Algebraic Expressions and Simple Equations

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Algebra uses letters and symbols to represent numbers and quantities in formulas, equations, and expressions. This allows us to generalise mathematical relationships and solve problems where the exact number is unknown.

Key Vocabulary:

Term	Meaning	Example
Variable	A symbol (letter) that represents an unknown value	$x, y, a, n$
Constant	A fixed number	$5, -3, \pi$
Coefficient	The number multiplying a variable	In $3x$, coefficient is 3
Term	A single number, variable, or product of numbers and variables	$5x, -3y^2, 7$
Expression	A combination of terms with + and − signs	$3x + 5$
Equation	A mathematical statement that two expressions are equal	$3x + 5 = 14$
Like Terms	Terms with the same variable(s) raised to the same power	$3x$ and $5x$ are like terms

Simplifying Expressions:

Combine like terms only — add or subtract the coefficients, keep the variable:

$3x + 5x = 8x$ $7y - 2y = 5y$ $4x + 3y$ cannot be combined (different variables)

Expanding Brackets:

Multiply every term inside the bracket by the term outside:

$3(x + 4) = 3x + 12$

$a(b + c) = ab + ac$

$(x + 3)(x + 5)$:

• $x \times x = x^2$
• $x \times 5 = 5x$
• $3 \times x = 3x$
• $3 \times 5 = 15$
• $x^2 + 5x + 3x + 15 = x^2 + 8x + 15$

⚡ Exam Tip (NCEE): When you see a question like “If $x = 3$, find the value of $2x + 5$”, simply substitute: $2(3) + 5 = 6 + 5 = 11$. This is called evaluation.

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🟡 Standard — Regular Study (2d–2mo)

For students who want genuine understanding of algebra.

Solving Simple Equations:

An equation says two things are equal. To solve, find the value of the unknown variable.

Golden Rule: Whatever you do to one side, you MUST do to the other.

Step-by-Step Method:

Solve: $3x + 7 = 22$

1. Isolate the term with the variable — remove constants from the left side $3x + 7 - 7 = 22 - 7$ $3x = 15$

2. Solve for the variable — divide by the coefficient $3x/3 = 15/3$ $x = 5$

Types of Equations:

Linear Equations (Unknown to power 1):

$x + 5 = 12$ → $x = 7$

$4x - 3 = 21$ → $4x = 24$ → $x = 6$

$7x + 2 = 5x + 14$: $7x - 5x = 14 - 2$ $2x = 12$ $x = 6$

Equations with Brackets:

$2(x + 3) = 16$ Divide both sides by 2 first: $x + 3 = 8$ Then: $x = 5$

OR expand first: $2x + 6 = 16$ → $2x = 10$ → $x = 5$

Equations with Fractions:

$\frac{x}{3} + 2 = 7$ $\frac{x}{3} = 5$ $x = 15$

Translating Word Problems into Equations:

Word Statement	Equation
A number plus 5 equals 12	$x + 5 = 12$
3 times a number is 24	$3x = 24$
A number decreased by 7 is 15	$x - 7 = 15$
The sum of twice a number and 6 is 20	$2x + 6 = 20$

Worked Example:

“Chidi has some sweets. If he receives 8 more sweets from his mother, he will have 25 sweets. How many sweets does Chidi have?”

Let $x$ = number of sweets Chidi has $x + 8 = 25$ $x = 25 - 8$ $x = 17$

Chidi has 17 sweets.

⚡ Common NCEE Error: When solving equations, students forget to apply the same operation to both sides. Always perform the same operation on both sides of the equals sign. If you subtract 3 from one side, you MUST subtract 3 from the other.

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Quadratic Equations (Unknown to power 2):

A quadratic equation has the form $ax^2 + bx + c = 0$ where $a \neq 0$.

Solving by Factorisation:

Example: $x^2 - 5x + 6 = 0$

Find two numbers that multiply to +6 (c) and add to -5 (b): $-2$ and $-3$

$(x - 2)(x - 3) = 0$

So $x - 2 = 0$ or $x - 3 = 0$ $x = 2$ or $x = 3$

Solving by the Quadratic Formula:

For $ax^2 + bx + c = 0$: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

Example: $x^2 - 5x + 6 = 0$ $a = 1, b = -5, c = 6$ $x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(6)}}{2(1)}$ $x = \frac{5 \pm \sqrt{25 - 24}}{2} = \frac{5 \pm 1}{2}$ $x = 3$ or $x = 2$

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

• If $b^2 - 4ac > 0$: Two distinct real roots
• If $b^2 - 4ac = 0$: One repeated root
• If $b^2 - 4ac < 0$: No real roots

Simultaneous Equations:

Two equations with two unknowns.

By Substitution:

$x + y = 7$ and $x - y = 3$

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

Then $x = 7 - 2 = 5$

Check: $5 + 2 = 7$ ✓, $5 - 2 = 3$ ✓

By Elimination:

$x + y = 7$ $x - y = 3$

Add the equations: $2x = 10$ → $x = 5$ Substitute: $5 + y = 7$ → $y = 2$

Simple Inequalities:

Solve $3x - 4 > 8$: $3x > 12$ $x > 4$

⚡ Extended Tip — Checking Your Answers: Always substitute your answer back into the original equation to check. For $x^2 - 5x + 6 = 0$, substituting $x = 2$: $4 - 10 + 6 = 0$ ✓. For simultaneous equations: substitute both values into the second equation to verify. This habit prevents losing marks to simple errors.

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