International Baccalaureate · MYP 2

Algebraic Expressions

⌚ ~194 min ▢ 22 blocks ✓ 97 exercises ➣ Prerequisites : Integers and Signed Numbers

I The Language of Algebra

Definition — Constant

A constant is a number.

Example

$0, \; 3, \; 2.5$

Definition — Variable

A variable is a quantity which we represent by a letter.

Example

The variable $x$ is the number of marbles inside the cup.

Definition — Expression

An expression is a calculation written with numbers, letters and operation signs such as $+$, $-$, $\times$, $\div$ and $\sqrt{\ }$, without an equal sign. A number or a letter on its own is also an expression.

Example

A cup contains $x$ marbles. Next to the cup, there are 2 marbles outside. The expression for the number of marbles is$$x+2$$

Definition — Equation

An equation is an equality between two expressions, the left-hand side and the right-hand side, separated by an equal sign $=$, and containing at least one unknown — a letter standing for a number we are looking for: the value that makes the equality true.

Example

A cup contains $x$ marbles. The equation for the number of marbles is$$x + 2 = 8$$

Definition — Formula

A formula is an equality that links quantities: it expresses one of them, the subject, in terms of the others. Its letters are variables: they can take many values. Formulas often come from the real world, physics or geometry.

Example

For a triangle:

, $A = \dfrac{b \times h}{2}$ is the formula for the area.

Proposition — Standard Expressions

For any number $x$:

twice $x$ is $2 \times x = 2x$, and three times $x$ is $3 \times x = 3x$;
half of $x$ is $x \div 2 = \dfrac{x}{2}$, and a third of $x$ is $x \div 3 = \dfrac{x}{3}$;
the square of $x$ is $x \times x = x^2$, and its cube is $x \times x \times x = x^3$.

For any whole number $n$: the successor of $n$ is $n + 1$, and, when $n \geq 1$, its predecessor is $n - 1$.

Example

Express in terms of the letter: twice $x$, a third of $t$, the square of $y$, the successor of the whole number $n$.

Answer

Twice $x$ is $2x$, a third of $t$ is $\dfrac{t}{3}$, the square of $y$ is $y^2$ and the successor of $n$ is $n + 1$.

Skills to practice

Identifying Expressions, Equations, Variables and Constants
Writing Expressions
Writing an Expression from Words
Reading an Expression in Words

II Algebraic Notation

Definition — Product Notation

We can omit the $\times$ sign when it is followed by a variable or by a bracket $($ (also called a parenthesis).

Example

$2 \times x = 2x$
$2 \times (l+w) = 2(l+w)$

Definition — Terms and Coefficients

In the expression $3x + 5$, the terms are $3x$ and $5$: the parts joined by the signs $+$ or $-$. In the term $3x$, the number $3$ is the coefficient of $x$. The term $5$, which has no letter, is a constant term.

Three classic traps of the first encounter with letters:

$4x$ is a hidden multiplication: $4x = 4 \times x$. The $4$ and the $x$ are not “glued”: if $x = 3$, then $4x = 12$, not $43$.
A letter on its own carries an invisible coefficient: $x = 1x$.
$x + 2$ is a finished answer. Nothing forces the sum of a letter and a number into one symbol: $x + 2$ cannot be written $2x$.

By convention, the number is written in front of the letter: $n \times 5 = 5n$.

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Definition — Repeated Addition

$$\overbrace{x + x + \ldots + x}^{n\text{ terms}} = n\times x$$

Example

Each cup contains $x$ marbles. Simplify the expression for the number of marbles:$$x + x + 1+1+1$$

Answer

$$x + x + 1 + 1 + 1 = 2x +3$$

Definition — Repeated Multiplication

For a whole number $n \geq 1$: $$\overbrace{x \times x \times \ldots \times x}^{n\text{ factors}} = x^n$$

Skills to practice

Writing Without the Multiplication Sign
Simplifying Repeated Addition
Simplifying Repeated Multiplication
Understanding Literal Notation

III Identities and Like Terms

Discover

Three students were asked to find the formula for the perimeter of the rectangle of length $L$ and width $l$:

They wrote:

Su: $P = 2(l + L)$
Louis: $P = l + L + l + L$
Hugo: $P = 2l + 2L$

Which students are correct?

Answer

They are all correct. These three expressions $2(l + L)$, $l + L + l + L$ and $2l + 2L$ produce the same result for the perimeter of the rectangle for all values of $l$ and $L$. The equalities between them, such as $2(l + L) = 2l + 2L$, are called identities.

Definition — Identity

An identity is an equality between two expressions such that their evaluations produce the same value for all values of the variables.

Identities are fundamental in algebra: they allow us to transform and simplify expressions and are the foundation for solving equations and manipulating formulas.

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Proposition — Properties of Multiplying by 1 and 0

$$1 \times x = x \qquad \text{and} \qquad 0 \times x = 0$$

Proposition — The Commutative Laws

$$\textcolor{colordef}{a} + \textcolor{colorprop}{b} = \textcolor{colorprop}{b} + \textcolor{colordef}{a}\qquad \text{and} \qquad\textcolor{colordef}{a} \times \textcolor{colorprop}{b} = \textcolor{colorprop}{b} \times \textcolor{colordef}{a}$$

Proposition — The Associative Laws

$$(\textcolor{colordef}{a} + \textcolor{colorprop}{b}) + \textcolor{olive}{c} = \textcolor{colordef}{a} + (\textcolor{colorprop}{b} + \textcolor{olive}{c})\qquad \text{and} \qquad(\textcolor{colordef}{a} \times \textcolor{colorprop}{b}) \times \textcolor{olive}{c} = \textcolor{colordef}{a} \times (\textcolor{colorprop}{b} \times \textcolor{olive}{c})$$

Example

Show that $l + L + l + L = 2l + 2L$.

Answer

$$\begin{aligned}\textcolor{colordef}{l} + \textcolor{colorprop}{L} + \textcolor{colordef}{l} + \textcolor{colorprop}{L}&= \textcolor{colordef}{l + l} + \textcolor{colorprop}{L + L} && (\text{collecting terms}) \\ &= \textcolor{colordef}{2l} + \textcolor{colorprop}{2L} && (\text{repeated addition})\end{aligned}$$

Method — Simplifying by Collecting Like Terms

Simplifying an expression by collecting like terms involves combining terms that have the same variables raised to the same powers.

Identify like terms: Like terms are terms that have the same variable(s) raised to the same power. For example, $3x$ and $5x$ are like terms, but $3x$ and $3x^2$ are not.
Combine like terms: Add or subtract the coefficients (numerical parts) of the like terms. The variable part remains the same.

Example

Simplify the expression: $2x + 4 + x - 2$

Answer

$$\begin{aligned}[t]2x + 4 + x - 2 &= \textcolor{colordef}{2x} \textcolor{colorprop}{+4} \textcolor{colordef}{+x} \textcolor{colorprop}{-2} &&(\text{identifying like terms}) \\ &= \textcolor{colordef}{(2+1)x} + \textcolor{colorprop}{4-2} &&(\text{combining like terms}) \\ &= \textcolor{colordef}{3x} + \textcolor{colorprop}{2} &&(\text{simplifying})\end{aligned}$$

Example

Simplify the products $3x \times 5x$ and $6 \times 2x$.

Answer

$$\begin{aligned}[t]3x \times 5x &= 3 \times x \times 5 \times x &&(\text{hidden multiplications}) \\ &= (3 \times 5) \times (x \times x) &&(\text{commutative and associative laws}) \\ &= 15x^2\end{aligned}$$ $$\begin{aligned}[t]6 \times 2x &= 6 \times 2 \times x &&(\text{hidden multiplication}) \\ &= 12x\end{aligned}$$ The numbers are multiplied together, and the letters together.

Skills to practice

Combining Like Terms
Simplifying Expressions
Simplifying using Commutativity and Associativity
Simplifying by Multiplying by Zero
Simplifying with Decimal Coefficients

IV Substituting and Testing an Equality

Definition — Substituting

Substituting is replacing a variable in an expression or equation with a specific value.

To avoid confusion with signs, especially when substituting negative values, we usually write substitutions in brackets.

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Method — Evaluating

To evaluate an expression, substitute a number for each variable and perform the arithmetic operations.

Example

Each cup contains $x$ marbles. The expression for the total number of marbles is$$2x + 4$$ Evaluate this expression when $x = 5$ (meaning there are 5 marbles in each cup):

Answer

$$\begin{aligned}2x + 4 &= 2 \times \textcolor{olive}{(5)} + 4 \quad (\text{substituting } \textcolor{olive}{x} \text{ by } \textcolor{olive}{5}) \\ &= 10 + 4 \\ &= 14\end{aligned}$$ There are 14 marbles.

Example

Evaluate the expressions $3 - 2x$ and $x^2 + x$ for $x = -3$.

Answer

$$\begin{aligned}[t]3 - 2x &= 3 - 2 \times \textcolor{olive}{(-3)} &&(\text{replacing } x \text{ with } \textcolor{olive}{(-3)}) \\ &= 3 - (-6) \\ &= 3 + 6 \\ &= 9\end{aligned}$$ $$\begin{aligned}[t]x^2 + x &= \textcolor{olive}{(-3)} \times \textcolor{olive}{(-3)} + \textcolor{olive}{(-3)} &&(\text{replacing } x \text{ with } \textcolor{olive}{(-3)}) \\ &= 9 + (-3) \\ &= 6\end{aligned}$$ The brackets keep the sign of $-3$ attached to the number.

Method — Testing an Equality

To test whether an equality between two expressions is true for a given value of the letter:

Substitute the value into each side separately — the same value everywhere the letter appears.
Compute each side.
Compare: same result, the equality is true for that value; different results, it is false for that value.

A test speaks only for the value tested: an equality true for one value may fail for another.

Example

Test the equality $4n = 15 + n$ for $n = 5$, then for $n = 8$.

Answer

For $n = 5$, each side separately: $$4 \times 5 = 20 \qquad\text{and}\qquad 15 + 5 = 20.$$ Same result: the equality is true for $n = 5$.
For $n = 8$: $$4 \times 8 = 32 \qquad\text{and}\qquad 15 + 8 = 23.$$ Different results: the equality is false for $n = 8$. One equality, two verdicts — a test only ever speaks for its own value.

Skills to practice

Evaluating Expressions
Testing an Equality: Level 1
Testing an Equality: Level 2

V Proving and Refuting

Some equalities are true for every value of the letter — they are identities. A claim that begins with “for every number” cannot be proved by tests: a thousand favourable values prove nothing — though one unfavourable value is enough to destroy it. Two tools decide such claims: algebra proves, and a counter-example refutes.

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Method — Proving a General Property with Algebra

Name the numbers with letters: a whole number is $n$, its successor is $n + 1$, twice the number is $2n$.
Translate the property into an expression.
Transform the expression — collect like terms, and expand brackets with the distributive law — until the announced form appears.
Conclude: the equality holds for every value of the letter at once — which no list of examples can ever do.

Example

Prove that the sum of a whole number and its successor is always an odd number.

Answer

Let $n$ be a whole number; its successor is $n + 1$. Their sum: $$\begin{aligned}[t]n + (n + 1) &= 2n + 1. && (\text{collect like terms})\end{aligned}$$ Now $2n$ is twice $n$, an even number — so $2n + 1$, one more than an even number, is odd. This is proved for every whole number $n$ at once. Illustration (not a proof): $n = 4$ gives $4 + 5 = 9$, odd indeed.

Method — Refuting with a Counter-Example

A claim of the form “for every number…” is refuted by one single value for which it fails:

Test the claim on values of the letter.
Exhibit one value that fails, computing both sides — this value is a counter-example.
Conclude: the claim is false. One counter-example suffices; checking many favourable values never proves.

Example

Zoé claims: “for every whole number $n$, we have $2n > n + 2$”. She checked $n = 5$, $n = 10$ and $n = 100$. Refute her claim.

Answer

Test $n = 1$: $$2 \times 1 = 2 \qquad\text{and}\qquad 1 + 2 = 3,$$ and $2 > 3$ is false: $n = 1$ is a counter-example, so the claim is refuted. Zoé's three checks were correct — $10 > 7$, $20 > 12$, $200 > 102$ — but favourable values, however many, never prove a “for every”: one failure destroys it.

Note the asymmetry: to prove “for every number”, checking examples is never enough — it takes a reasoning that covers all values at once, and algebra is such a tool; to refute, one single number is enough. Tests can only ever do the second job.

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Method — Conjecturing with a Spreadsheet or an Algorithm

Tests never prove a claim about every number, but they are how such a claim is found. Running a calculation program for many starting numbers can reveal a regularity; stated for every starting number, this regularity is a conjecture — a claim that seems true and is not yet proved.

Run the program for many starting numbers, with a spreadsheet or with an algorithm.
Read the regularity in the results: always the same number, always twice the starting number, always an odd number…
State the conjecture for every starting number: “whatever the starting number, the result is…”.
Prove it or refute it: however many rows agree, a conjecture is not a proof. Algebra, with a letter for the starting number, proves it; one single starting number for which it fails, a counter-example, refutes it.

With a spreadsheet: column A receives the starting numbers, and cell B2 the formula of the program, written with the cell A2 in place of the starting number. A formula always starts with the sign =, and * is the multiplication sign. For the program “multiply by $2$, add $3$”, cell B2 receives =2*A2+3; copying it downwards changes A2 into A3, A4…, so that each row runs the program on its own starting number:

	A	B
1	Starting number	Result
2	1	`=2*A2+3`
3	2	`=2*A3+3`
4	3	`=2*A4+3`

and column B displays $5$, $7$, $9$.
With an algorithm: a few lines in natural language do the same work.

1	For $n$ from $1$ to $10$, repeat lines 2 and 3.
2	Compute the result $r$ of the program for $n$.
3	Display $n$ and $r$.

Example

Consider the program: “choose a number; add $3$; multiply by $2$; subtract twice the starting number”.

Run the program in a spreadsheet for several starting numbers.
State a conjecture.
Prove it with algebra.

Answer

Cell B2 receives the formula =(A2+3)*2-2*A2, copied downwards — the brackets make the addition come before the multiplication, as in the program. With starting numbers chosen small, large and decimal:

A B

1 Starting number Result

2 $0$ $6$

3 $1$ $6$

4 $2$ $6$

5 $5$ $6$

6 $10$ $6$

7 $2.5$ $6$

One row checked by hand: $(2.5 + 3) \times 2 - 2 \times 2.5 = 11 - 5 = 6$.
Every row displays $6$. Conjecture: whatever the starting number, the result is $6$. Six rows are six tests, not a proof.
Call $n$ the starting number. The program builds, step by step: $$\begin{aligned}[t] n &\longrightarrow n + 3 && \text{(add 3)}\\ &\longrightarrow 2(n + 3) && \text{(multiply by 2)}\\ &\longrightarrow 2(n + 3) - 2n && \text{(subtract twice $n$)}. \end{aligned}$$ Then: $$\begin{aligned}[t] 2(n + 3) - 2n &= 2n + 6 - 2n && \text{(distributivity)}\\ &= 0 + 6 && (2n - 2n = 0)\\ &= 6. \end{aligned}$$ The result is $6$ for every starting number $n$: the conjecture is proved. The spreadsheet showed it for six numbers; algebra shows it for all of them at once, and explains it: the multiplication by $2$ doubles the starting number, the last step subtracts twice the starting number, and only $2 \times 3 = 6$ remains.

Skills to practice

Deciding What a Test Proves
Conjecturing from a Table of Values
Proving a General Property
Refuting with a Counter-Example
Conjecturing Then Proving

VI Going further

Skills to practice

Reasoning and Problem-Solving
Quiz

	A	B
1	Starting number	Result
2	\(0\)	\(6\)
3	\(1\)	\(6\)
4	\(2\)	\(6\)
5	\(5\)	\(6\)
6	\(10\)	\(6\)
7	\(2.5\)	\(6\)

1	For \(n\) from \(1\) to \(10\), repeat lines 2 and 3.
2	Compute the result \(r\) of the program for \(n\).
3	Display \(n\) and \(r\).