Simple Equations and Inequalities

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

An equation is a mathematical statement that two expressions are equal. An inequality shows that two expressions are not equal — one is greater or less than the other.

Equations:

• Linear equation: Highest power of unknown is 1 (e.g., 2x + 3 = 7)
• Quadratic equation: Highest power is 2 (e.g., x² − 5x + 6 = 0)
• Simultaneous equations: Two or more equations with two or more unknowns

Solving Linear Equations: $$2x + 5 = 13$$ $$2x = 13 - 5 \quad \text{(subtract 5 from both sides)}$$ $$2x = 8$$ $$x = \frac{8}{2} = 4$$

Inequality Symbols:

• $>$ : Greater than
• $<$ : Less than
• $\geq$ : Greater than or equal to
• $\leq$ : Less than or equal to
• $\neq$ : Not equal to

Solving Inequalities: $$3x - 4 > 5$$ $$3x > 9$$ $$x > 3$$

⚡ WAEC Tip: When multiplying or dividing an inequality by a NEGATIVE number, you must REVERSE the inequality sign. Example: −2x > 6 → dividing by −2 → x < −3 (sign reversed!).

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🟡 Standard — Regular Study (2d–2mo)

For students who want genuine understanding.

Linear Equations in One Variable:

General form: $ax + b = 0$, where $a \neq 0$

Steps to solve:

1. Simplify both sides (expand brackets, combine like terms)
2. Collect all terms with x on one side, numbers on the other
3. Divide by the coefficient of x

Example with brackets: $$3(2x - 1) - 2(x + 4) = 7$$ $$6x - 3 - 2x - 8 = 7$$ $$4x - 11 = 7$$ $$4x = 18$$ $$x = \frac{18}{4} = \frac{9}{2} = 4.5$$

Equations with Fractions:

Multiply both sides by the LCM of denominators:

$$\frac{x}{3} + \frac{x}{4} = 7$$ $$\frac{4x + 3x}{12} = 7$$ $$\frac{7x}{12} = 7$$ $$7x = 84$$ $$x = 12$$

Word Problems:

Problem: The sum of three consecutive integers is 72. Find the integers.

Let integers be: $n$, $n+1$, $n+2$ $$n + (n+1) + (n+2) = 72$$ $$3n + 3 = 72$$ $$3n = 69$$ $$n = 23$$

Integers: 23, 24, 25

Linear Inequalities:

Problem: Solve and illustrate on a number line: $2x + 3 \leq 11$

$$2x \leq 8$$ $$x \leq 4$$

Number line: closed circle at 4, shading leftwards ←●────→

Quadratic Equations:

General form: $ax^2 + bx + c = 0$

Solution by factorisation: $$x^2 - 5x + 6 = 0$$ $$(x - 2)(x - 3) = 0$$ $$x = 2 \text{ or } x = 3$$

Solution by formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

Problem: Solve $2x^2 + 5x - 3 = 0$ $$x = \frac{-5 \pm \sqrt{25 - 4(2)(-3)}}{2(2)} = \frac{-5 \pm \sqrt{25 + 24}}{4} = \frac{-5 \pm \sqrt{49}}{4} = \frac{-5 \pm 7}{4}$$ $$x = \frac{2}{4} = \frac{1}{2} \quad \text{or} \quad x = \frac{-12}{4} = -3$$

Discriminant ($b^2 - 4ac$):

• If $> 0$: Two distinct real roots
• If $= 0$: Two equal real roots
• If $< 0$: No real roots (complex roots)

⚡ Common Student Mistakes: Forgetting to reverse the inequality sign when multiplying/dividing by negative. Losing negative signs when expanding brackets. Not writing the final answer clearly.

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🔴 Extended — Deep Study (3mo+)

Comprehensive theory for thorough preparation.

Simultaneous Equations:

Two equations with two unknowns.

Method 1: Substitution $$2x + y = 7 \quad \text{…(1)}$$ $$x - y = 2 \quad \text{…(2)}$$

From (2): $x = y + 2$ Substitute into (1): $$2(y + 2) + y = 7$$ $$2y + 4 + y = 7$$ $$3y = 3$$ $$y = 1$$

Then $x = 1 + 2 = 3$

Method 2: Elimination $$2x + y = 7 \quad \text{…(1)}$$ $$x - y = 2 \quad \text{…(2)}$$

Add (1) and (2): $$3x = 9$$ $$x = 3$$

Substitute back: $2(3) + y = 7$, so $y = 1$

Simultaneous Equations with Quadratics:

Problem: $$x + y = 5 \quad \text{…(1)}$$ $$x^2 + y^2 = 13 \quad \text{…(2)}$$

From (1): $y = 5 - x$ Substitute into (2): $$x^2 + (5-x)^2 = 13$$ $$x^2 + 25 - 10x + x^2 = 13$$ $$2x^2 - 10x + 25 = 13$$ $$2x^2 - 10x + 12 = 0$$ $$x^2 - 5x + 6 = 0$$ $$(x-2)(x-3) = 0$$ $$x = 2 \text{ or } x = 3$$

When $x = 2$, $y = 3$; When $x = 3$, $y = 2$

Quadratic Inequalities:

Problem: Solve $x^2 - x - 6 > 0$

Factor: $(x - 3)(x + 2) > 0$

Critical points: $x = 3$ and $x = -2$

Test intervals:

• $x < -2$: $(negative)(negative) > 0$ ✓
• $-2 < x < 3$: $(negative)(positive) < 0$ ✗
• $x > 3$: $(positive)(positive) > 0$ ✓

Solution: $x < -2$ or $x > 3$

Absolute Value Inequalities:

$$|x - 3| < 5$$ means $-5 < x - 3 < 5$ $$-2 < x < 8$$

$$|2x + 1| \geq 7$$ means $2x + 1 \leq -7$ or $2x + 1 \geq 7$ $$x \leq -4 \text{ or } x \geq 3$$

Application Problems:

Problem: A rectangular garden is 10 m longer than it is wide. Its area is 144 m². Find its dimensions.

Let width = $w$, length = $w + 10$ $$w(w + 10) = 144$$ $$w^2 + 10w - 144 = 0$$ $$(w + 18)(w - 8) = 0$$ $$w = 8 \text{ m (rejecting } w = -18 \text{)}$$

Width = 8 m, Length = 18 m

Problem: Two numbers differ by 5. Their product is 84. Find the numbers.

Let $x$ and $y$ with $x - y = 5$ and $xy = 84$ $$x = y + 5$$ $$(y + 5)y = 84$$ $$y^2 + 5y - 84 = 0$$ $$(y + 12)(y - 7) = 0$$ $$y = 7 \text{ or } y = -12$$

Numbers: 12 and 7, or −7 and −12

Simultaneous Inequalities:

Find the range of values satisfying both: $$2x + 3 < 11 \quad \Rightarrow \quad x < 4$$ $$3x - 2 > 1 \quad \Rightarrow \quad x > 1$$

Solution: $1 < x < 4$

Indicial/Exponential Equations:

Problem: Solve $2^{x+1} = 16$ $$2^{x+1} = 2^4$$ $$x + 1 = 4$$ $$x = 3$$

Problem: Solve $3^{2x} = 27$ $$3^{2x} = 3^3$$ $$2x = 3$$ $$x = \frac{3}{2}$$

Logarithmic Equations:

Problem: Solve $\log_2(x + 3) = 5$ $$x + 3 = 2^5$$ $$x + 3 = 32$$ $$x = 29$$

⚡ WAEC Examination Patterns: Solve linear equations (including with brackets and fractions). Solve quadratic equations by formula and factorisation. Solve simultaneous linear equations. Set up and solve equations from word problems. Solve quadratic inequalities. Use the discriminant to determine the nature of roots.