Class 9 Mathematics Federal Board of Intermediate and Secondary Education (FBISE) (FBISE) 📚 Model Textbook of Mathematics Grade 9 (FBISE / NBF)

Mastery Guide: Linear Equations, Radicals, Absolute Values & Inequalities

📖 Chapter 5: Linear Equations and Inequalities 📅 Updated: Sep 18, 2026

FBISE Class 9 Mathematics • Chapter 5

Mastery Guide: Linear Equations, Radicals, Absolute Values & Inequalities

Single National Curriculum (SNC) • Complete Theory, Visual Models, Real-World Systems & 100% Solved Exercises

📖 1. Unit Overview & Target Learning Outcomes

Equations and inequalities form the analytical backbone of algebra. While an equation establishes an exact balance between two quantities, an inequality defines a allowable boundary or constraint. In this chapter, students master the foundational tools required to model, manipulate, and solve linear relationships, algebraic fractions, radical equations, absolute value models, and compound inequalities.

🎯 Core Learning Outcomes:

Linear Equations in One Variable: Define and solve the standard form $ax + b = 0$ ($a, b \in \mathbb{R}, a \neq 0$) using inverse arithmetic operations.
Fractional Equations: Clear algebraic and numerical denominators by multiplying all terms by the Least Common Multiple (LCM).
Radical Equations & Extraneous Roots: Isolate radical terms $\sqrt{ax + b} = c$, apply squaring properties, and rigorously verify candidate roots to identify and discard fake (extraneous) solutions.
Absolute Value Equations: Understand absolute value $|x|$ as geometric distance from zero on the real number line, formulating dual linear cases $|ax + b| = c \implies ax + b = \pm c$.
Linear Inequalities: Distinguish between strict ($<, >$) and non-strict ($\le, \ge$) relations, applying Trichotomy, Transitive, Additive, and Multiplicative properties.
The Golden Reversal Rule: Master the mandatory reversal of inequality orientation when multiplying or dividing by negative real numbers.
Compound Inequalities: Solve coupled inequalities joined by "and" (intersection $\cap$) and "or" (union $\cup$), expressing solutions in formal set-builder notation and 1D graphical number lines.

💡 2. Kid-Friendly Tips for Success & Memory Hooks

⚖️ The Two-Pan Balance

An equation is like a physical scale. Whatever you add, subtract, multiply, or divide on the left pan, you MUST perform on the right pan to keep it in perfect equilibrium!

🕵️ The Extraneous Impostor

Squaring both sides of an equation can manufacture "fake" roots because $(-5)^2 = 25$ just like $(+5)^2 = 25$. Always test candidate roots in the original equation!

📏 Absolute Value = Distance

$|x| = 5$ asks: "Which points on the number line are exactly 5 steps away from 0?" Answer: $+5$ and $-5$. Since distance is never negative, $|x| = -3$ has no solution ($\emptyset$)!

🔄 The Golden Inequality Flip

When you multiply or divide both sides of an inequality by a negative number, the comparison arrow reverses direction: $< \longleftrightarrow >$ and $\le \longleftrightarrow \ge$.

🌍 3. Real-World Connections

🚗 Highway Speed Radars

Motorway speed limits are expressed as inequalities: $60 \le v \le 120\text{ km/h}$. Radar sensors detect violations whenever $v > 120$ or $v < 60$.

🏗️ Precision Engineering Tolerances

Aircraft components must meet strict tolerance bounds: $|L - 100\text{ mm}| \le 0.05\text{ mm}$, ensuring parts fit safely inside turbine engines.

💰 Financial Budgeting & Break-Even

A startup business models operational viability using linear constraints: $\text{Total Revenue } R(x) \ge \text{Total Cost } C(x)$ to avoid bankruptcy.

🌡️ Climate Control Systems

Smart thermostats regulate indoor air quality by maintaining temperatures in comfortable safe intervals: $20^\circ\text{C} \le T \le 24^\circ\text{C}$.

🔑 4. Study Cues & Essential Inquiries

Inquiry 1: Why does multiplying both sides of an inequality by a negative number reverse the inequality sign?
→ On the real number line, multiplying by $-1$ reflects positions across the origin $0$. If $2 < 5$, their reflections are $-2$ and $-5$. Since $-2$ lies to the right of $-5$, $-2 > -5$.
Inquiry 2: Under what circumstances can a radical equation have an empty solution set ($\emptyset$)?
→ The principal radical $\sqrt{X}$ is by definition non-negative ($\sqrt{X} \ge 0$). Thus, an equation like $\sqrt{3x - 5} = -10$ has immediately zero real solutions.
Inquiry 3: What is the fundamental difference between compound inequalities joined by "AND" versus "OR"?
→ "AND" requires simultaneous satisfaction (intersection $\cap$), forming a closed or bounded overlapping segment. "OR" requires satisfaction of at least one condition (union $\cup$), typically producing disjoint, outward-pointing rays.

🌟 5. Section-by-Section Conceptual Explanations & Visual Models

5.1 Linear Equations in One Variable & Fractional Forms

A linear equation in one variable is an algebraic statement in which the highest exponent of the unknown variable is exactly $1$. The standard canonical form is:

$$ax + b = 0 \quad (a, b \in \mathbb{R}, \; a \neq 0)$$

Solving Fractional Linear Equations: When an equation contains fractions, determine the Least Common Multiple (LCM) of all denominators. Multiplying every single term on both sides by this LCM clears all fractions, converting the equation into a standard integer form.

Figure 5.1: The Two-Pan Balance Model — Invariance under equal inverse operations.

5.2 Linear Equations Involving Radicals & Extraneous Roots

An equation in which the variable appears under a radical sign is called a radical equation (e.g., $\sqrt{2x + 1} = 9$).

Standard 4-Step Solution Algorithm:

Isolate: Rearrange terms so that the radical expression stands alone on one side.
Square: Eliminate the square root by squaring both sides of the equation.
Solve: Solve the resulting linear equation for the unknown variable.
Check (Compulsory): Substitute candidate solutions into the original equation to catch and discard extraneous roots.

Figure 5.2: Radical Equation Verification Pipeline — Catching Extraneous Roots.

5.3 Equations Involving Absolute Value

The absolute value (or modulus) of a real number $x$, denoted $|x|$, represents the non-directed geometric distance between $x$ and $0$ on the real number line:

$$|x| = \begin{cases} x & \text{if } x \ge 0 \\ -x & \text{if } x < 0 \end{cases}$$

Key Absolute Value Principles:

$|ax + b| = c \quad (c > 0) \implies ax + b = c \quad \text{or} \quad ax + b = -c$ (Two real solutions).
$|ax + b| = 0 \implies ax + b = 0$ (Exactly one unique solution).
$|ax + b| = c \quad (c < 0) \implies \text{No real solution, Solution Set } = \emptyset$.
$|f(x)| = |g(x)| \implies f(x) = g(x) \quad \text{or} \quad f(x) = -g(x)$.

Figure 5.3: Absolute Value as Dual-Directional Distance from Origin: $|x| = 5 \implies x = \pm 5$.

5.4 Linear Inequalities & The Four Fundamental Axioms

An algebraic statement containing inequality symbols ($<, >, \le, \ge, \neq$) is called an inequality (or inequation).

📜 The Four Fundamental Properties of Real Inequalities:

Trichotomy Property: For any $a, b \in \mathbb{R}$, exactly one of three relations holds: $a < b$, $a = b$, or $a > b$.
Transitive Property: If $a < b$ and $b < c$, then $a < c$. (Similarly for $>$).
Additive Property: If $a < b$, then $a + c < b + c$ and $a - c < b - c$ for any real number $c$.
Multiplicative Property:
- If $c > 0$ (positive): $a < b \implies ac < bc$ (Sign remains unchanged).
- If $c < 0$ (negative): $a < b \implies ac > bc$ (SIGN FLIPS!).

Figure 5.4: Graphical Representation of Compound Inequalities on the 1D Coordinate Line.

🎯 6. Unit Synthesis Summary

Chapter 5 provides the essential analytical architecture for algebraic problem solving. A linear equation represents a point balance solved via inverse operations and LCM clearing. A radical equation requires squaring and mandatory verification to discard extraneous roots created by non-invertible squaring. An absolute value equation models distance on the real number line, bifurcating into dual positive/negative linear pathways. Finally, linear inequalities establish boundary conditions, where multiplying or dividing by negative numbers flips the order relation, and compound systems model combined physical constraints through intersection (AND) and union (OR).

📝 7. Solved Textbook Exercises & Step-by-Step Manual

Exercise 5.1 • Linear Equations in One Variable & Fractions

Q1. Solve $5x - 2 - x = 4 - 3x - 27$

Step 1 (Combine like terms): LHS: $(5x - x) - 2 = 4x - 2$. RHS: $(4 - 27) - 3x = -23 - 3x$.
Step 2 (Equation): $4x - 2 = -3x - 23$.
Step 3 (Collect variables on LHS): $4x + 3x = -23 + 2 \implies 7x = -21$.
Step 4 (Divide by 7): $x = \frac{-21}{7} = -3$.

Final Answer: $\mathbf{x = -3}$ (Solution Set = $\mathbf{\{-3\}}$)

Q2. Solve $4a - 3(5a - 14) = 5(7 + a) - 9$

Step 1 (Expand parentheses): $4a - 15a + 42 = 35 + 5a - 9$.
Step 2 (Combine like terms): $-11a + 42 = 5a + 26$.
Step 3 (Rearrange): $-11a - 5a = 26 - 42 \implies -16a = -16$.
Step 4 (Divide by -16): $a = \frac{-16}{-16} = 1$.

Final Answer: $\mathbf{a = 1}$ (Solution Set = $\mathbf{\{1\}}$)

Q3. Solve $7(2 - 5x) + 27 = 18x - 3(8 - 4x)$

Step 1 (Expand parentheses): $14 - 35x + 27 = 18x - 24 + 12x$.
Step 2 (Combine like terms): $41 - 35x = 30x - 24$.
Step 3 (Transpose terms): $41 + 24 = 30x + 35x \implies 65 = 65x$.
Step 4 (Divide by 65): $x = 1$.

Final Answer: $\mathbf{x = 1}$ (Solution Set = $\mathbf{\{1\}}$)

Q4. Solve $\frac{5x}{4} + \frac{1}{2} = 0$

Step 1 (Isolate term with variable): $\frac{5x}{4} = -\frac{1}{2}$.
Step 2 (Multiply by 4): $5x = -\frac{1}{2} \times 4 = -2$.
Step 3 (Divide by 5): $x = -\frac{2}{5}$.

Final Answer: $\mathbf{x = -\frac{2}{5}}$ (Solution Set = $\mathbf{\{-\frac{2}{5}\}}$)

Q5. Solve $\frac{x - 2}{2} + \frac{x + 10}{9} = 5$

Step 1 (Multiply by LCM = 18): $18 \left(\frac{x - 2}{2}\right) + 18 \left(\frac{x + 10}{9}\right) = 18(5)$.
Step 2 (Simplify terms): $9(x - 2) + 2(x + 10) = 90$.
Step 3 (Expand and collect): $9x - 18 + 2x + 20 = 90 \implies 11x + 2 = 90$.
Step 4 (Solve for x): $11x = 88 \implies x = 8$.

Final Answer: $\mathbf{x = 8}$ (Solution Set = $\mathbf{\{8\}}$)

Q6. Solve $\frac{4(x + 2)}{3} - \frac{6(x - 7)}{7} = 12$

Step 1 (Multiply by LCM = 21): $21 \left[\frac{4(x + 2)}{3}\right] - 21 \left[\frac{6(x - 7)}{7}\right] = 21(12)$.
Step 2 (Simplify): $7 \times 4(x + 2) - 3 \times 6(x - 7) = 252 \implies 28(x + 2) - 18(x - 7) = 252$.
Step 3 (Expand): $28x + 56 - 18x + 126 = 252 \implies 10x + 182 = 252$.
Step 4 (Solve): $10x = 252 - 182 = 70 \implies x = 7$.

Final Answer: $\mathbf{x = 7}$ (Solution Set = $\mathbf{\{7\}}$)

Q7. Solve $\frac{x}{2} + \frac{x}{3} - \frac{x}{4} + \frac{x}{5} = 7\frac{5}{6}$

Step 1 (Convert mixed fraction): $7\frac{5}{6} = \frac{47}{6}$.
Step 2 (Multiply by LCM of 2, 3, 4, 5, 6 = 60): $60\left(\frac{x}{2}\right) + 60\left(\frac{x}{3}\right) - 60\left(\frac{x}{4}\right) + 60\left(\frac{x}{5}\right) = 60\left(\frac{47}{6}\right)$.
Step 3 (Simplify): $30x + 20x - 15x + 12x = 10(47) \implies 47x = 470$.
Step 4 (Divide by 47): $x = \frac{470}{47} = 10$.

Final Answer: $\mathbf{x = 10}$ (Solution Set = $\mathbf{\{10\}}$)

Q8. Solve $\frac{y + 1}{3} + \frac{y + 1}{2} = 2 - \frac{y + 3}{2}$

Step 1 (Multiply by LCM = 6): $6\left(\frac{y + 1}{3}\right) + 6\left(\frac{y + 1}{2}\right) = 6(2) - 6\left(\frac{y + 3}{2}\right)$.
Step 2 (Simplify): $2(y + 1) + 3(y + 1) = 12 - 3(y + 3)$.
Step 3 (Expand): $2y + 2 + 3y + 3 = 12 - 3y - 9 \implies 5y + 5 = 3 - 3y$.
Step 4 (Rearrange): $5y + 3y = 3 - 5 \implies 8y = -2 \implies y = -\frac{2}{8} = -\frac{1}{4}$.

Final Answer: $\mathbf{y = -\frac{1}{4}}$ (Solution Set = $\mathbf{\{-\frac{1}{4}\}}$)

Q9. Solve $\frac{1}{5}(x - 8) + \frac{4 + x}{7} = 7 - \frac{23 - x}{5}$

Step 1 (Multiply by LCM of 5 and 7 = 35): $35\left[\frac{x - 8}{5}\right] + 35\left[\frac{4 + x}{7}\right] = 35(7) - 35\left[\frac{23 - x}{5}\right]$.
Step 2 (Simplify): $7(x - 8) + 5(4 + x) = 245 - 7(23 - x)$.
Step 3 (Expand): $7x - 56 + 20 + 5x = 245 - 161 + 7x \implies 12x - 36 = 84 + 7x$.
Step 4 (Solve for x): $12x - 7x = 84 + 36 \implies 5x = 120 \implies x = 24$.

Final Answer: $\mathbf{x = 24}$ (Solution Set = $\mathbf{\{24\}}$)

Q10. Solve $\frac{1}{2y} - \frac{1}{6} = \frac{1}{4y} + 1 - \frac{1}{y} \quad (y \neq 0)$

Step 1 (Multiply by LCM = 12y): $12y\left(\frac{1}{2y}\right) - 12y\left(\frac{1}{6}\right) = 12y\left(\frac{1}{4y}\right) + 12y(1) - 12y\left(\frac{1}{y}\right)$.
Step 2 (Simplify): $6 - 2y = 3 + 12y - 12 \implies 6 - 2y = 12y - 9$.
Step 3 (Transpose): $6 + 9 = 12y + 2y \implies 15 = 14y$.
Step 4 (Divide by 14): $y = \frac{15}{14}$.

Final Answer: $\mathbf{y = \frac{15}{14}}$ (Solution Set = $\mathbf{\{\frac{15}{14}\}}$)

Q11. Solve $4 - 0.3(1 - x) = 7$

Step 1 (Expand decimal): $4 - 0.3 + 0.3x = 7 \implies 3.7 + 0.3x = 7$.
Step 2 (Subtract 3.7): $0.3x = 7 - 3.7 = 3.3$.
Step 3 (Divide by 0.3): $x = \frac{3.3}{0.3} = 11$.

Final Answer: $\mathbf{x = 11}$ (Solution Set = $\mathbf{\{11\}}$)

Q12. Solve $0.5x = 6.3 - 0.2x$

Step 1 (Collect variable terms on LHS): $0.5x + 0.2x = 6.3 \implies 0.7x = 6.3$.
Step 2 (Divide by 0.7): $x = \frac{6.3}{0.7} = 9$.

Final Answer: $\mathbf{x = 9}$ (Solution Set = $\mathbf{\{9\}}$)

Q13. Solve $1.3x - 0.2 = 0.3x - 1.5$

Step 1 (Collect variable terms): $1.3x - 0.3x = -1.5 + 0.2$.
Step 2 (Simplify): $1.0x = -1.3 \implies x = -1.3$.

Final Answer: $\mathbf{x = -1.3}$ (Solution Set = $\mathbf{\{-1.3\}}$)

Exercise 5.2 • Radical Equations & Extraneous Root Verification

Q1. Solve $\sqrt{2x} = 4$

Step 1: Square both sides: $(\sqrt{2x})^2 = 4^2 \implies 2x = 16 \implies x = 8$.
Check: $\sqrt{2(8)} = \sqrt{16} = 4 = 4$ (True).

Solution Set = $\mathbf{\{8\}}$

Q2. Solve $\sqrt{x - 3} = 2$

Step 1: Square both sides: $x - 3 = 2^2 = 4 \implies x = 4 + 3 = 7$.
Check: $\sqrt{7 - 3} = \sqrt{4} = 2 = 2$ (True).

Solution Set = $\mathbf{\{7\}}$

Q3. Solve $\sqrt{x - 5} = 3$

Step 1: Square both sides: $x - 5 = 3^2 = 9 \implies x = 9 + 5 = 14$.
Check: $\sqrt{14 - 5} = \sqrt{9} = 3 = 3$ (True).

Solution Set = $\mathbf{\{14\}}$

Q4. Solve $\sqrt{2x + 1} = 9$

Step 1: Square both sides: $2x + 1 = 9^2 = 81 \implies 2x = 80 \implies x = 40$.
Check: $\sqrt{2(40) + 1} = \sqrt{81} = 9 = 9$ (True).

Solution Set = $\mathbf{\{40\}}$

Q5. Solve $\sqrt{5x - 4} = 14$

Step 1: Square both sides: $5x - 4 = 14^2 = 196 \implies 5x = 200 \implies x = 40$.
Check: $\sqrt{5(40) - 4} = \sqrt{196} = 14 = 14$ (True).

Solution Set = $\mathbf{\{40\}}$

Q6. Solve $\sqrt{3x - 5} = -10$

Key Fact: The principal square root $\sqrt{3x - 5}$ cannot be negative. If squared, $3x - 5 = 100 \implies 3x = 105 \implies x = 35$.
Check: $\sqrt{3(35) - 5} = \sqrt{100} = 10 \neq -10$ (Extraneous root).

Solution Set = $\mathbf{\phi}$ (or $\mathbf{\{\}}$)

Q7. Solve $\sqrt{y + 4} - 3 = 2$

Step 1: Isolate radical: $\sqrt{y + 4} = 2 + 3 = 5$.
Step 2: Square both sides: $y + 4 = 5^2 = 25 \implies y = 21$.
Check: $\sqrt{21 + 4} - 3 = 5 - 3 = 2 = 2$ (True).

Solution Set = $\mathbf{\{21\}}$

Q8. Solve $5 - \sqrt{2x - 1} = 0$

Step 1: Isolate radical: $\sqrt{2x - 1} = 5$.
Step 2: Square both sides: $2x - 1 = 25 \implies 2x = 26 \implies x = 13$.
Check: $5 - \sqrt{2(13) - 1} = 5 - \sqrt{25} = 5 - 5 = 0 = 0$ (True).

Solution Set = $\mathbf{\{13\}}$

Q9. Solve $\sqrt{y + 1} - 12 = -10$

Step 1: Isolate radical: $\sqrt{y + 1} = -10 + 12 = 2$.
Step 2: Square both sides: $y + 1 = 2^2 = 4 \implies y = 3$.
Check: $\sqrt{3 + 1} - 12 = 2 - 12 = -10 = -10$ (True).

Solution Set = $\mathbf{\{3\}}$

Q10. Solve $\sqrt{5t - 2} = \sqrt{3t + 4}$

Step 1: Square both sides: $5t - 2 = 3t + 4 \implies 5t - 3t = 4 + 2 \implies 2t = 6 \implies t = 3$.
Check: $\sqrt{5(3) - 2} = \sqrt{13}$; $\sqrt{3(3) + 4} = \sqrt{13}$ (True).

Solution Set = $\mathbf{\{3\}}$

Q11. Solve $\sqrt{9 - 2x} = \sqrt{5x - 12}$

Step 1: Square both sides: $9 - 2x = 5x - 12 \implies 9 + 12 = 5x + 2x \implies 21 = 7x \implies x = 3$.
Check: $\sqrt{9 - 6} = \sqrt{3}$; $\sqrt{15 - 12} = \sqrt{3}$ (True).

Solution Set = $\mathbf{\{3\}}$

Q12. Solve $12 - \sqrt{y + 1} = 14$

Step 1: Isolate radical: $-\sqrt{y + 1} = 14 - 12 = 2 \implies \sqrt{y + 1} = -2$.
Since a square root cannot equal a negative number, squaring produces $y + 1 = 4 \implies y = 3$, which fails verification ($12 - \sqrt{4} = 10 \neq 14$).

Solution Set = $\mathbf{\phi}$ (or $\mathbf{\{\}}$)

Q13. Solve $4\sqrt{z} + 8 = 40$

Step 1: $4\sqrt{z} = 40 - 8 = 32 \implies \sqrt{z} = \frac{32}{4} = 8$.
Step 2: Square both sides: $z = 8^2 = 64$.
Check: $4\sqrt{64} + 8 = 4(8) + 8 = 32 + 8 = 40 = 40$ (True).

Solution Set = $\mathbf{\{64\}}$

Q14. Solve $\sqrt{\frac{a + 6}{a + 2}} = \sqrt{\frac{a + 2}{a - 1}}$

Step 1: Square both sides: $\frac{a + 6}{a + 2} = \frac{a + 2}{a - 1}$.
Step 2: Cross multiply: $(a + 6)(a - 1) = (a + 2)^2 \implies a^2 + 5a - 6 = a^2 + 4a + 4$.
Step 3: $5a - 4a = 4 + 6 \implies a = 10$.
Check: $\sqrt{\frac{16}{12}} = \sqrt{\frac{4}{3}} = \sqrt{\frac{12}{9}} = \sqrt{\frac{4}{3}}$ (True).

Solution Set = $\mathbf{\{10\}}$

Q15. Solve $\sqrt{\frac{z}{z + 3}} = \sqrt{\frac{z + 2}{z + 6}}$

Step 1: Square both sides: $\frac{z}{z + 3} = \frac{z + 2}{z + 6}$.
Step 2: Cross multiply: $z(z + 6) = (z + 3)(z + 2) \implies z^2 + 6z = z^2 + 5z + 6$.
Step 3: $6z - 5z = 6 \implies z = 6$.
Check: $\sqrt{\frac{6}{9}} = \sqrt{\frac{2}{3}} = \sqrt{\frac{8}{12}} = \sqrt{\frac{2}{3}}$ (True).

Solution Set = $\mathbf{\{6\}}$

Q16. Solve $\sqrt{5x - 4} = \sqrt{7x + 2}$

Step 1: Square both sides: $5x - 4 = 7x + 2 \implies -2x = 6 \implies x = -3$.
Check: For $x = -3$, $\sqrt{5(-3) - 4} = \sqrt{-19}$, which is not a real number. Hence, $x = -3$ is extraneous.

Solution Set = $\mathbf{\phi}$ (or $\mathbf{\{\}}$)

Exercise 5.3 • Equations Involving Absolute Value

Q1. Solve $|x| = \frac{5}{3}$

By definition of absolute value: $x = \frac{5}{3} \quad \text{or} \quad x = -\frac{5}{3}$.

Solution Set = $\mathbf{\{-\frac{5}{3}, \frac{5}{3}\}}$

Q2. Solve $|x + 2| = 6$

$x + 2 = 6 \implies x = 4 \quad \text{or} \quad x + 2 = -6 \implies x = -8$.

Solution Set = $\mathbf{\{-8, 4\}}$

Q3. Solve $|5y - 1| = 9$

$5y - 1 = 9 \implies 5y = 10 \implies y = 2 \quad \text{or} \quad 5y - 1 = -9 \implies 5y = -8 \implies y = -\frac{8}{5}$.

Solution Set = $\mathbf{\{-\frac{8}{5}, 2\}}$

Q4. Solve $|x + 1| = 2$

$x + 1 = 2 \implies x = 1 \quad \text{or} \quad x + 1 = -2 \implies x = -3$.

Solution Set = $\mathbf{\{-3, 1\}}$

Q5. Solve $|6 - 3y| = 0$

$6 - 3y = 0 \implies 3y = 6 \implies y = 2$.

Solution Set = $\mathbf{\{2\}}$

Q6. Solve $3|z - 2| - 4 = -2$

$3|z - 2| = -2 + 4 = 2 \implies |z - 2| = \frac{2}{3}$.
$z - 2 = \frac{2}{3} \implies z = 2 + \frac{2}{3} = \frac{8}{3} \quad \text{or} \quad z - 2 = -\frac{2}{3} \implies z = 2 - \frac{2}{3} = \frac{4}{3}$.

Solution Set = $\mathbf{\{\frac{4}{3}, \frac{8}{3}\}}$

Q7. Solve $|2x - 1| = 5$

$2x - 1 = 5 \implies 2x = 6 \implies x = 3 \quad \text{or} \quad 2x - 1 = -5 \implies 2x = -4 \implies x = -2$.

Solution Set = $\mathbf{\{-2, 3\}}$

Q8. Solve $|3x + 2| = 7$

$3x + 2 = 7 \implies 3x = 5 \implies x = \frac{5}{3} \quad \text{or} \quad 3x + 2 = -7 \implies 3x = -9 \implies x = -3$.

Solution Set = $\mathbf{\{-3, \frac{5}{3}\}}$

Q9. Solve $\frac{|4x|}{3} = 12$

$|4x| = 36 \implies 4|x| = 36 \implies |x| = 9 \implies x = 9 \quad \text{or} \quad x = -9$.

Solution Set = $\mathbf{\{-9, 9\}}$

Q10. Solve $|5x| + 10 = 5$

$|5x| = 5 - 10 = -5$. Since the absolute value of a real expression can never be negative, there is no real solution.

Solution Set = $\mathbf{\phi}$ (or $\mathbf{\{\}}$)

Q11. Solve $\frac{|1 - 2y|}{4} = 3$

$|1 - 2y| = 12$.
Case 1: $1 - 2y = 12 \implies -2y = 11 \implies y = -\frac{11}{2}$.
Case 2: $1 - 2y = -12 \implies -2y = -13 \implies y = \frac{13}{2}$.

Solution Set = $\mathbf{\{-\frac{11}{2}, \frac{13}{2}\}}$

Q12. Solve $\frac{|x + 1|}{2} = \frac{|2x - 1|}{3}$

$3|x + 1| = 2|2x - 1| \implies |3x + 3| = |4x - 2|$.
Case 1: $3x + 3 = 4x - 2 \implies -x = -5 \implies x = 5$.
Case 2: $3x + 3 = -(4x - 2) \implies 3x + 3 = -4x + 2 \implies 7x = -1 \implies x = -\frac{1}{7}$.

Solution Set = $\mathbf{\{-\frac{1}{7}, 5\}}$

Q13. Solve $|5x - 3| = |x + 7|$

Case 1: $5x - 3 = x + 7 \implies 4x = 10 \implies x = \frac{10}{4} = \frac{5}{2}$.
Case 2: $5x - 3 = -(x + 7) \implies 5x - 3 = -x - 7 \implies 6x = -4 \implies x = -\frac{4}{6} = -\frac{2}{3}$.

Solution Set = $\mathbf{\{-\frac{2}{3}, \frac{5}{2}\}}$

Q14. Solve $|z + 3| - 3 = 5 - |z + 3|$

$|z + 3| + |z + 3| = 5 + 3 \implies 2|z + 3| = 8 \implies |z + 3| = 4$.
Case 1: $z + 3 = 4 \implies z = 1$.
Case 2: $z + 3 = -4 \implies z = -7$.

Solution Set = $\mathbf{\{-7, 1\}}$

Exercise 5.4 • Linear Inequalities & Compound Relations

1.(a) Check whether the given value of each variable satisfies the inequality:

(i) $5y - 12 > 0; \quad y = 3$: $5(3) - 12 = 15 - 12 = 3 > 0$ → Satisfies (True)
(ii) $4 - 2x \le 9; \quad x = -3$: $4 - 2(-3) = 4 + 6 = 10 \le 9$ → Does not satisfy (False)
(iii) $5 - 2x > -4x + 5; \quad x = 4$: LHS $= 5 - 8 = -3$; RHS $= -16 + 5 = -11$. Since $-3 > -11$ → Satisfies (True)
(iv) $3(z + 4) \le 6; \quad z = -2$: $3(-2 + 4) = 3(2) = 6 \le 6$ → Satisfies (True)
(v) $5(x - 2) \ge 9x - 3(2x - 4); \quad x = 11$: LHS $= 5(9) = 45$; RHS $= 99 - 3(18) = 99 - 54 = 45$. Since $45 \ge 45$ → Satisfies (True)

1.(b) Write each solution in set notation form:

(i) $2 \le x \le 5$, where $x \in \mathbb{N}$: Tabular form: $\mathbf{\{2, 3, 4, 5\}}$ or $\{x \mid x \in \mathbb{N} \wedge 2 \le x \le 5\}$
(ii) $y < 7$, where $y \in \mathbb{N}$: Tabular form: $\mathbf{\{1, 2, 3, 4, 5, 6\}}$ or $\{y \mid y \in \mathbb{N} \wedge y < 7\}$
(iii) $z \le -3$, where $z \in \mathbb{R}$: $\mathbf{\{z \mid z \in \mathbb{R} \wedge z \le -3\}}$
(iv) $x \le 4$, where $x \in \mathbb{W}$: Tabular form: $\mathbf{\{0, 1, 2, 3, 4\}}$ or $\{x \mid x \in \mathbb{W} \wedge x \le 4\}$
(v) $-4 < x < -\frac{3}{2}$, where $x \in \mathbb{R}$: $\mathbf{\{x \mid x \in \mathbb{R} \wedge -4 < x < -\frac{3}{2}\}}$

Q2. Solve $3x - 2 < 7, \quad x \in \mathbb{N}$

$3x < 9 \implies x < 3$. Since $x \in \mathbb{N} = \{1, 2, 3, \dots\}$:

Solution Set = $\mathbf{\{1, 2\}}$

Q3. Solve $6x - 5 \le 35 - 2x, \quad x \in \mathbb{W}$

$6x + 2x \le 35 + 5 \implies 8x \le 40 \implies x \le 5$. Since $x \in \mathbb{W} = \{0, 1, 2, \dots\}$:

Solution Set = $\mathbf{\{0, 1, 2, 3, 4, 5\}}$

Q4. Solve $16 - 5y < 4(y - 1) - 7, \quad y \in \mathbb{R}$

$16 - 5y < 4y - 4 - 7 \implies 16 - 5y < 4y - 11 \implies -9y < -27 \implies y > 3$ (Dividing by $-9$ flips $<$ to $>$).

Solution Set = $\mathbf{\{y \mid y \in \mathbb{R} \wedge y > 3\}}$

Q5. Solve $10 - (7 - y) \ge 3y - 9, \quad y \in \mathbb{R}$

$10 - 7 + y \ge 3y - 9 \implies 3 + y \ge 3y - 9 \implies -2y \ge -12 \implies y \le 6$.

Solution Set = $\mathbf{\{y \mid y \in \mathbb{R} \wedge y \le 6\}}$

Q6. Solve compound inequality: $5 - 3x < 11 \quad \text{or} \quad 2x + 3 < -9, \quad x \in \mathbb{R}$

Left: $-3x < 6 \implies x > -2$.
Right: $2x < -12 \implies x < -6$.
Union (OR): $x < -6 \text{ or } x > -2$.

Solution Set = $\mathbf{\{x \mid x \in \mathbb{R} \wedge (x < -6 \vee x > -2)\}}$

Q7. Solve compound inequality: $2x + 3 \le 9 \quad \text{and} \quad x - 5 > -6, \quad x \in \mathbb{R}$

Left: $2x \le 6 \implies x \le 3$.
Right: $x > -6 + 5 \implies x > -1$.
Intersection (AND): $-1 < x \le 3$.

Solution Set = $\mathbf{\{x \mid x \in \mathbb{R} \wedge -1 < x \le 3\}}$

Q8. Solve $1 \le 7 - 3x \le 22, \quad x \in \mathbb{R}$

Subtract 7 from all three parts: $1 - 7 \le -3x \le 22 - 7 \implies -6 \le -3x \le 15$.
Divide all parts by $-3$ (reversing inequality signs): $\frac{-6}{-3} \ge x \ge \frac{15}{-3} \implies 2 \ge x \ge -5 \implies -5 \le x \le 2$.

Solution Set = $\mathbf{\{x \mid x \in \mathbb{R} \wedge -5 \le x \le 2\}}$

Q9. Solve $3x + 21 < 1 - x \quad \text{or} \quad 3x + 8 > 3 - 2x, \quad x \in \mathbb{R}$

Left: $3x + x < 1 - 21 \implies 4x < -20 \implies x < -5$.
Right: $3x + 2x > 3 - 8 \implies 5x > -5 \implies x > -1$.
Union: $x < -5 \text{ or } x > -1$.

Solution Set = $\mathbf{\{x \mid x \in \mathbb{R} \wedge (x < -5 \vee x > -1)\}}$

Q10. Solve $1 - 5x > 16 \quad \text{and} \quad 3 - \frac{3x}{2} \le 9, \quad x \in \mathbb{R}$

Left: $-5x > 15 \implies x < -3$.
Right: $-\frac{3x}{2} \le 9 - 3 = 6 \implies -3x \le 12 \implies x \ge -4$.
Intersection: $-4 \le x < -3$.

Solution Set = $\mathbf{\{x \mid x \in \mathbb{R} \wedge -4 \le x < -3\}}$

Q11. The sum of five times a number $x$ and 10 is less than $-35$ or greater than $-5$. What real numbers does $x$ represent?

Equation Formulation: $5x + 10 < -35 \quad \text{or} \quad 5x + 10 > -5$.
Case 1: $5x < -45 \implies x < -9$.
Case 2: $5x > -15 \implies x > -3$.

Final Answer: $\mathbf{\{x \mid x \in \mathbb{R} \wedge (x < -9 \vee x > -3)\}}$

Q12. Two times a number decreased by 5 is greater than or equal to the number increased by 8. Find the possible values for the number.

Let the number be $x$.
Inequality: $2x - 5 \ge x + 8 \implies 2x - x \ge 8 + 5 \implies x \ge 13$.

Final Answer: $\mathbf{\{x \mid x \in \mathbb{R} \wedge x \ge 13\}}$ (The number is at least 13)

Review Exercise 5 • Comprehensive Examination Problem Set

Q2. Solve $\frac{2x - 11}{12} = \frac{2x + 10}{12} - \left(\frac{28 - 2x}{4} - \frac{1}{4}\right)$

Simplify parentheses: $\frac{28 - 2x - 1}{4} = \frac{27 - 2x}{4}$.
Equation becomes: $\frac{2x - 11}{12} = \frac{2x + 10}{12} - \frac{27 - 2x}{4}$.
Multiply entire equation by LCM = 12: $2x - 11 = (2x + 10) - 3(27 - 2x)$.
Expand: $2x - 11 = 2x + 10 - 81 + 6x \implies 2x - 11 = 8x - 71$.
Transpose: $2x - 8x = -71 + 11 \implies -6x = -60 \implies x = 10$.

Final Answer: $\mathbf{x = 10}$

Q3. Solve $\sqrt{a - \frac{1}{2}} = \sqrt{\frac{2a}{5} + \frac{2}{5}}$

Square both sides: $a - \frac{1}{2} = \frac{2a}{5} + \frac{2}{5}$.
Multiply by LCM = 10: $10a - 5 = 4a + 4$.
Solve: $10a - 4a = 4 + 5 \implies 6a = 9 \implies a = \frac{9}{6} = \frac{3}{2} = 1.5$.
Verification: $\sqrt{\frac{3}{2} - \frac{1}{2}} = \sqrt{1} = 1$; $\sqrt{\frac{2(1.5)}{5} + \frac{2}{5}} = \sqrt{\frac{5}{5}} = 1$ (True).

Final Answer: $\mathbf{a = \frac{3}{2}}$ (Solution Set = $\mathbf{\{\frac{3}{2}\}}$)

Q4. Solve $\frac{\sqrt{5x - 4} - 4}{10} = -1$

Multiply by 10: $\sqrt{5x - 4} - 4 = -10 \implies \sqrt{5x - 4} = -6$.
Since principal square root cannot be negative, there is no real solution (Squaring gives $5x - 4 = 36 \implies x = 8$, but checking yields $\frac{6 - 4}{10} = \frac{2}{10} \neq -1$).

Final Answer: Solution Set = $\mathbf{\phi}$ (or $\mathbf{\{\}}$)

Q5. Solve $5 - |5y + 1| = -9$