LESSON 15 / 100 | 15% COMPLETE | STAGE I — FOUNDATIONS | FINAL LESSON
Stage I — Foundations · Capstone Lesson
Evaluating Expressions
The complete skill — from words to expression to answer — in one unbroken workflow.
Section 01
The Capstone Lesson
This is the final lesson of Stage I. Rather than introducing new mechanics, it consolidates everything learned across lessons 1–14 into a single, unified skill: reading a situation, building an expression, and evaluating it to produce a meaningful result.
Words
Real situation or problem
→
Define
Choose & name variable
→
Build
Write the expression
→
Substitute
Replace variables
→
Evaluate
Answer
Think of the pipeline above as a translation engine. At one end enters a sentence about the world; at the other end emerges a number that describes it precisely. Every step between is a skill you have already built. This lesson is the first time you run the engine at full speed.
Section 02
Stage I Skills — The Complete Set
Lessons 01–03
Variables, Terms & Expressions
The language of algebra — naming parts, reading structure
Lessons 04–07
The Number System
Integers, absolute value, directed number arithmetic
Lessons 08–09
Fractions, Decimals & Percentages
The three forms of rational numbers and their operations
Replacing variables with values; words into algebra
Section 03
Pure Evaluation — No Context
The most direct form: an expression is given, values are given, evaluate. No translation required — only substitution and BODMAS.
Worked Example A — Evaluate (3a − b)² + 2ab when a = 2, b = −3
Write with brackets:(3(2) − (−3))² + 2(2)(−3)
Inner brackets:(6 + 3)² + 2(2)(−3)→ (9)²
Orders:81 + 2(2)(−3)
Multiplication (L→R):81 + 4(−3) = 81 + (−12)
Addition:81 − 12= 69
Worked Example B — Evaluate [2(x + y) − x²] ÷ y when x = 3, y = −1
Substitute:[2((3) + (−1)) − (3)²] ÷ (−1)
Inner brackets:[2(2) − 9] ÷ (−1)
Square brackets:[4 − 9] ÷ (−1) = [−5] ÷ (−1)
Division:(−5) ÷ (−1)= 5
Section 04
Translate, Then Evaluate
The full pipeline: a situation is described in words, you construct the expression, then evaluate it for given values. This is the standard form of applied mathematics problems.
Full Pipeline A — “The sum of a number and its square, decreased by four times the number.” Evaluate when n = 5.
1
Define: Let n = the number
2
Translate: “sum of n and n²” → n + n² | “decreased by four times n” → − 4n
3
Expression:n² + n − 4n = n² − 3n (simplified)
4
Substitute n = 5: (5)² − 3(5) = 25 − 15
5
Answer:10
Full Pipeline B — Fuel Cost. A car travels at v km/h and uses (v² / 400 + 2) litres per 100 km. Petrol costs R p per litre. Write an expression for the cost of a 300 km trip, then evaluate when v = 120, p = 24.
1
Consumption per 100 km:v²/400 + 2 litres
2
For 300 km (3 × 100 km):3(v²/400 + 2) litres total
When an expression can be simplified before substitution, always do so. A simplified expression is faster to evaluate and reduces arithmetic errors.
Worked Example C — Evaluate 4(2x + 1) − 3(x − 5) when x = −2
1
Expand first:8x + 4 − 3x + 15
2
Collect like terms:5x + 19
3
Substitute x = −2: 5(−2) + 19 = −10 + 19
4
Answer:9
Why Simplify First?
Substituting directly into 4(2(−2) + 1) − 3((−2) − 5) produces a longer, more error-prone chain of arithmetic. Simplifying to 5x + 19 first reduces the problem to a single multiplication and addition. Both methods give the same answer — the simplified route is simply safer at speed.
Section 06
Comparing Two Expressions
A common task: evaluate two different expressions for the same values and determine which is greater, or find the value(s) where they are equal. This is a direct preview of equation-solving in Stage II.
Worked Example D — Which is greater when x = 3: Expression A = 2x² − 5 or Expression B = 7x − 4?
1
Evaluate A:2(3)² − 5 = 2(9) − 5 = 18 − 5 = 13
2
Evaluate B:7(3) − 4 = 21 − 4 = 17
3
Compare: A = 13, B = 17. Expression B is greater when x = 3.
Worked Example E — At what integer value of x do the two expressions first become equal? A = x + 6, B = 2x − 1 (check x = 1, 2, 3 …)
x = 1:A = 7, B = 1A > B
x = 5:A = 11, B = 9A > B
x = 7:A = 13, B = 13A = B ✓
The expressions are equal when x = 7. In Stage II, you will solve this directly using an equation.
Section 07
Error Spotting
A key skill in examination settings: identify the error in a worked solution and correct it. Below is a flawed evaluation — find and fix every mistake.
Flawed Solution — Find the Errors
Evaluate3x² − 2(x − 4) when x = −3
Student writes:3 × −3² − 2(−3 − 4)
Step 2:3 × −9 − 2(−7)
Step 3:−27 − (−14) = −27 − 14 = −41
Corrected Solution
Error 1 fixed:3(−3)² − 2((−3) − 4) — bracket the substituted value
Error 2 fixed:(−3)² = +9, not −9 — squared negative is positive
4. Evaluate √(a² + b² + c²) when a = 2, b = 4, c = 4.
√(4 + 16 + 16) = √36 = 6
5. A rectangle has length (3x − 1) cm and width (x + 4) cm. Write a simplified expression for the area, then evaluate when x = 5. (Hint: area = length × width — expand the product.)
6. Evaluate P = 2l + 2w (perimeter formula) and A = lw (area formula) for a rectangle with l = 3x + 2 and w = x − 1, when x = 4. State both the perimeter and area.
When x = 4: l = 3(4) + 2 = 14, w = 4 − 1 = 3. P = 2(14) + 2(3) = 28 + 6 = 34 cm. A = 14 × 3 = 42 cm²
7. Error Spotting: A student evaluates −2x² + 5x when x = 3 and writes: −2 × 3² + 5 × 3 = −2 × 9 + 15 = −18 + 15 = −3. Is this correct? If not, identify the error.
The student’s answer of −3 is actually correct in this case. The expression −2x² applies the exponent to x only (not −2x), so −2(3)² = −2(9) = −18 is right. The process is valid: −18 + 15 = −3 ✓. (Common trap: this question tests whether you recognise a correct solution even when the method looks suspicious.)
8. Challenge — Full Pipeline: A mobile data plan charges a fixed fee of R f per month plus R c per GB used. A 15% VAT is added to the total. Write a fully simplified expression for the monthly bill, then calculate it when f = 150, c = 12 and the customer used 8 GB.
Base cost: f + cg (where g = GB used). With 15% VAT: 1.15(f + cg). Substitute f = 150, c = 12, g = 8: 1.15(150 + 12 × 8) = 1.15(150 + 96) = 1.15(246) = R282.90
✦
Stage I Complete
Foundations · Lessons 1–15 · 15% of the Series
You have built the full foundation of algebraic reasoning — from the meaning of a variable to translating language into expressions and producing numerical results. Every technique encountered in Stages II through X rests on what you have practised here.
Summary
Stage I — Master Checklist
You Can Now — The Complete Stage I Skill Set
Define variables, identify coefficients, constants, terms, and expressions
Place integers on a number line and compute absolute value
Add, subtract, multiply, and divide directed numbers correctly
Operate on fractions: simplify, multiply, divide, add, subtract
Convert fluently between fractions, decimals, and percentages
Apply percentage increase, decrease, and change calculations
Evaluate any arithmetic or algebraic expression using BODMAS
Collect like terms and simplify multi-variable expressions
Expand brackets using the distributive property; factorise by HCF
Substitute values into expressions using the bracket-first habit
Translate English phrases into algebraic expressions
Run the complete pipeline: words → expression → evaluation → answer
Up Next → Stage II · Lesson 16
What is an Equation?
Stage II introduces the equation — the moment algebra gains the power to find unknowns, not merely describe them. The expression becomes a statement; the statement becomes solvable.