \): We have \( |x - 3| = -(x - 3) = 3 - x \), so the inequality becomes \( \frac{1}{3 - 2x} < 2 \). Simplifying: \( \frac{3 - 4x}{x} < 0 \). The solution for this interval is \( x < 0 \)

For \( 0 < x < 3 \): We still have \( |x - 3| = 3 - x \), giving \( \frac{3 - 4x}{x} < 0 \). The solution is \( \frac{3}{4} < x < 3 \)

For \( x \geq 3 \): We have \( |x - 3| = x - 3 \), so \( \frac{1}{-3 - 2x} < 2 \). This simplifies to \( \frac{-3 - 2x}{x} < 0 \), giving \( x < 0 \) or \( x > \frac{3}{4} \). Combined with \( x \geq 3 \), we get \( x \geq 3 \)

Therefore, \( x \in (-\infty, 0) \cup \left(\frac{3}{4}, 3\right) \cup [3, \infty) \)
In simple words: Split the absolute value into cases based on when the expression inside is positive or negative. In each case, solve the resulting inequality and keep only the values that fit that case. Combine all valid regions.

Exam Tip: Remember that absolute value has two cases. Always check which case applies to your current interval before substituting.

Question 26. Solve each of the following in equations and represent the solution set on the number line.
\( \left| \frac{2x - 1}{x - 1} \right| < 2, x \in \mathbb{R} \)
Answer: Given: \( \left| \frac{2x - 1}{x - 1} \right| < 2, x \in \mathbb{R} \)

The inequality \( \left| \frac{2x - 1}{x - 1} \right| < 2 \) is equivalent to \( -2 < \frac{2x - 1}{x - 1} < 2 \)

This splits into two conditions:
(1) \( \frac{2x - 1}{x - 1} > -2 \)
(2) \( \frac{2x - 1}{x - 1} < 2 \)

For condition (1): Adding 2 to both sides and simplifying: \( \frac{2x - 1 + 2(x - 1)}{x - 1} > 0 \), which gives \( \frac{4x - 3}{x - 1} > 0 \)

Signs of \( 4x - 3 \): equals zero at \( x = \frac{3}{4} \); positive when \( x > \frac{3}{4} \); negative when \( x < \frac{3}{4} \)

Signs of \( x - 1 \): equals zero at \( x = 1 \); positive when \( x > 1 \); negative when \( x < 1 \)

From condition (1), the solution is \( x < \frac{3}{4} \) or \( x > 1 \)

For condition (2): Subtracting 2 and simplifying: \( \frac{2x - 1 - 2(x - 1)}{x - 1} < 0 \), which gives \( \frac{1}{x - 1} < 0 \)

This holds when \( x - 1 < 0 \), so \( x < 1 \)

Combining both conditions: \( \left( x < \frac{3}{4} \text{ or } x > 1 \right) \) and \( x < 1 \) gives \( x < \frac{3}{4} \)

Therefore, \( x \in \left( -\infty, \frac{3}{4} \right) \)
In simple words: An absolute value inequality breaks into two separate inequalities. Solve each one carefully by analyzing the signs of the numerator and denominator. The final answer is where both conditions hold at the same time.

Exam Tip: When solving \( |f(x)| < a \), always split it into \( -a < f(x) < a \). Mark critical points where numerator and denominator equal zero to build sign charts.

Question 27. Solve each of the following in equations and represent the solution set on the number line.
\( \frac{|x - 3|}{x - 3} < 0, x \in \mathbb{R} \)
Answer: Given: \( \frac{|x - 3|}{x - 3} < 0, x \in \mathbb{R} \)

For the inequality to be defined, we need \( x \neq 3 \)

When \( x > 3 \): We have \( |x - 3| = x - 3 \), so \( \frac{|x - 3|}{x - 3} = \frac{x - 3}{x - 3} = 1 > 0 \). This does not satisfy the inequality.

When \( x < 3 \): We have \( |x - 3| = -(x - 3) = 3 - x \), so \( \frac{|x - 3|}{x - 3} = \frac{3 - x}{x - 3} = \frac{-(x - 3)}{x - 3} = -1 < 0 \). This satisfies the inequality.

Therefore, the solution is \( x < 3 \)

Thus, \( x \in (-\infty, 3) \)
In simple words: When the numerator is an absolute value and the denominator is the expression inside it, the fraction equals 1 when positive and -1 when negative. Find which values make it negative.

Exam Tip: Recognize that \( \frac{|A|}{A} = 1 \) if \( A > 0 \) and \( \frac{|A|}{A} = -1 \) if \( A < 0 \). This simplifies such problems instantly.

Question 28. Solve each of the following in equations and represent the solution set on the number line.
\( \frac{|x| - 1}{|x| - 2} \geq 0, x \in \mathbb{R} - \{-2, 2\} \)
Answer: Given: \( \frac{|x| - 1}{|x| - 2} \geq 0, x \in \mathbb{R} - \{-2, 2\} \)

For absolute value: \( |x| = x \) when \( x \geq 0 \) and \( |x| = -x \) when \( x < 0 \)

The expression is undefined when \( |x| = 2 \), i.e., \( x = -2 \) or \( x = 2 \). The domain is \( x < -2 \) or \( -2 < x < 2 \) or \( x > 2 \)

For \( x < -2 \): We have \( |x| = -x \), so \( \frac{-x - 1}{-x - 2} \geq 0 \). The numerator \( -x - 1 = 0 \) at \( x = -1 \); it is positive when \( x < -1 \) and negative when \( x > -1 \). The denominator \( -x - 2 = 0 \) at \( x = -2 \); it is positive when \( x < -2 \) and negative when \( x > -2 \). The solution for this interval is \( x < -2 \) or \( x \geq -1 \). Since we are in \( x < -2 \), we get \( x < -2 \)

For \( -2 < x < 0 \): Similarly, \( \frac{-x - 1}{-x - 2} \geq 0 \) gives \( x < -2 \) or \( x \geq -1 \). Within \( -2 < x < 0 \), this is \( -1 \leq x < 0 \)

For \( 0 \leq x < 2 \): We have \( |x| = x \), so \( \frac{x - 1}{x - 2} \geq 0 \). The numerator equals zero at \( x = 1 \) (positive when \( x > 1 \), negative when \( x < 1 \)). The denominator equals zero at \( x = 2 \) (positive when \( x > 2 \), negative when \( x < 2 \)). The solution is \( x \leq 1 \) or \( x > 2 \). Within \( 0 \leq x < 2 \), this is \( 0 \leq x \leq 1 \)

For \( x > 2 \): Using \( |x| = x \), we get \( \frac{x - 1}{x - 2} \geq 0 \), which gives \( x \leq 1 \) or \( x > 2 \). Within \( x > 2 \), this is \( x > 2 \)

Combining all intervals: \( x < -2 \) or \( -1 \leq x \leq 1 \) or \( x > 2 \)

Therefore, \( x \in (-\infty, -2) \cup [-1, 1] \cup (2, \infty) \)
In simple words: Split based on whether \( x \) is negative or non-negative, since that changes the absolute value. In each case, find where the fraction is non-negative by analyzing the signs of numerator and denominator separately. Then bring all solutions together.

Exam Tip: Use a sign chart: mark zeros of numerator and denominator on a number line, then test the sign of the fraction in each interval. Include zeros of the numerator (they make the fraction zero, which satisfies \( \geq 0 \)), but exclude zeros of the denominator (undefined points).

Question 29. Solve each of the following in equations and represent the solution set on the number line.
\( \frac{1}{2-|x|} \geq 1, x \in \mathbb{R} - \{-2, 2\} \)
Answer: Given the inequality \( \frac{1}{2-|x|} \geq 1 \) where x is a real number excluding -2 and 2.

The absolute value function has two cases:
When x ≥ 0, |x| = x; when x < 0, |x| = -x

Domain: The denominator is undefined when 2 - |x| = 0, which means |x| = 2. This gives x = -2 or x = 2. Thus, the domain is x < -2 or x > 2, combined with the given restriction x < -2 or x > 2.

For x < -2 (where |x| = -x):
\( \frac{1}{2-(-x)} \geq 1 \)
\( \frac{1}{2+x} \geq 1 \)
\( \frac{1}{2+x} - 1 \geq 0 \)
\( \frac{1-(2+x)}{2+x} \geq 0 \)
\( \frac{-1-x}{2+x} \geq 0 \)

Signs of -1 - x: This equals zero when x = -1, is positive when x < -1, and is negative when x > -1.
Signs of 2 + x: This equals zero when x = -2, is positive when x > -2, and is negative when x < -2.

For the fraction to be non-negative in the interval x < -2, we need the numerator and denominator to have the same sign. When x < -2: numerator is positive (since -1 - x > 0), denominator is negative. This gives no solution in this interval.

For -2 < x < 0 (where |x| = -x):
Following similar analysis, we get -2 < x ≤ 1.

For 0 ≤ x < 2 (where |x| = x):
\( \frac{1}{2-x} \geq 1 \)
\( \frac{1-(2-x)}{2-x} \geq 0 \)
\( \frac{x-1}{2-x} \geq 0 \)

Signs of x - 1: equals zero at x = 1, positive when x > 1, negative when x < 1.
Signs of 2 - x: equals zero at x = 2, positive when x < 2, negative when x > 2.

For 0 ≤ x < 2, the fraction is non-negative when 1 ≤ x < 2.

For x > 2 (where |x| = x):
\( \frac{1}{2-x} \geq 1 \) becomes \( \frac{x-1}{2-x} \geq 0 \)

When x > 2: numerator is positive, denominator is negative, so no solution here.

Combining all valid intervals: -2 < x ≤ 1 or 1 ≤ x < 2 = -2 < x < 2.

Therefore, \( x \in (-2, -1] \cup [1, 2) \)
In simple words: The solution consists of two separate sections: from just after -2 up to and including -1, plus from 1 up to (but not including) 2. These are the x-values that satisfy the original inequality while respecting the domain restrictions.

Exam Tip: Always identify domain restrictions first by finding where the denominator equals zero. Handle absolute value by splitting into separate cases based on the sign of the expression inside.

Question 30. Solve each of the following in equations and represent the solution set on the number line.
\( |x + a| + |x| > 3, x \in \mathbb{R} \)
Answer: Given the inequality \( |x + a| + |x| > 3 \) where x is any real number.

The absolute value expressions can be written as:
\( |x + a| = -(x+a) \text{ or } (x+a) \)
\( |x| = -x \text{ or } x \)

This gives us four cases to consider:

Case 1: When \( |x + a| = -(x+a) \) and \( |x| = -x \)
\( -(x+a) + (-x) > 3 \)
\( -x - a - x > 3 \)
\( -2x - a > 3 \)
\( -2x > 3 + a \)
\( x < -\frac{3+a}{2} \)

Case 2: When \( |x + a| = -(x+a) \) and \( |x| = x \)
\( -(x+a) + x > 3 \)
\( -a > 3 \)
This is a constant condition that doesn't depend on x. If -a > 3 (i.e., a < -3), all x satisfy this. Otherwise, no solution from this case.

Case 3: When \( |x + a| = (x+a) \) and \( |x| = -x \)
\( (x+a) + (-x) > 3 \)
\( a > 3 \)
Again, this doesn't depend on x. If a > 3, all x satisfy this case. Otherwise, no solution.

Case 4: When \( |x + a| = (x+a) \) and \( |x| = x \)
\( (x+a) + x > 3 \)
\( 2x + a > 3 \)
\( 2x > 3 - a \)
\( x > \frac{3-a}{2} \)

Combining the results from all four cases, the solution depends on the value of a. When considering the conditions without additional constraints on a:

Solution: \( x < -\frac{3+a}{2} \text{ or } x > \frac{3-a}{2} \)
In simple words: The answer has two parts separated by a gap: values of x smaller than a certain threshold involving a, or values larger than another threshold. The exact boundaries shift depending on the parameter a.

Exam Tip: When dealing with compound absolute value inequalities, systematically work through all possible sign combinations. The final answer often consists of multiple disjoint intervals or depends on parameters.

Question 31. Solve each of the following in equations and represent the solution set on the number line.
\( \frac{x}{x-4} > 1, x \neq 4 \)
Answer: Given \( \frac{x}{x-4} > 1 \) where \( x \neq 4 \).

Rearranging the inequality:
\( \frac{x}{x-4} - 1 > 0 \)
\( \frac{x - (x-4)}{x-4} > 0 \)
\( \frac{x - x + 4}{x-4} > 0 \)
\( \frac{4}{x-4} > 0 \)

For this fraction to be positive, the denominator must be positive (since the numerator 4 is always positive):
\( x - 4 > 0 \)
\( x > 4 \)

Therefore, \( x \in (4, \infty) \)

Note: The inequality cannot be satisfied when x ≤ 4 because either the denominator is zero (making the fraction undefined at x = 4) or negative (making the fraction negative when x < 4).
In simple words: We need the fraction to be positive. Since the top is always 4 (positive), the bottom must also be positive. This means x must be larger than 4.

Exam Tip: When solving rational inequalities, rewrite by moving all terms to one side and combining over a common denominator. Analyze the sign of numerator and denominator separately to determine where the fraction is positive or negative.

Question 32. Solve the following systems of linear in equations:
\( \frac{4}{x+1} \leq 3 \leq \frac{6}{x+1}, x > 0 \)
Answer: The compound inequality splits into two parts: \( \frac{4}{x+1} \leq 3 \) and \( 3 \leq \frac{6}{x+1} \), both with x > 0.

For the first inequality \( \frac{4}{x+1} \leq 3 \):
Subtracting 3 from both sides:
\( \frac{4}{x+1} - 3 \leq 0 \)
\( \frac{4 - 3(x+1)}{x+1} \leq 0 \)
\( \frac{4 - 3x - 3}{x+1} \leq 0 \)
\( \frac{1 - 3x}{x+1} \leq 0 \)

The numerator 1 - 3x equals zero when \( x = \frac{1}{3} \), is positive when \( x < \frac{1}{3} \), and is negative when \( x > \frac{1}{3} \).

The denominator x + 1 equals zero when x = -1, is positive when x > -1, and is negative when x < -1.

For the fraction to be non-positive with x > 0, we need \( x \geq \frac{1}{3} \) (numerator non-positive, denominator positive).

For the second inequality \( 3 \leq \frac{6}{x+1} \):
Subtracting 3 from both sides:
\( 0 \leq \frac{6}{x+1} - 3 \)
\( 0 \leq \frac{6 - 3(x+1)}{x+1} \)
\( 0 \leq \frac{6 - 3x - 3}{x+1} \)
\( 0 \leq \frac{3 - 3x}{x+1} \)
Dividing by 3: \( 0 \leq \frac{1-x}{x+1} \)

The numerator 1 - x equals zero when x = 1, is positive when x < 1, and is negative when x > 1.
The denominator x + 1 is positive when x > -1.

For the fraction to be non-negative with x > 0, we need \( 0 < x \leq 1 \) (numerator non-negative, denominator positive).

Combining both conditions: \( x \geq \frac{1}{3} \) and \( x \leq 1 \) with x > 0.

Therefore, \( x \in \left[\frac{1}{3}, 1\right] \)
In simple words: The first part requires x to be at least one-third, and the second part requires x to be at most 1. Together with x > 0, the solution is all x between one-third and 1, including both endpoints.

Exam Tip: For compound inequalities, solve each part separately then find the intersection. Use sign analysis tables to track where numerators and denominators are positive or negative.

Question 33. Solve the following systems of linear in equations:
\( -11 \leq 4x - 3 \leq 13 \)
Answer: This compound inequality breaks into two parts: \( -11 \leq 4x - 3 \) and \( 4x - 3 \leq 13 \).

For \( -11 \leq 4x - 3 \):
Rewrite as \( 4x - 3 \geq -11 \)
Adding 3 to both sides:
\( 4x - 3 + 3 \geq -11 + 3 \)
\( 4x \geq -8 \)
Dividing by 4:
\( x \geq -2 \)

For \( 4x - 3 \leq 13 \):
Adding 3 to both sides:
\( 4x - 3 + 3 \leq 13 + 3 \)
\( 4x \leq 16 \)
Dividing by 4:
\( x \leq 4 \)

Combining both conditions: \( x \geq -2 \) and \( x \leq 4 \)

Therefore, \( x \in [-2, 4] \)
In simple words: Add 3 to all parts of the inequality, then divide all parts by 4. The result shows that x can range from -2 up to 4, including both boundary values.

Exam Tip: When solving compound linear inequalities, apply the same operations to all three parts simultaneously. Remember that dividing or multiplying by a negative number reverses the inequality sign, but here we divide by positive 4 so signs stay the same.

Question 34. Solve the following systems of linear in equations:
\( 5x - 7 < (x + 3), 1 - \frac{3x}{2} \geq x - 4 \)
Answer: This system consists of two separate inequalities that must both be satisfied.

For the first inequality \( 5x - 7 < x + 3 \):
Adding 7 to both sides:
\( 5x - 7 + 7 < x + 3 + 7 \)
\( 5x < x + 10 \)
Subtracting x from both sides:
\( 5x - x < x + 10 - x \)
\( 4x < 10 \)
Dividing by 4:
\( x < \frac{5}{2} \)

For the second inequality \( 1 - \frac{3x}{2} \geq x - 4 \):
Subtracting 1 from both sides:
\( 1 - \frac{3x}{2} - 1 \geq x - 4 - 1 \)
\( -\frac{3x}{2} \geq x - 5 \)
Multiplying both sides by 2:
\( -3x \geq 2x - 10 \)
Subtracting 2x from both sides:
\( -3x - 2x \geq 2x - 10 - 2x \)
\( -5x \geq -10 \)
Multiplying both sides by -1 (reversing the inequality):
\( 5x \leq 10 \)
Dividing by 5:
\( x \leq 2 \)

Combining both conditions: \( x < \frac{5}{2} \) and \( x \leq 2 \)

Since \( \frac{5}{2} = 2.5 > 2 \), the more restrictive condition is \( x \leq 2 \).

Therefore, \( x \in (-\infty, 2] \)
In simple words: Solve each inequality separately. The first gives x less than 2.5, and the second gives x less than or equal to 2. Since x must satisfy both, we take the stricter limit, which is x ≤ 2.

Exam Tip: When solving a system of inequalities, find the solution to each inequality separately, then find the intersection (values satisfying all conditions simultaneously). Always reverse the inequality sign when multiplying or dividing by a negative number.

Question 35. Solve the following systems of linear in equations:
\( -2 < \frac{6-5x}{4} < 7 \)
Answer: This compound inequality splits into two parts: \( -2 < \frac{6-5x}{4} \) and \( \frac{6-5x}{4} < 7 \).

For the first inequality \( -2 < \frac{6-5x}{4} \):
Rewrite as \( \frac{6-5x}{4} > -2 \)
Multiplying both sides by 4:
\( 6 - 5x > -8 \)
Subtracting 6 from both sides:
\( 6 - 5x - 6 > -8 - 6 \)
\( -5x > -14 \)
Multiplying both sides by -1 (reversing the inequality):
\( 5x < 14 \)
Dividing by 5:
\( x < \frac{14}{5} \)

For the second inequality \( \frac{6-5x}{4} < 7 \):
Multiplying both sides by 4:
\( 6 - 5x < 28 \)
Subtracting 6 from both sides:
\( 6 - 5x - 6 < 28 - 6 \)
\( -5x < 22 \)
Multiplying both sides by -1 (reversing the inequality):
\( 5x > -22 \)
Dividing by 5:
\( x > -\frac{22}{5} \)

Combining both conditions: \( x > -\frac{22}{5} \) and \( x < \frac{14}{5} \)

Therefore, \( x \in \left(-\frac{22}{5}, \frac{14}{5}\right) \)
In simple words: Multiply all parts by 4 to clear the fraction, then isolate x in both resulting inequalities. The solution is the range of x-values that satisfy both conditions simultaneously, which is between -4.4 and 2.8.

Exam Tip: To solve compound inequalities with fractions, multiply all parts by the same positive number to eliminate denominators. Remember to reverse inequality signs when multiplying or dividing by negative values, and always check your final answer by testing a value in the solution interval.

Question 36. Solve the following systems of linear inequations: 3x - x > x + (4-x)/3 and x + (4-x)/3 > 3
Answer: We treat each inequality separately. For the first inequality, \( 3x - x > x + \frac{4-x}{3} \): \( 2x > x + \frac{4-x}{3} \) Removing \( x \) from both sides: \( x > \frac{4-x}{3} \) Multiplying both sides by 3: \( 3x > 4 - x \) Adding \( x \) to both sides: \( 4x > 4 \) Dividing both sides by 4: \( x > 1 \) For the second inequality, \( x + \frac{4-x}{3} > 3 \): Multiplying both sides by 3: \( 3x + (4-x) > 9 \) \( 2x + 4 > 9 \) Removing 4 from both sides: \( 2x > 5 \) Dividing both sides by 2: \( x > \frac{5}{2} \) Taking the intersection of both conditions (\( x > 1 \) and \( x > \frac{5}{2} \)): Therefore, \( x > \frac{5}{2} \)

Exam Tip: When solving compound inequalities, always solve each part separately and then combine using intersection (AND) or union (OR) as required by the problem.

Question 37. Solve the following systems of linear inequations: (7x-1)/2 < -3, (3x+8)/5 + 11 < 0
Answer: We address each inequality separately. For the first inequality, \( \frac{7x-1}{2} < -3 \): Multiplying both sides by 2: \( 7x - 1 < -6 \) Adding 1 to both sides: \( 7x < -5 \) Dividing both sides by 7: \( x < -\frac{5}{7} \) For the second inequality, \( \frac{3x+8}{5} + 11 < 0 \): Subtracting 11 from both sides: \( \frac{3x+8}{5} < -11 \) Multiplying both sides by 5: \( 3x + 8 < -55 \) Subtracting 8 from both sides: \( 3x < -63 \) Dividing both sides by 3: \( x < -21 \) Taking the intersection of both conditions (\( x < -\frac{5}{7} \) and \( x < -21 \)): The more restrictive condition is \( x < -21 \). Therefore, \( x < -21 \)

Exam Tip: After solving inequalities, verify by substituting a value that satisfies your solution back into the original inequalities to check correctness.

Question 38. Solve the following systems of linear inequations: -12 < 4 - (3x)/5 ≤ 2
Answer: We treat this as two linked inequalities. For the first part, \( -12 < 4 - \frac{3x}{5} \): Rearranging: \( 4 - \frac{3x}{5} > -12 \) Subtracting 4 from both sides: \( -\frac{3x}{5} > -16 \) Multiplying both sides by 5: \( -3x > -80 \) Dividing both sides by -3 (inequality flips): \( x < \frac{80}{3} \) For the second part, \( 4 - \frac{3x}{5} \leq 2 \): Subtracting 4 from both sides: \( -\frac{3x}{5} \leq -2 \) Multiplying both sides by 5: \( -3x \leq -10 \) Dividing both sides by -3 (inequality flips): \( x \geq \frac{10}{3} \) Taking the intersection of both conditions (\( x < \frac{80}{3} \) and \( x \geq \frac{10}{3} \)): Therefore, \( \frac{10}{3} \leq x < \frac{80}{3} \)

Exam Tip: Remember that when multiplying or dividing by a negative number, the inequality sign must be reversed - this is a critical rule that many students forget.

Question 39. Solve the following systems of linear inequations: 1 ≤ |x - 2| ≤ 3
Answer: We treat this as two separate absolute value inequalities. For \( |x - 2| \geq 1 \): This means either \( x - 2 \geq 1 \) or \( x - 2 \leq -1 \) From \( x - 2 \geq 1 \): \( x \geq 3 \) From \( x - 2 \leq -1 \): \( x \leq 1 \) So the solution is \( x \leq 1 \) or \( x \geq 3 \) For \( |x - 2| \leq 3 \): This means \( -3 \leq x - 2 \leq 3 \) Adding 2 throughout: \( -1 \leq x \leq 5 \) Taking the intersection of both conditions: - From the first: \( x \leq 1 \) or \( x \geq 3 \) - From the second: \( -1 \leq x \leq 5 \) The overlap gives us \( -1 \leq x \leq 1 \) or \( 3 \leq x \leq 5 \) Therefore, \( x \in [-1, 1] \cup [3, 5] \)

Exam Tip: For absolute value inequalities, always split into two cases and remember that \( |a| \geq b \) leads to "or" while \( |a| \leq b \) leads to "and".

Question 40. Find all pairs of consecutive even positive integers both of which are larger than 8 such that their sum is less than 25.
Answer: Let the two consecutive even positive integers be \( x \) and \( x + 2 \). From the condition that both are larger than 8: \( x > 8 \) and \( x + 2 > 8 \) The second condition simplifies to \( x > 6 \), so the binding constraint is \( x > 8 \). From the condition that their sum is less than 25: \( x + (x + 2) < 25 \) \( 2x + 2 < 25 \) Subtracting 2 from both sides: \( 2x < 23 \) Dividing both sides by 2: \( x < 11.5 \) Combining the constraints \( x > 8 \) and \( x < 11.5 \): Since \( x \) must be an even integer, the only possible value is \( x = 10 \). Therefore, \( x + 2 = 12 \). The required pair is \( (10, 12) \)

Exam Tip: When variables must be integers, always identify the constraint interval first, then check which integers fall within it rather than guessing.

Question 41. Find all pairs of consecutive even positive integers both of which are larger than 8 such that their sum is less than 25.
Answer: Let the two consecutive even positive integers be represented as \( x \) and \( x + 2 \). Given that both integers exceed 8: \( x > 8 \) and \( x + 2 > 8 \) Simplifying the second inequality gives \( x > 6 \), so the stronger condition is \( x > 8 \). Given that their sum is less than 25: \( x + (x + 2) < 25 \) \( 2x + 2 < 25 \) Subtracting 2 from both sides: \( 2x < 23 \) Dividing by 2: \( x < 11.5 \) Merging both constraints (\( x > 8 \) and \( x < 11.5 \)): The only even integer satisfying this range is \( x = 10 \). Thus \( x + 2 = 12 \). The required pair is \( (10, 12) \)

Exam Tip: Always verify your answer by substituting back: 10 > 8 ✓, 12 > 8 ✓, and 10 + 12 = 22 < 25 ✓.

Question 42. A company manufactures cassettes. Its cost and revenue function are C(x) = 25000 + 30x and R(x) = 43x respectively, where x is the number of cassettes produced and sold in a week. How many cassettes must be sold by the company to realize some profit?
Answer: For the company to gain profit, revenue must exceed cost: \( R(x) > C(x) \) \( 43x > 25000 + 30x \) Subtracting \( 30x \) from both sides: \( 13x > 25000 \) Dividing both sides by 13: \( x > \frac{25000}{13} \approx 1923.08 \) Since the number of cassettes must be a whole number, the minimum number of cassettes that must be sold is 1924. Therefore, the company must sell at least 1924 cassettes to achieve profit.

Exam Tip: In real-world applications, always round up when dealing with quantities that must be whole numbers - you cannot sell a fractional cassette.

Question 43. The watering acidity in a pool is considered normal when the average pH reading of three daily measurements is between 8.2 and 8.5. If the first two pH readings are 8.48 and 8.35, find the range of the pH values for the third reading that will result in the acidity level being normal.
Answer: Let \( x \) denote the third pH reading. The average of the three readings must fall between 8.2 and 8.5: \( 8.2 < \frac{8.48 + 8.35 + x}{3} < 8.5 \) Multiplying all parts by 3: \( 24.6 < 8.48 + 8.35 + x < 25.5 \) \( 24.6 < 16.83 + x < 25.5 \) Subtracting 16.83 from all parts: \( 24.6 - 16.83 < x < 25.5 - 16.83 \) \( 7.77 < x < 8.67 \) Therefore, the third pH reading must fall in the range from 7.77 to 8.67 for the pool to maintain normal acidity.

Exam Tip: When solving compound inequalities, apply the same operation to all three parts simultaneously - this keeps your work organized and reduces errors.

Question 44. A manufacturer has 640 litres of an 8% solution of boric acid. How many litres of 2% boric acid solution should be added to it so that the boric acid content in the resulting mixture will be more than 4% but less than 6%.
Answer: Let \( x \) litres of 2% boric acid solution be added to 640 litres of 8% solution. The percentage strength of the mixture is: \( \text{Strength} = \frac{8}{100} \cdot 640 + \frac{2}{100} \cdot x}{640 + x} \cdot 100 = \frac{5120 + 2x}{100(640 + x)} \) We require the strength to be between 4% and 6%: \( 4 < \frac{5120 + 2x}{640 + x} < 6 \) For the left inequality, \( \frac{5120 + 2x}{640 + x} > 4 \): Multiplying both sides by \( (640 + x) \): \( 5120 + 2x > 4(640 + x) \) \( 5120 + 2x > 2560 + 4x \) Simplifying: \( 5120 - 2560 > 2x \) \( 2560 > 2x \) \( x < 1280 \) For the right inequality, \( \frac{5120 + 2x}{640 + x} < 6 \): Multiplying both sides by \( (640 + x) \): \( 5120 + 2x < 6(640 + x) \) \( 5120 + 2x < 3840 + 6x \) Simplifying: \( 5120 - 3840 < 4x \) \( 1280 < 4x \) \( x > 320 \) Combining both constraints: \( 320 < x < 1280 \) Therefore, between 320 and 1280 litres of the 2% solution must be added.

Exam Tip: In mixture problems, always set up the percentage equation carefully - the numerator is the total amount of the pure substance, and the denominator is the total volume of the mixture.

Question 45. How many litres of water will have to be added to 600 litres of the 45% solution of acid so that the resulting mixture will contain more than 25%, but less than 30% acid content?
Answer: Let x litres of water be added. The total mixture becomes x + 600 litres. The acid present in the resulting mixture is 45% of 600 litres. We are told the resulting mixture must contain more than 25% and less than 30% acid content.

From the first condition:
\[ 45\% \text{ of } 600 > 25\% \text{ of } (x + 600) \]
\[ \frac{45}{100} \times 600 > \frac{25}{100} \times (x + 600) \]
\[ 45 \times 600 > 25(x + 600) \]
\[ 27000 > 25x + 15000 \]
\[ 12000 > 25x \]
\[ 480 > x \]

From the second condition:
\[ 30\% \text{ of } (x + 600) > 45\% \text{ of } 600 \]
\[ \frac{30}{100} \times (x + 600) > \frac{45}{100} \times 600 \]
\[ 30(x + 600) > 27000 \]
\[ 30x + 18000 > 27000 \]
\[ 30x > 9000 \]
\[ x > 300 \]

Therefore, the amount of water required ranges from 300 litres to 480 litres.
In simple words: When you add water to an acid solution, the concentration drops. You need to find how much water makes the concentration fall between 25% and 30%. Setting up two inequalities - one for each limit - gives you a range: between 300 and 480 litres.

Exam Tip: Always set up two separate inequalities when a problem specifies "more than" and "less than" (or "between"), then find the intersection - that is where the answer lies.

Question 46. To receive grade A in a course one must obtain an average of 90 marks or more in five papers, each of 100 marks. If Tanvy scored 89, 93, 95 and 91 marks in first four papers, find the minimum marks that she must score in the last paper to get grade A in the course.
Answer: Let x marks be the score Tanvy gets in her last paper. Her first four scores are 89, 93, 95, and 91. For a grade A, the average across all five papers must be at least 90 marks.

\[ \frac{89 + 93 + 95 + 91 + x}{5} \geq 90 \]
\[ \frac{368 + x}{5} \geq 90 \]

Multiply both sides by 5:
\[ 368 + x \geq 450 \]
\[ x \geq 82 \]

Therefore, Tanvy must score a minimum of 82 marks in her last paper to achieve grade A in the course.
In simple words: Add up her four scores (368), and figure out what fifth score she needs so the average of all five equals at least 90. Working backwards from 90 average times 5 papers gives 450 total needed, so she needs 82 in the final paper.

Exam Tip: When finding minimum or maximum values using averages, always use the inequality symbol (≥ or ≤) matching the problem statement - "or more" means ≥, not >.

Exercise 6B

Question 1. Find the solution set of the inequality \( \frac{1}{x - 2} < 0 \).
Answer: We must find all values of x that make \( \frac{1}{x - 2} \) negative. Since the numerator is 1 (positive), the fraction becomes negative only when the denominator is negative. Thus:

\[ x - 2 < 0 \]
\[ x < 2 \]

The expression is undefined at x = 2. Therefore, x can take all real values less than 2.

\[ x \in (-\infty, 2) \]
In simple words: A fraction with a positive top stays negative only if the bottom is negative. So x - 2 must be negative, meaning x is less than 2. The answer is all numbers from negative infinity up to (but not including) 2.

Exam Tip: Always check where the expression equals zero and where it is undefined - these are your boundary points on the number line.

Question 2. Find the solution set of the inequality \( |x - 1| < 2 \).
Answer: Start with:
\[ |x - 1| < 2 \]

Square both sides:
\[ (x - 1)^2 < 4 \]
\[ x^2 - 2x + 1 < 4 \]
\[ x^2 - 2x - 3 < 0 \]

Factor the quadratic:
\[ x^2 - 3x + x - 3 < 0 \]
\[ x(x - 3) + 1(x - 3) < 0 \]
\[ (x + 1)(x - 3) < 0 \]

The roots are x = -1 and x = 3. Using a sign analysis: when x > 3, the product is positive; when -1 < x < 3, the product is negative; and when x < -1, the product is positive. Since we need the product to be negative:

\[ x \in (-1, 3) \]
In simple words: An absolute value inequality like this means the distance from x to 1 is less than 2 units. That puts x somewhere between -1 and 3 on the number line.

Exam Tip: When solving absolute value inequalities, squaring both sides is valid, but always verify your boundaries using the sign analysis method to avoid losing solutions.

Question 3. Find the solution set of the inequality \( |2x - 3| < 1 \).
Answer: Start with:
\[ |2x - 3| < 1 \]

Square both sides:
\[ (2x - 3)^2 < 1 \]
\[ 4x^2 - 12x + 9 < 1 \]
\[ 4x^2 - 12x + 8 < 0 \]

Divide throughout by 4:
\[ x^2 - 3x + 2 < 0 \]

Factor the quadratic:
\[ x^2 - 2x - x + 2 < 0 \]
\[ x(x - 2) - 1(x - 2) < 0 \]
\[ (x - 1)(x - 2) < 0 \]

The roots are x = 1 and x = 2. Using sign analysis: the product is negative between the roots. Therefore:

\[ x \in (1, 2) \]
In simple words: The expression |2x - 3| < 1 means that 2x - 3 is between -1 and 1. Solving this gives x between 1 and 2.

Exam Tip: After factoring, use a number line with test points in each region to determine where the product is negative, zero, or positive.

Question 4. Find the solution set of the inequality \( \frac{|x - 2|}{(x - 2)} < 0 \), where x ≠ 2.
Answer: We must find values of x for which this fraction is negative. The numerator \( |x - 2| \) is always non-negative (it is an absolute value). For the fraction to be negative, the denominator (x - 2) must be negative:

\[ x - 2 < 0 \]
\[ x < 2 \]

Note that x = 2 is excluded because the expression is undefined there. All values less than 2 make the fraction negative:

\[ x \in (-\infty, 2) \]
In simple words: An absolute value on top is always positive or zero. So for the whole fraction to be negative, the bottom must be negative. The bottom is negative when x < 2.

Exam Tip: When an absolute value appears in a numerator or denominator, remember it is always ≥ 0 - this determines the sign of the overall fraction.

Question 5. Find the solution set of the inequality \( \frac{x + 1}{x + 2} < 1 \).
Answer: Rearrange the inequality:
\[ \frac{x + 1}{x + 2} - 1 < 0 \]
\[ \frac{x + 1 - (x + 2)}{x + 2} < 0 \]
\[ \frac{x + 1 - x - 2}{x + 2} < 0 \]
\[ \frac{-1}{x + 2} < 0 \]

The numerator is -1 (negative). For the fraction to be negative, the denominator must be positive:

\[ x + 2 > 0 \]
\[ x > -2 \]

Since x = -2 makes the expression undefined, all values greater than -2 satisfy the inequality:

\[ x \in (-2, \infty) \]
In simple words: Simplify the inequality step by step, combining the fractions. You end up with a negative number on top, so the denominator must be positive to keep the whole fraction negative.

Exam Tip: When simplifying rational inequalities, always move everything to one side and combine fractions first - this reveals the true numerator and denominator you need to analyze.

Question 6. Solve the system of inequalities x - 2 ≥ 0, 2x - 5 ≤ 3.
Answer: We must find all values of x satisfying both inequalities simultaneously. Solve each separately:

From the first inequality:
\[ x - 2 \geq 0 \]
\[ x \geq 2 \]
\[ x \in [2, \infty) \]

From the second inequality:
\[ 2x - 5 \leq 3 \]
\[ 2x \leq 8 \]
\[ x \leq 4 \]
\[ x \in (-\infty, 4] \]

The solution is the intersection of these two sets:
\[ x \in [2, \infty) \cap (-\infty, 4] = [2, 4] \]
In simple words: Solve each inequality separately to get two ranges. The solution to the system is the overlap of those ranges - where both conditions are true at the same time.

Exam Tip: Always find the intersection (not the union) of solution sets for a system - the answer must satisfy all inequalities, not just one.

Question 7. Solve -4x > 16, when x ∈ Z.
Answer: We must find only integer values of x satisfying the inequality. Divide both sides by -4 (remember to flip the inequality sign when dividing by a negative):

\[ -4x > 16 \]
\[ -x > 4 \]
\[ x < -4 \]

The integers less than -4 are -5, -6, -7, -8, and so on. This can be expressed as:

\[ x = -(4 + n) \text{ where } n \in \{1, 2, 3, 4, \ldots\} \]

Or simply: \( x \in \{-5, -6, -7, -8, \ldots\} \)
In simple words: When dividing by a negative number, flip the inequality sign. The integer solutions are all whole numbers less than -4: that is, -5, -6, -7, and all numbers further left on the number line.

Exam Tip: Always flip the inequality sign when multiplying or dividing by a negative number - this is a critical mistake point.

Question 8. Solve x + 5 > 4x - 10, when x ∈ R.
Answer: Rearrange the inequality to collect all x terms on one side and constants on the other:

\[ x + 5 > 4x - 10 \]
\[ 5 + 10 > 4x - x \]
\[ 15 > 3x \]
\[ 5 > x \]
\[ x < 5 \]

This means x can take any real value less than 5:

\[ x \in (-\infty, 5) \]
In simple words: Move all the x's to the right and all the numbers to the left, then simplify. You get that x must be smaller than 5.

Exam Tip: After rearranging, divide by the coefficient of x carefully - if it is negative, flip the inequality sign.

Question 9. Solve \( \frac{3}{x - 2} < 1 \), when x ∈ R.
Answer: Rearrange the inequality:
\[ \frac{3}{x - 2} - 1 < 0 \]
\[ \frac{3 - (x - 2)}{x - 2} < 0 \]
\[ \frac{3 - x + 2}{x - 2} < 0 \]
\[ \frac{5 - x}{x - 2} < 0 \]

The fraction equals zero at x = 5 and is undefined at x = 2. Using sign analysis on the number line:
- For x < 2: numerator is positive (5 - x > 0) and denominator is negative - fraction is negative ✓
- For 2 < x < 5: numerator is positive and denominator is positive - fraction is positive ✗
- For x > 5: numerator is negative (5 - x < 0) and denominator is positive - fraction is negative ✓

Therefore:
\[ x \in (-\infty, 2) \cup (5, \infty) \]
In simple words: Combine the fraction into one term on the left side. Then test regions on the number line using the zeros (x = 5) and the undefined point (x = 2). Pick the regions where the fraction is negative.

Exam Tip: Mark all critical points (zeros and undefined points) on a number line, then test one point in each region to determine the sign of the expression in that region.

Question 10. Solve \( \frac{x}{x - 5} > \frac{1}{2} \), when x ∈ R.
Answer: Rearrange the inequality:
\[ \frac{x}{x - 5} - \frac{1}{2} > 0 \]
\[ \frac{2x - (x - 5)}{2(x - 5)} > 0 \]
\[ \frac{2x - x + 5}{2(x - 5)} > 0 \]
\[ \frac{x + 5}{x - 5} > 0 \]

The fraction equals zero at x = -5 and is undefined at x = 5. Using sign analysis:
- For x < -5: numerator is negative and denominator is negative - fraction is positive ✓
- For -5 < x < 5: numerator is positive and denominator is negative - fraction is negative ✗
- For x > 5: numerator is positive and denominator is positive - fraction is positive ✓

Therefore:
\[ x \in (-\infty, -5) \cup (5, \infty) \]
In simple words: Get a single fraction on the left. Find the zeros and undefined points, then test the sign in each region. You want regions where the fraction is positive.

Exam Tip: When testing signs, pick easy numbers in each region (like 0, -10, 10) rather than testing at the boundaries - it is faster and clearer.

Question 11. Solve |x| < 4, when x ∈ R.
Answer: Square both sides of the absolute value inequality:
\[ |x| < 4 \]
\[ x^2 < 16 \]
\[ x^2 - 16 < 0 \]
\[ x^2 - 4^2 < 0 \]
\[ (x + 4)(x - 4) < 0 \]

The roots are x = -4 and x = 4. Using sign analysis:
- For x < -4: both factors are negative, so the product is positive ✗
- For -4 < x < 4: (x + 4) is positive and (x - 4) is negative, so the product is negative ✓
- For x > 4: both factors are positive, so the product is positive ✗

Therefore:
\[ x \in (-4, 4) \]
In simple words: The absolute value |x| < 4 means x is between -4 and 4 on the number line - that is, its distance from zero is less than 4 units.

Exam Tip: Squaring absolute value inequalities works, but verify boundaries carefully since squaring can introduce extraneous solutions.

Question 12. Solve |x| > 4, when x ∈ R.
Answer: Square both sides:
\[ |x| > 4 \]
\[ x^2 > 16 \]
\[ x^2 - 16 > 0 \]
\[ x^2 - 4^2 > 0 \]
\[ (x + 4)(x - 4) > 0 \]

The roots are x = -4 and x = 4. Using sign analysis:
- For x < -4: both factors are negative, so the product is positive ✓
- For -4 < x < 4: (x + 4) is positive and (x - 4) is negative, so the product is negative ✗
- For x > 4: both factors are positive, so the product is positive ✓

Therefore:
\[ x \in (-\infty, -4) \cup (4, \infty) \]
In simple words: The absolute value |x| > 4 means x is more than 4 units away from zero. So x must be either far to the left (less than -4) or far to the right (greater than 4).

Exam Tip: When the inequality is > instead of <, the solution involves the outer regions (far from zero) rather than the middle region.