Linear Equations & Inequalities — SL 1.7 Linear equations and inequalities in context (intro) - IB Mathematics Applications and Interpretation

SECTION 1

Historical Context & Motivation

Humans have been setting up and solving equations for thousands of years, long before anyone called them "linear." Ancient merchants needed to figure out how many goods they could trade, builders had to calculate materials, and tax collectors tried to determine fair payments. The idea of expressing an unknown quantity with a symbol and then manipulating a balanced relationship to find that quantity is one of the oldest and most powerful tools in all of mathematics.

The concept of linear equations — equations where the variable appears only to the first power — evolved over centuries as mathematicians from Babylon, Egypt, Greece, India, and the Islamic world refined notation and methods. Similarly, linear inequalities arose naturally whenever people needed to express constraints: "I can spend at most this much" or "I need at least that many." Today, these tools underpin everything from budgeting to engineering.

c. 1800 BCE • Babylonian Clay Tablets

Babylonian scribes recorded problems equivalent to linear equations on cuneiform tablets, solving them using step-by-step "recipes" — the earliest known algebra.

c. 300 BCE • Euclid's Geometric Algebra

Greek mathematicians expressed relationships between quantities using geometry. A line segment represented an unknown, and proportional reasoning solved for it.

c. 820 CE • Al-Khwārizmī's Al-Jabr

The Persian mathematician al-Khwārizmī systematized equation-solving techniques in his book, giving us the word "algebra" and the idea of balancing both sides of an equation.

1637 • Descartes & Symbolic Notation

René Descartes introduced the use of letters like x and y for unknowns, connecting algebra to coordinate geometry and making linear equations look the way we write them today.

1827 • Inequalities Formalized

The symbols < and > had been used since 1631, but formal work on inequalities as mathematical objects grew in the 19th century, enabling optimization and constraint-based modeling.

In SL 1.7 of the IB Applications and Interpretation course, the focus is on contextual problems — situations described in words that you must translate into mathematical statements. The central question is: how do we convert a real-world scenario into an equation or inequality, solve it, and then interpret the answer back in context?

SECTION 2

Core Principles & Definitions

Before diving into problems, you need a solid grip on the vocabulary and core ideas that make this topic work. A linear equation is any equation that can be written in the form ax + b = c, where a, b, and c are constants and x is the unknown variable raised to the first power. A linear inequality replaces the equals sign with an inequality symbol (< , > , ≤ , or ≥), expressing a range of acceptable values rather than a single solution.

Variable Identification

Read the problem carefully and decide what the unknown quantity is. Assign it a letter (often x or n). Every other quantity should be expressed in terms of this variable or given as a constant.

Equation vs. Inequality

If the problem says "equals," "is," or "costs exactly," use an equation (=). If it says "at most," "no more than," "at least," or "must exceed," use an inequality (≤ or ≥).

Balance Principle

Whatever operation you perform on one side of an equation or inequality, you must perform the same operation on the other side. For inequalities, multiplying or dividing by a negative number reverses the direction of the inequality symbol.

Contextual Interpretation

After solving, always translate your answer back into the real-world context. Check whether the answer makes sense — for instance, the number of people can't be negative or fractional.

Units & Reasonableness

Keep track of units (dollars, hours, kilograms) throughout the problem. A final reasonableness check — does this answer make practical sense? — can catch algebraic mistakes before they cost you marks.

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KEY TAKEAWAY

Think of a linear equation like a perfectly balanced seesaw: whatever you add to or remove from one side, you must do the same to the other to keep it level. An inequality is like a seesaw with a weight limit on one side — you're looking for all the loads that keep the heavy end from touching the ground, not just the one load that balances perfectly.

SECTION 3

Visual Explanation — From Words to Algebra

The diagram below walks through the translation process that sits at the heart of SL 1.7. Starting from a word problem, you identify the unknown, build an algebraic model, solve it, and then interpret the result. This four-stage pipeline is the same whether you are dealing with an equation or an inequality.

The four-stage pipeline: Read & Identify → Translate → Solve → Interpret. Every contextual linear problem follows this sequence.

Notice how each stage produces a clear output that feeds into the next. Stage 1 outputs a variable definition ("let x = km"). Stage 2 outputs a mathematical statement (1.50x + 3 = 18). Stage 3 outputs a numerical answer (x = 10). Stage 4 outputs a sentence answering the original question in plain language. If you get stuck on a contextual problem, ask yourself: which stage am I in, and what is my output supposed to look like?

SECTION 4

Mathematical Framework

The algebra behind contextual linear problems is straightforward once you recognize the standard forms. Below are the key equation and inequality structures you will encounter, along with the rules for solving them.

STANDARD LINEAR EQUATION

ax + b = c

where a is the rate (cost per unit, speed, etc.), x is the unknown quantity, b is the fixed/initial value, and c is the total or target value. Solve by subtracting b from both sides, then dividing both sides by a.

TWO-EXPRESSION EQUATION

ax + b = dx + e

Arises when two scenarios are compared (e.g., "when does Plan A cost the same as Plan B?"). Collect x-terms on one side and constants on the other: (a − d)x = e − b, then divide.

LINEAR INEQUALITY

ax + b ≤ c (or ≥ , < , >)

Solve exactly like an equation except: if you multiply or divide both sides by a negative number, you must reverse the inequality sign. The solution is a range of values, not a single number.

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The Inequality Flip Rule

Consider −2x > 6. To isolate x, divide both sides by −2. Because you are dividing by a negative, flip the sign: x < −3. A quick check confirms: if x = −4, then −2(−4) = 8, and 8 > 6 ✓. If x = −2, then −2(−2) = 4, and 4 > 6 ✗. The flip is essential.

In IB context problems, you will also need to consider domain restrictions. If x represents the number of people on a bus, x must be a non-negative integer — even if the algebra gives you x = 12.7, the contextual answer might be 12 or 13 depending on the inequality direction. Always re-read the question to decide whether to round up or down.

SECTION 5

Translating Key Phrases

One of the trickiest parts of contextual problems is recognizing which English phrases map to which mathematical symbols. The table below is a reference guide you can use whenever you encounter a word problem. Pay special attention to inequality phrases — the IB frequently tests whether students can distinguish between strict (< , >) and non-strict (≤ , ≥) inequalities.

English Phrase	Math Symbol	Example
is, equals, the same as, results in	=	Total cost is $40 → C = 40
more than, exceeds, greater than	>	Speed exceeds 60 → v > 60
at least, no fewer than, minimum	≥	At least 5 people → n ≥ 5
less than, fewer than, under	<	Under 100 g → m < 100
at most, no more than, maximum	≤	Budget at most $200 → C ≤ 200
per, for each, every	× (multiply)	$5 per hour → 5h
total, combined, altogether	+ (add)	Fixed fee plus variable → b + ax

Common English-to-algebra translations for contextual linear problems

An equation produces a single point (filled dot). A non-strict inequality (≤ or ≥) uses a filled dot at the boundary. A strict inequality (< or >) uses an open circle to show the boundary value is excluded.

When you solve an inequality and need to graph the solution, remember: a filled circle (●) means the endpoint is included (≤ or ≥), while an open circle (○) means it is not (< or >). In contextual problems, the distinction often determines whether the boundary value itself is a valid answer — for example, whether you can spend exactly $200 when the budget says "at most $200."

SECTION 6

Worked Example — Mobile Phone Plans

Let's work through a full contextual problem that combines both an equation and an inequality, as the IB commonly does.

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Problem Statement

Two mobile phone plans are available. Plan A charges a fixed monthly fee of $25 plus $0.10 per minute of calls. Plan B charges no monthly fee but costs $0.35 per minute. (a) For how many minutes of calls would both plans cost the same? (b) Aisha's monthly budget for phone calls is at most $55. If she chooses Plan A, what is the maximum number of whole minutes she can use?

Part (a) — When do the plans cost the same?

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Step 1 — Define the variable

Let x = the number of minutes of calls in a month.
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Step 2 — Write expressions for each plan

Cost of Plan A = 25 + 0.10x (fixed fee plus per-minute charge). Cost of Plan B = 0.35x (per-minute charge only, no fixed fee).
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Step 3 — Set up the equation

For the plans to cost the same: 25 + 0.10x = 0.35x.
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Step 4 — Solve

Subtract 0.10x from both sides: 25 = 0.25x. Divide both sides by 0.25: x = 100.

The plans cost the same at x = 100 minutes.
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Step 5 — Interpret

If Aisha uses exactly 100 minutes per month, both plans cost $25 + $10 = $35. For fewer than 100 minutes, Plan B is cheaper; for more than 100 minutes, Plan A is cheaper.

Part (b) — Budget constraint with Plan A

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Step 1 — Set up the inequality

Aisha's budget is at most $55, so: 25 + 0.10x ≤ 55.
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Step 2 — Solve the inequality

Subtract 25 from both sides: 0.10x ≤ 30. Divide both sides by 0.10: x ≤ 300.
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Step 3 — Apply domain restriction

Since x represents whole minutes and x ≤ 300, the maximum number of whole minutes is 300.

Aisha can use at most 300 minutes per month on Plan A within her $55 budget.
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Step 4 — Verify

At 300 minutes: 25 + 0.10(300) = 25 + 30 = $55 ✓. At 301 minutes: 25 + 0.10(301) = $55.10 > $55 ✗. The boundary value itself is included because the inequality is ≤ ("at most").

SECTION 7

Equations vs. Inequalities — Strengths & Limitations

Equations and inequalities are closely related, but they serve different purposes and behave differently in a few critical ways. Understanding when to use each — and the pitfalls unique to inequalities — is essential for IB exam success.

Feature	Linear Equation	Linear Inequality
Solution type	A single value (e.g., x = 10)	A range of values (e.g., x ≤ 10)
When to use	"How many?" / "What value exactly?"	"At most," "at least," "no more than"
Multiplying/dividing by negative	No special rule	Must flip the inequality sign
Graphical representation	Single point on number line	Shaded ray or segment on number line
Contextual answer format	"The answer is 10 km."	"She can travel up to 10 km."
Common IB trap	Forgetting to state the answer in context	Forgetting the flip rule or using wrong boundary type

Key differences between linear equations and linear inequalities in contextual problems

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KEY TAKEAWAY

Equations are like GPS navigation — they tell you the exact destination. Inequalities are like guardrails on a highway — they define the safe zone you can operate within. Both are essential: the GPS gets you somewhere specific, but the guardrails keep you from going too far.

SECTION 8

Connection to Advanced Topics

The skills you build in SL 1.7 are foundational for several topics you will encounter later in the IB course and beyond. Setting up equations from context is the backbone of mathematical modeling, while inequalities lead directly into optimization — finding the best possible outcome under constraints.

SL 1.7 Concept	Where It Leads	What Changes
Setting up linear equations from context	Systems of linear equations (SL 1.8)	Two unknowns instead of one; two equations needed
Solving linear inequalities	Linear programming (HL)	Multiple inequalities graphed on a coordinate plane; finding optimal solutions
Rate × quantity + fixed cost model	Linear functions and modeling (SL 2.5)	Same equation form, but analyzed as y = mx + c with slope and intercept
Interpreting solutions in context	Statistical modeling, chi-squared tests	Interpretation becomes more nuanced with probability and uncertainty

How SL 1.7 concepts connect to later IB topics

If you master the art of translating words into algebra in SL 1.7, you will find that nearly every subsequent topic in the IB AI course asks you to do the same thing — just with more complex mathematical structures. The four-stage pipeline (Read → Translate → Solve → Interpret) remains your most reliable strategy throughout the entire course.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

A student sets up the inequality 3x + 7 > 22 to model a word problem. Explain, in your own words, what the phrase "more than" tells us about the solution — specifically, does x = 5 satisfy the inequality, and why or why not?

Show answer

Substituting x = 5 gives 3(5) + 7 = 22, which equals 22 but is not greater than 22. Because the inequality uses the strict symbol > (not ≥), the value x = 5 does not satisfy it. The phrase "more than" implies strictly greater, so the boundary value is excluded. Any value of x greater than 5 (such as x = 6, which gives 25 > 22 ✓) would satisfy the inequality.

Question 2: BASIC CALCULATION

A gym charges a one-time registration fee of $40 plus $12 per month. If Marco has saved $160 for gym expenses, write and solve an equation to find how many full months he can afford.

Show answer

Let x = number of months. The total cost is 40 + 12x. Setting this equal to 160: 40 + 12x = 160. Subtract 40: 12x = 120. Divide by 12: x = 10. Marco can afford exactly 10 months of gym membership with his $160. Note that if the problem had said "at most $160," we would use ≤ and the answer would be the same (x ≤ 10), but any whole number from 0 to 10 would be valid.

Question 3: INTERMEDIATE

A school is organizing a field trip. Bus Company X charges $200 plus $5 per student. Bus Company Y charges $8 per student with no fixed fee. (a) For how many students do both companies charge the same? (b) The school has a transport budget of at most $600. Using Company X, find the maximum number of students who can go.

Show answer

(a) Let n = number of students. Set 200 + 5n = 8n. Subtract 5n: 200 = 3n. Divide: n = 66.67. Since we need a whole number and the costs are equal at n ≈ 66.67, neither company is exactly cheaper at 66 or 67 students — the crossover point is between them. However, for n ≤ 66, Company Y is cheaper, and for n ≥ 67, Company X is cheaper. (b) Using Company X: 200 + 5n ≤ 600. Subtract 200: 5n ≤ 400. Divide: n ≤ 80. The maximum number of students is 80, which would cost exactly $600.

Question 4: APPLIED

A small bakery makes custom cakes. Each cake requires $8 in ingredients and 1.5 hours of labour at $20 per hour. The bakery also pays $500 per month in fixed costs (rent, utilities). If each cake sells for $50, write and solve an inequality to find the minimum number of cakes the bakery must sell per month to avoid a loss (i.e., revenue ≥ costs).

Show answer

Let n = number of cakes sold per month. Revenue = 50n. Cost per cake = 8 + 1.5(20) = 8 + 30 = $38. Total costs = 38n + 500. For no loss: 50n ≥ 38n + 500. Subtract 38n: 12n ≥ 500. Divide by 12: n ≥ 41.67. Since n must be a whole number and we need revenue to be at least equal to costs, round up to n = 42. The bakery must sell at least 42 cakes per month. At 42 cakes: revenue = $2100, costs = 38(42) + 500 = $2096, giving a small profit of $4.

Question 5: CRITICAL THINKING

A student solves the inequality 15 − 3x ≥ 6 and writes x ≥ −3 as the answer. (a) Identify and explain the error. (b) Find the correct solution. (c) A delivery service says packages must weigh at least 6 kg less than their 15 kg limit. Using this inequality, determine the range of acceptable package weights and explain whether 3 kg is acceptable.

Show answer

(a) The student forgot to reverse the inequality sign when dividing by −3. Starting from 15 − 3x ≥ 6: subtract 15 to get −3x ≥ −9, then divide by −3 (flip the sign!) to get x ≤ 3, not x ≥ −3. (b) The correct solution is x ≤ 3. (c) If x represents the package weight, then x must satisfy 15 − 3x ≥ 6, which gives x ≤ 3. However, weight must also be positive (x > 0), so the acceptable range is 0 < x ≤ 3. A 3 kg package is acceptable because the inequality is non-strict (≤), so the boundary is included. This problem highlights two key skills: applying the flip rule and considering real-world domain restrictions.

SUMMARY

Lesson Summary

Summary

SL 1.7 is all about connecting real-world situations to algebra. You learned the four-stage pipeline — Read & Identify, Translate, Solve, Interpret — which guides you from a word problem to a contextual answer. Linear equations (ax + b = c) produce a single solution, while linear inequalities (ax + b ≤ c) produce a range of values. Key English phrases like "at most," "at least," "more than," and "fewer than" tell you which inequality symbol to use.

Remember the critical flip rule: when multiplying or dividing both sides of an inequality by a negative number, reverse the direction of the inequality sign. Always check your answer for domain restrictions — quantities like the number of people, items, or months must be non-negative whole numbers, which may require rounding. Finally, always state your answer in context using a complete sentence that refers back to the original situation. These skills form the foundation for systems of equations, linear functions, and optimization problems that come later in the IB course.