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Algebra & Equations — Grade 10

Build the toolkit: simplify, factorise, solve and check. Learn the algebra moves behind every Grade 10 question.

1. Guided Course 2. Mastery Bank 3. Test Your Knowledge
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Grade 10 CAPS Mathematics

Algebra & Equations

Build the toolkit • show each step • practise and check.

The 3-Year Algebra Journey

You are building the tools every later topic will use.

10
BuildExpressions • factorising • equations • inequalities
11
ExtendSurds • quadratics • formula • nature of roots
12
ApplyChoose a method quickly in full Paper 1 riders
Teacher cue
Do not rush factorising. It unlocks fractions, quadratics, functions and calculus.

What Is Algebra, Really?

Before any rule or formula: algebra is just a shorter way to write reasoning you already do.

Starting from something you already know — no algebra needed yet

You already know how to fill in a missing number: \(3+\,?\,=10\). The missing number is obviously \(\boxed{7}\) — you didn't need algebra to see that.

Algebra just replaces that question mark with a letter. Instead of \(?\), we write \(x\): \(3+x=10\). The letter \(x\) means exactly the same thing as the question mark — "the number that goes here" — and you already know how to find it: \(x=7\).

That's the whole idea behind algebra: use a letter instead of a blank or a question mark for a number you don't know yet, then use the rules on the next few slides to find it — even when it's not as obvious as this one.

The big idea

Algebra is the toolbox every other topic borrows from — factorising, exponents and equation-solving show up inside Patterns & Sequences, Functions & Graphs, Finance, and every grade that follows. Get this year's toolbox solid.

Where we're headed

Numbers & factorising → algebraic fractions → exponents → equations → inequalities. Each section only needs what came before it, and two of the factorising formulas here (sum and difference of two cubes) are completely new — not a revision of Grade 9.

Real Numbers and Surds

Every number you'll use this year is a real number — rational or irrational.

Rational numbers

Can be written as a fraction \(\dfrac{a}{b}\) with integers \(a,b\) (\(b\ne0\)): whole numbers, fractions, and decimals that terminate or repeat, e.g. \(\dfrac34\), \(-5\), \(0{,}333\ldots\)

Irrational numbers

Cannot be written as a fraction of integers — the decimal goes on forever with no repeating pattern, e.g. \(\sqrt2\), \(\sqrt5\), \(\pi\). Most square roots that aren't perfect squares are irrational.

Use anchor squares

For this week, use only the closest anchors: \(5^2=25,\ 7^2=49,\ 8^2=64,\ 10^2=100\).

Goal: choose the two nearest squares; do not memorise a long wall of numbers.

Locating a surd
A surd is an irrational root, like \(\sqrt{52}\). To find which two consecutive integers it lies between, find the two closest perfect squares: since \(7^2=49\) and \(8^2=64\), and \(49<52<64\), it follows that \(7<\sqrt{52}<8\).
Worked example — classify each number Level 1

State whether each of the following is rational or irrational: (a) \(\sqrt{49}\) (b) \(\sqrt{50}\) (c) \(0{,}375\)

Show solution
  1. 1(a) \(49\) is a perfect square: \(\sqrt{49}=7\), a whole number — \(\boxed{\text{rational}}\)
  2. 2(b) \(50\) is not a perfect square, so \(\sqrt{50}=5\sqrt2\) cannot be written as a whole number or fraction — \(\boxed{\text{irrational}}\)
  3. 3(c) \(0{,}375\) is a terminating decimal, so it can be written as a fraction (\(0{,}375=\dfrac38\)) — \(\boxed{\text{rational}}\)
Worked example Level 1

Between which two consecutive integers does \(\sqrt{52}\) lie?

Show solution
  1. 1Find the perfect squares closest to 52 on either side: \(49=7^2\) and \(64=8^2\).
  2. 2Since \(49<52<64\), taking square roots (all positive) preserves the order: \(7<\sqrt{52}<8\).
  3. 3\(\boxed{\sqrt{52}\text{ lies between }7\text{ and }8}\)
Worked example — rounding Level 1

Write \(\dfrac57\) as a decimal, rounded to 3 decimal places.

Show solution
  1. 1Divide: \(5\div7=0{,}714285714\ldots\) (this repeats forever, so \(\dfrac57\) is rational but its decimal form never terminates).
  2. 2To round to 3 decimal places, look at the 4th decimal digit to decide whether to round up or down: the digits are \(0{,}\underline{714}2\ldots\), and that 4th digit is \(2\).
  3. 3\(2<5\), so round down — keep the 3rd decimal digit as it is: \(\boxed{0{,}714}\)
Quick Check

Which of these is irrational?

Between which two integers does \(\sqrt{20}\) lie?

Before We Start: Multiplying Two Brackets

Factorising is expanding in reverse — so a quick recap of expanding first.

Quick recall

Multiply out: \((x+4)(x-2)\)

Show the four products
  1. 1Multiply every term in the first bracket by every term in the second: \(x\times x=x^2\), \(x\times(-2)=-2x\), \(4\times x=4x\), \(4\times(-2)=-8\)
  2. 2Add all four products together: \(x^2-2x+4x-8\)
  3. 3Combine the two middle (like) terms: \(\boxed{x^2+2x-8}\)
Your turn

Multiply out: \((x-3)(x+5)\)

Show solution
  1. 1\(x\times x=x^2\), \(x\times5=5x\), \((-3)\times x=-3x\), \((-3)\times5=-15\)
  2. 2Add all four products: \(x^2+5x-3x-15\)
  3. 3Combine like terms: \(\boxed{x^2+2x-15}\)
Why this matters next
Factorising a trinomial is the exact same process, run backward: starting from \(x^2+2x-8\), we're hunting for the two brackets — \((x+4)\) and \((x-2)\) — that multiply together to give it back. Every factorising method on the next few slides is really just a faster way to find those brackets than guessing.

Factorising — Revision Plus Trinomials

The Grade 9 types, extended to messier trinomials.

Revision from Grade 9
  • Common factor: \(6x^2+9x=3x(2x+3)\)
  • Difference of two squares: \(x^2-16=(x-4)(x+4)\)
  • Ordinary trinomial (two numbers that multiply to the constant and add to the middle coefficient): \(x^2+8x+15=(x+3)(x+5)\), since \(3\times5=15\) and \(3+5=8\)
  • Perfect square trinomial, a special case of the same idea: \(x^2-2x+1=(x-1)^2\)
New this year: leading coefficient not 1

The Grade 9 method above only needs two numbers that multiply to the constant. When the leading coefficient isn't 1, e.g. \(2x^2-x-3\), it generalises: find two numbers that multiply to give (leading coefficient) × (constant) and add to give the middle coefficient, then split the middle term and factorise by grouping.

Grade 9 Revision, Worked in Full

Quick, but every step shown — don't skip these even if they look familiar.

Worked example — common factor Level 1

Factorise fully: \(4x^3-8x^2+12x\)

Show solution
  1. 1Find the highest common factor of all three terms: each term has a factor of \(4x\) (\(4x^3,\,-8x^2,\,12x\) all divide by \(4x\))
  2. 2Divide each term by \(4x\) and write it outside a bracket: \(\boxed{4x(x^2-2x+3)}\)
  3. 3Check by expanding: \(4x(x^2-2x+3)=4x^3-8x^2+12x\)✓ (the trinomial left behind doesn't factorise further — that's fine, not every factorisation goes further than one step)
Worked example — difference of two squares Level 2

Factorise fully: \(4x^2-25\)

Show solution
  1. 1Recognise both terms as perfect squares: \(4x^2=(2x)^2\) and \(25=5^2\)
  2. 2Apply \(a^2-b^2=(a-b)(a+b)\) with \(a=2x,\,b=5\): \(\boxed{(2x-5)(2x+5)}\)
  3. 3Check: \((2x-5)(2x+5)=4x^2+10x-10x-25=4x^2-25\)✓
Worked example — perfect square trinomial Level 2

Factorise fully: \(x^2+6x+9\)

Show solution
  1. 1Find two numbers that multiply to \(9\) and add to \(6\): \(3\) and \(3\) — the same number twice
  2. 2\(\boxed{(x+3)^2}\) — a repeated factor is written as a square
  3. 3Check: \((x+3)^2=x^2+3x+3x+9=x^2+6x+9\)✓
Recognise it when...
Common factor: every term shares a factor — always check this first, on every question. Difference of squares: exactly two terms, both perfect squares, joined by a minus sign. Perfect square trinomial: the two numbers you find happen to be identical.
Worked example — leading coefficient 1 Level 1

Factorise fully: \(x^2-2x-15\)

Show solution
  1. 1Find two numbers that multiply to \(-15\) (the constant) and add to \(-2\) (the middle coefficient): \(-5\) and \(3\)
  2. 2Write them straight into two brackets: \(\boxed{(x-5)(x+3)}\)
  3. 3Check by expanding: \((x-5)(x+3)=x^2+3x-5x-15=x^2-2x-15\)✓
Worked example — leading coefficient not 1 Level 2

Factorise fully: \(2x^2-x-3\)

Show solution
  1. 1Multiply the leading coefficient by the constant: \(2\times(-3)=-6\). Find two numbers that multiply to \(-6\) and add to \(-1\) (the middle coefficient): \(2\) and \(-3\).
  2. 2Split the middle term using these numbers: \(2x^2+2x-3x-3\).
  3. 3Group in pairs and factorise each: \(2x(x+1)-3(x+1)\).
  4. 4Both groups share the factor \((x+1)\): \(\boxed{(2x-3)(x+1)}\)
Worked example — more practice Level 2

Factorise fully: \(3x^2+2x-5\)

Show solution
  1. 1Multiply the leading coefficient by the constant: \(3\times(-5)=-15\). Find two numbers that multiply to \(-15\) and add to \(2\): \(5\) and \(-3\).
  2. 2Split the middle term: \(3x^2+5x-3x-5\)
  3. 3Group in pairs and factorise each: \(x(3x+5)-1(3x+5)\)
  4. 4Both groups share the factor \((3x+5)\): \(\boxed{(x-1)(3x+5)}\)
Check your answer
Always expand your factorised answer back out to confirm it matches the original: \((2x-3)(x+1)=2x^2+2x-3x-3=2x^2-x-3\). ✓

Factorising by Grouping in Pairs

For expressions with four terms and no single common factor.

Worked example Level 1

Factorise fully: \(xy-3x+2y-6\)

Show solution
  1. 1There's no factor common to all four terms, but the first two share \(x\) and the last two share \(2\): \(x(y-3)+2(y-3)\).
  2. 2Both groups now share the factor \((y-3)\), so factor it out: \(\boxed{(x+2)(y-3)}\)
Worked example Level 2

Factorise fully: \(ab+3a-2b-6\)

Show solution
  1. 1The first two terms share \(a\); the last two share \(-2\): \(a(b+3)-2(b+3)\)
  2. 2Both groups share the factor \((b+3)\): \(\boxed{(a-2)(b+3)}\)
Worked example — a negative common factor Level 3

Factorise fully: \(-xy+3x+2y-6\)

Show solution
  1. 1The first two terms share \(-x\); the last two share \(2\): \(-x(y-3)+2(y-3)\)
  2. 2Both groups share the factor \((y-3)\): \(\boxed{(2-x)(y-3)}\)
  3. 3Check: \((2-x)(y-3)=2y-6-xy+3x=-xy+3x+2y-6\) ✓
What to look for
Grouping works when the two pairs, once each has its own common factor removed, leave behind the same bracket. If they don't match, try reordering the four terms first.

Sum and Difference of Two Cubes

Two brand-new formulas — not a revision of anything from Grade 9.

Formulas to know
\[a^3+b^3=(a+b)(a^2-ab+b^2)\qquad a^3-b^3=(a-b)(a^2+ab+b^2)\]
Spotting it
Check whether every term is a perfect cube (\(1,8,27,64,125,\ldots\) or \(x^3,y^3,\ldots\)). Note the sign pattern: the bracket with the matching sign \((a+b)\) or \((a-b)\) comes first; the second bracket's middle term always has the opposite sign, and its outer terms are always positive.
Common trap
It's tempting to keep the same sign in both brackets. For \(x^3-8\), that would wrongly give \((x-2)(x^2-2x+4)\) — but expanding this back out gives \(x^3-4x^2+8x-8\), not \(x^3-8\). The correct factorisation, \((x-2)(x^2+2x+4)\), expands to exactly \(x^3-8\)✓ — always check by expanding if you're unsure of the sign.
Worked example — simplest case, \(a=x,b=1\) Level 1

Factorise fully: \(x^3+1\)

Show solution
  1. 1Recognise both terms as perfect cubes: \(x^3\) and \(1=1^3\), so \(a=x\), \(b=1\).
  2. 2Apply \(a^3+b^3=(a+b)(a^2-ab+b^2)\): \(x^3+1^3=(x+1)(x^2-x+1)\)
  3. 3\(\boxed{(x+1)(x^2-x+1)}\)
Worked example — difference of two cubes Level 2

Factorise fully: \(x^3-8\)

Show solution
  1. 1Recognise both terms as perfect cubes: \(x^3\) and \(8=2^3\), so \(a=x\), \(b=2\).
  2. 2Apply \(a^3-b^3=(a-b)(a^2+ab+b^2)\): \(x^3-2^3=(x-2)(x^2+2x+4)\)
  3. 3\(\boxed{(x-2)(x^2+2x+4)}\)
Worked example — sum of two cubes with coefficients Level 3

Factorise fully: \(8x^3+27\)

Show solution
  1. 1Recognise both terms as perfect cubes: \(8x^3=(2x)^3\) and \(27=3^3\), so \(a=2x\), \(b=3\).
  2. 2Apply \(a^3+b^3=(a+b)(a^2-ab+b^2)\): \((2x)^3+3^3=(2x+3)\big((2x)^2-(2x)(3)+3^2\big)\)
  3. 3\(\boxed{(2x+3)(4x^2-6x+9)}\)

Combining Techniques

The hardest factorising questions need more than one method, in the right order.

Worked example — common factor, then difference of squares Level 1

Factorise fully: \(2x^3-2x\)

Show solution
  1. 1Common factor first: every term shares \(2x\): \(2x(x^2-1)\)
  2. 2\(x^2-1\) is a difference of two squares: \(x^2-1=(x-1)(x+1)\)
  3. 3\(\boxed{2x(x-1)(x+1)}\)
Worked example — common factor, then a trinomial Level 3

Factorise fully: \(2x^3+2x^2-24x\)

Show solution
  1. 1Common factor first: every term shares \(2x\): \(2x(x^2+x-12)\)
  2. 2\(x^2+x-12\) is a trinomial: \(x^2+x-12=(x+4)(x-3)\)
  3. 3\(\boxed{2x(x+4)(x-3)}\)
Worked example Level 4

Factorise fully: \(3x^3-12x\)

Show solution
  1. 1Always check for a common factor first, even when a question looks like it needs a different method: every term shares a factor of \(3x\)
  2. 2Take out the common factor: \(3x(x^2-4)\)
  3. 3Look again at what's left in the bracket — \(x^2-4\) is a difference of two squares, so it factorises further: \(x^2-4=(x-2)(x+2)\)
  4. 4\(\boxed{3x(x-2)(x+2)}\) — check: \(3x(x-2)(x+2)=3x(x^2-4)=3x^3-12x\)✓
The habit that matters most
"Factorise fully" means keep going until nothing inside any bracket can be factorised any further. After every step, look at what's left behind and ask: common factor? Difference of squares? Trinomial? Only stop when the answer to all three is no.
Quick Check

Factorise \(3x^2+5x-2\).

Factorise \(27x^3-1\).

Simplifying Algebraic Fractions

Factorise fully first — then cancel common factors, never individual terms.

Just like a numeric fraction
You already do this with plain numbers: to simplify \(\dfrac69\), write both as a product with a shared factor, \(\dfrac{2\times3}{3\times3}\), then cancel the shared factor \(3\) to get \(\dfrac23\). Algebraic fractions use the exact same idea — the only difference is the shared factor is an expression instead of a number.
Method
  • Factorise the numerator and denominator completely.
  • Cancel any factor that appears in both (never cancel a single term inside an unfactorised expression).
  • State the value(s) of the variable that would make the original denominator zero — these are excluded.
Common mistake
You can only cancel factors (things multiplied), never terms (things added or subtracted). \(\dfrac{x+2}{x}\) does not simplify to \(1+2\) or to \(2\) — \(x\) is a term inside the numerator's sum, not a common factor of the whole numerator.
Worked example — single common factor Level 1

Simplify fully: \(\dfrac{3x+6}{x+2}\)

Show solution
  1. 1Factorise the numerator: \(3x+6=3(x+2)\)
  2. 2\(\dfrac{3(x+2)}{x+2}\) — cancel the common factor \((x+2)\), valid provided \(x\ne-2\)
  3. 3\(\boxed{3,\quad x\ne-2}\)
Worked example Level 2

Simplify fully: \(\dfrac{2x^2-8}{x^2-2x}\)

Show solution
  1. 1Factorise the numerator (common factor, then difference of two squares): \(2x^2-8=2(x^2-4)=2(x-2)(x+2)\)
  2. 2Factorise the denominator (common factor): \(x^2-2x=x(x-2)\)
  3. 3\(\dfrac{2(x-2)(x+2)}{x(x-2)}\) — cancel the common factor \((x-2)\), valid provided \(x\ne2\)
  4. 4\(\boxed{\dfrac{2(x+2)}{x},\quad x\ne0,2}\)
Worked example — cube denominator Level 3

Simplify fully: \(\dfrac{x^2-2x-8}{x^3-64}\)

Show solution
  1. 1Factorise the numerator (a trinomial): \(x^2-2x-8=(x-4)(x+2)\).
  2. 2Factorise the denominator (a difference of two cubes, \(x^3-4^3\)): \(x^3-64=(x-4)(x^2+4x+16)\).
  3. 3\(\dfrac{(x-4)(x+2)}{(x-4)(x^2+4x+16)}\) — cancel the common factor \((x-4)\), valid provided \(x\ne4\).
  4. 4\(\boxed{\dfrac{x+2}{x^2+4x+16},\quad x\ne4}\)
Quick Check

Simplify \(\dfrac{x^2-9}{x^2+3x}\).

Why can't \(\dfrac{x+5}{x}\) be simplified by cancelling the \(x\)'s?

Laws of Exponents

Revision from Grade 9, plus zero and negative exponents.

\(2^3\)\(2^2\)\(2^1\)\(2^0\)\(2^{-1}\)\(2^{-2}\)
\(8\)\(4\)\(2\)\(1\)\(0{,}5\)\(0{,}25\)
Spot the pattern first
Every time the exponent drops by 1, the value halves. Continuing that pattern from \(2^1=2\) forces \(2^0=1\) (half of 2), and continuing past that forces \(2^{-1}=\dfrac12\) (half of 1) — the rules below aren't arbitrary, they're exactly what keeps this pattern consistent.
The laws (\(x,y>0\), \(m,n\in\mathbb{Z}\))
\[x^m\cdot x^n=x^{m+n}\qquad x^m\div x^n=x^{m-n}\qquad(x^m)^n=x^{mn}\qquad x^m\cdot y^m=(xy)^m\]\[x^0=1\ (x\ne0)\qquad x^{-n}=\dfrac{1}{x^n}\ (x\ne0)\]
Reading a negative exponent
A negative exponent means "reciprocal," not "negative number": \(2^{-3}=\dfrac{1}{2^3}=\dfrac18\), which is a small positive fraction — never \(-8\).
Worked example — one law at a time Level 1

Simplify: \(x^3\cdot x^4\)

Show solution
  1. 1Same base, multiplying — add the exponents: \(x^m\cdot x^n=x^{m+n}\)
  2. 2\(\boxed{x^7}\)
Worked example Level 2

Simplify fully: \((2x^3y^2)^2\times(3xy^3)\)

Show solution
  1. 1Apply the power of a product to the bracket first: \((2x^3y^2)^2=2^2x^{3\times2}y^{2\times2}=4x^6y^4\).
  2. 2Multiply by the remaining factor, combining like bases with \(x^m\cdot x^n=x^{m+n}\): \(4x^6y^4\times3xy^3=(4\times3)\,x^{6+1}y^{4+3}\)
  3. 3\(\boxed{12x^7y^7}\)
Worked example — negative exponents Level 2

Simplify fully, leaving no negative exponents: \(\dfrac{x^{-2}\cdot x^5}{x^{-1}}\)

Show solution
  1. 1Combine the numerator: \(x^{-2}\cdot x^5=x^{-2+5}=x^3\).
  2. 2Divide by \(x^{-1}\), subtracting exponents: \(x^3\div x^{-1}=x^{3-(-1)}=x^4\).
  3. 3\(\boxed{x^4}\)

Solving Exponential Equations

Get both sides to the same base, then equate the exponents.

Method
  • Write every term as a power of the same base.
  • Once the bases match, the equation is true only if the exponents are equal — set them equal and solve.
Why this works
If \(b^m=b^n\) for the same base \(b>0,\,b\ne1\), then \(m=n\) — an exponential function never gives the same output twice for different inputs, so matching bases forces matching exponents.
Worked example — exponent already alone Level 1

Solve for \(x\): \(2^x=16\)

Show solution
  1. 1Write \(16\) as a power of \(2\): \(16=2^4\).
  2. 2\(2^x=2^4\) — same base, so equate the exponents: \(\boxed{x=4}\)
Worked example Level 2

Solve for \(x\): \(3^{x+1}=81\)

Show solution
  1. 1Write \(81\) as a power of \(3\): \(81=3^4\).
  2. 2\(3^{x+1}=3^4\) — same base, so equate the exponents: \(x+1=4\)
  3. 3\(\boxed{x=3}\)
Worked example — different starting bases Level 3

Solve for \(x\): \(9^x=27\)

Show solution
  1. 1Neither \(9\) nor \(27\) is a power of the other directly, but both are powers of \(3\): \(9=3^2\), \(27=3^3\).
  2. 2Rewrite: \((3^2)^x=3^3\ \Rightarrow\ 3^{2x}=3^3\)
  3. 3Equate exponents: \(2x=3\)
  4. 4\(\boxed{x=1{,}5}\)
Quick Check

Simplify: \(\dfrac{x^7}{x^{-2}}\)

Solve for \(x\): \(2^{2x-1}=8\)

Linear and Quadratic Equations

Linear is revision; quadratic by factorisation is new this year.

Linear (revision)

One unknown, no squared terms. Collect like terms on each side, then isolate the variable, e.g. \(3x-5=2x+7\Rightarrow x=12\).

Quadratic by factorisation

Set the equation to zero, factorise the resulting trinomial, then use the fact that if a product is zero, at least one factor must be zero.

Why "product = 0" forces a factor to be 0
Try a concrete check: \(3\times4=12\), and no matter what two nonzero numbers you pick, their product is never \(0\) (e.g. \(3\times4=12\ne0\)). So if \((x-1)(x-2)=0\) really is true, one of those brackets genuinely must equal \(0\) — there's no other way for the product to vanish.
Never divide by the variable
Given \(x^2=5x\), do not divide both sides by \(x\) — that silently discards the solution \(x=0\). Instead move everything to one side and factorise: \(x^2-5x=0\Rightarrow x(x-5)=0\Rightarrow x=0\) or \(x=5\).
Worked example — linear Level 1

Solve for \(x\): \(2x+7=3x-4\)

Show solution
  1. 1Collect the \(x\)-terms on one side: subtract \(2x\) from both sides: \(7=x-4\)
  2. 2Add \(4\) to both sides: \(\boxed{x=11}\)
  3. 3Check: \(2(11)+7=29\) and \(3(11)-4=29\)✓
Worked example — quadratic, leading coefficient 1 Level 1

Solve for \(x\): \(x^2-5x+6=0\)

Show solution
  1. 1Already set to zero. Find two numbers that multiply to \(6\) and add to \(-5\): \(-2\) and \(-3\)
  2. 2Factorise: \((x-2)(x-3)=0\)
  3. 3Zero-product law: \(x-2=0\) or \(x-3=0\)
  4. 4\(\boxed{x=2\text{ or }x=3}\)
x=2x=3

The graph of \(y=x^2-5x+6\) crosses the x-axis exactly at the two solutions.

Worked example — leading coefficient not 1 Level 2

Solve for \(m\): \(2m^2-m=1\)

Show solution
  1. 1Move every term to one side so the equation equals zero: \(2m^2-m-1=0\)
  2. 2Factorise: \(2\times(-1)=-2\); two numbers multiplying to \(-2\) and adding to \(-1\) are \(-2\) and \(1\): \(2m^2-2m+m-1=2m(m-1)+1(m-1)=(2m+1)(m-1)\)
  3. 3Set each factor to zero: \(2m+1=0\) or \(m-1=0\)
  4. 4\(\boxed{m=-0{,}5\text{ or }m=1}\)
Worked example Level 2

Solve for \(x\): \(3x^2+2x=8\)

Show solution
  1. 1Move every term to one side: \(3x^2+2x-8=0\)
  2. 2\(3\times(-8)=-24\); two numbers multiplying to \(-24\) and adding to \(2\) are \(6\) and \(-4\): \(3x^2+6x-4x-8=3x(x+2)-4(x+2)=(3x-4)(x+2)\)
  3. 3Set each factor to zero: \(3x-4=0\) or \(x+2=0\)
  4. 4\(\boxed{x=\dfrac43\text{ or }x=-2}\)

Simultaneous Linear Equations

Two equations, two unknowns — solve by substitution or elimination.

Worked example — already isolated Level 1

Solve simultaneously: \(y=2x-3\) and \(3x+y=12\)

Show solution
  1. 1The first equation already has \(y\) as the subject, so substitute it straight into the second equation: \(3x+(2x-3)=12\)
  2. 2\(5x-3=12\Rightarrow5x=15\Rightarrow x=3\)
  3. 3Substitute back: \(y=2(3)-3=3\)
  4. 4\(\boxed{x=3,\ y=3}\) — check: \(3(3)+3=12\)✓
Substitution method
Make one variable the subject of whichever equation makes that easiest, substitute it into the other equation, solve for the remaining variable, then substitute back to find the first one. Always check both original equations at the end.
Worked example — isolate first Level 2

Solve simultaneously: \(2x+y=7\) and \(x-y=2\)

Show solution
  1. 1Neither equation has a variable already isolated, so choose one: from the second equation, make \(x\) the subject: \(x=y+2\)
  2. 2Substitute into the first equation: \(2(y+2)+y=7\)
  3. 3\(2y+4+y=7\Rightarrow3y=3\Rightarrow y=1\)
  4. 4Substitute back: \(x=1+2=3\)
  5. 5\(\boxed{x=3,\ y=1}\) — check: \(2(3)+1=7\)✓ and \(3-1=2\)✓
Two lines intersecting at a single pointThe lines 2x+y=7 and x-y=2 crossing at the point (3,1), which is the solution to the system.(3,1)2x+y=7x−y=2
Each equation is a straight line. The two lines cross at exactly one point, \((3,1)\) — that shared point is the only \((x,y)\) pair satisfying both equations at once, which is exactly what "solving simultaneously" means.
What if the lines don't cross like this?
Two parallel lines (same gradient, different intercepts) never meet — the system has no solution. Two lines that are actually the same line overlap everywhere — the system has infinitely many solutions. Every pair of straight lines falls into exactly one of these three cases.

Simultaneous Equations — Elimination

Add or subtract the two equations directly to cancel one variable.

Worked example — elimination Level 1

Solve simultaneously: \(2x+3y=12\) and \(x-3y=-3\)

Show solution
  1. 1Look at the \(y\)-coefficients: \(+3y\) and \(-3y\) are already opposites, so adding the two equations will cancel \(y\) completely.
  2. 2Add the equations, left side to left side and right side to right side: \((2x+3y)+(x-3y)=12+(-3)\)
  3. 3\(3x=9\Rightarrow x=3\)
  4. 4Substitute \(x=3\) into either original equation, e.g. \(x-3y=-3\): \(3-3y=-3\Rightarrow-3y=-6\Rightarrow y=2\)
  5. 5\(\boxed{x=3,\ y=2}\) — check: \(2(3)+3(2)=6+6=12\)✓ and \(3-3(2)=3-6=-3\)✓
Worked example — elimination Level 2

Solve simultaneously: \(4x+y=11\) and \(2x-y=1\)

Show solution
  1. 1The \(y\)-coefficients \(+y\) and \(-y\) are already opposite, so adding cancels \(y\).
  2. 2Add the equations: \((4x+y)+(2x-y)=11+1\)
  3. 3\(6x=12\Rightarrow x=2\)
  4. 4Substitute \(x=2\) into \(2x-y=1\): \(4-y=1\Rightarrow y=3\)
  5. 5\(\boxed{x=2,\ y=3}\) — check: \(4(2)+3=11\)✓ and \(2(2)-3=1\)✓
Worked example — multiply first Level 3

Solve simultaneously: \(3x+2y=16\) and \(x+y=7\)

Show solution
  1. 1No coefficient matches or is already opposite, so multiply the second equation by \(2\) to line up the \(y\)-coefficients: \(2x+2y=14\)
  2. 2Subtract this from the first equation: \((3x+2y)-(2x+2y)=16-14\)
  3. 3\(x=2\)
  4. 4Substitute \(x=2\) into \(x+y=7\): \(2+y=7\Rightarrow y=5\)
  5. 5\(\boxed{x=2,\ y=5}\) — check: \(3(2)+2(5)=6+10=16\)✓ and \(2+5=7\)✓
Elimination method
Elimination works best when one variable's coefficients are already equal (add or subtract to cancel it) or already opposite (add to cancel it). If neither is true, first multiply one or both equations by a constant so that one variable's coefficients line up — substitution is often quicker when nothing lines up neatly.

Literal Equations

Changing the subject of a formula — the same algebra, a different goal.

Worked example — division only Level 1

Solve for \(h\) in terms of \(V\), \(\pi\) and \(r\): \(V=\pi r^2h\)

Show solution
  1. 1\(h\) is multiplied by \(\pi r^2\), so divide both sides by \(\pi r^2\) to isolate it: \(\boxed{h=\dfrac{V}{\pi r^2}}\)
Worked example — with a square root Level 2

Solve for \(r\) in terms of \(V\), \(\pi\) and \(h\): \(V=\pi r^2h\)

Show solution
  1. 1Isolate \(r^2\) by dividing both sides by \(\pi h\): \(r^2=\dfrac{V}{\pi h}\)
  2. 2Take the square root of both sides: \(r=\pm\sqrt{\dfrac{V}{\pi h}}\)
  3. 3A radius can't be negative, so reject the negative root: \(\boxed{r=\sqrt{\dfrac{V}{\pi h}}}\)
It's the same algebra
A literal equation just has more than one letter. The steps — isolate the variable you want, undo operations in reverse order — are identical to any other equation; the only difference is the "answer" is an expression in other letters, not a number.
Quick Check

Solve for x: \(x^2=7x\)

Make \(b\) the subject of \(A=\dfrac12bh\).

Solve simultaneously using elimination: \(3x+2y=16\) and \(x-2y=0\). What is \(x\)?

Solve for x: \(3x^2+5x-2=0\)

Linear Inequalities

Solved exactly like equations — with one crucial extra rule.

Why the sign flips — check it concretely
\(2<5\) is true. Multiply both sides by \(-1\): naively you might expect \(-2<-5\), but check a number line — \(-2\) sits to the right of \(-5\), so actually \(-2>-5\). Multiplying by a negative reverses the order of every number on the line, so the inequality sign must flip to stay true.
The one rule that's different

Multiplying or dividing both sides of an inequality by a negative number flips the direction of the inequality sign. Adding, subtracting, or multiplying/dividing by a positive number never flips it.

Interval notationMeaning
\((a,b)\)\(a
\([a,b]\)\(a\le x\le b\), endpoints included
\([a,b)\)\(a\le x
\((-\infty,a)\)\(x
\((a,\infty)\)\(x>a\), unbounded on the right
Worked example — one step Level 1

Solve for \(x\): \(3x+2>11\)

Show solution
  1. 1Subtract \(2\) from both sides (a positive-number step, no flip): \(3x>9\)
  2. 2Divide both sides by \(3\) (still positive, still no flip): \(\boxed{x>3}\)
Worked example — one step, with a sign flip Level 2

Solve for \(x\): \(-3x+2>11\)

Show solution
  1. 1Subtract \(2\) from both sides (positive step, no flip): \(-3x>9\)
  2. 2Divide both sides by \(-3\) — a negative number, so flip the sign: \(\boxed{x<-3}\)
Worked example — compound, with a sign flip Level 3

Solve for \(x\) and show the solution graphically: \(-1\le2-3x\le8\)

Show solution
  1. 1Subtract 2 from all three parts: \(-1-2\le-3x\le8-2\ \Rightarrow\ -3\le-3x\le6\)
  2. 2Divide all three parts by \(-3\) — a negative number, so flip both inequality signs: \(1\ge x\ge-2\)
  3. 3Rewrite from smallest to largest: \(\boxed{-2\le x\le1}\)
A number line showing the solution -2 less than or equal to x less than or equal to 1A number line from -3 to 2 with a thick segment between -2 and 1, and closed (filled) circles at both -2 and 1, showing the solution set.−3−2−1012
Closed circles at both \(-2\) and \(1\) show the endpoints are included (\(\le\), not \(<\)), matching the solid line joining them.

Word Problem: Putting It Together

Translate words into an equation, then solve and check it makes sense.

Worked example — single step Level 1

A number, increased by 5, gives 12. Find the number.

Show solution
  1. 1Let the number be \(x\). "Increased by 5" is \(x+5\); "gives 12" means \(=12\)
  2. 2\(x+5=12\)
  3. 3\(\boxed{x=7}\) — check: \(7+5=12\)✓
Worked example — linear Level 2

Twice a number, decreased by 7, is 15. Find the number.

Show solution
  1. 1Let the number be \(x\). "Twice a number" is \(2x\); "decreased by 7" is \(2x-7\); "is 15" means \(=15\)
  2. 2\(2x-7=15\)
  3. 3\(2x=22\)
  4. 4\(\boxed{x=11}\) — check: twice \(11\) is \(22\), decreased by \(7\) is \(15\)✓
Translating words to symbols
"Sum"/"more than"/"increased by" → \(+\). "Difference"/"less than"/"decreased by" → \(-\). "Product"/"twice"/"times" → \(\times\). "Is"/"gives"/"equals" → \(=\). Read the sentence piece by piece and translate each piece in order.
A rectangle labelled with width x and length x plus 3A rectangle with its width labelled x on the left side and its length labelled x plus 3 along the top, with the area 40 square centimetres written inside.x + 3xArea = 40 cm²
The length is 3 cm more than the width — call the width \(x\), so the length is \(x+3\).
Worked example Level 3

The length of a rectangle is 3 cm more than its width. If the area is \(40\text{ cm}^2\), determine the width.

Show solution
  1. 1Let the width be \(x\) cm, so the length is \((x+3)\) cm.
  2. 2Area = length × width: \(x(x+3)=40\)
  3. 3Expand and set to zero: \(x^2+3x-40=0\)
  4. 4Factorise: \((x+8)(x-5)=0\Rightarrow x=-8\) or \(x=5\)
  5. 5A length can't be negative, so reject \(x=-8\): \(\boxed{\text{width}=5\text{ cm}}\) (length \(=8\) cm, area \(=5\times8=40\)✓)
Worked example — consecutive integers Level 3

The product of two consecutive positive integers is 72. Find the two integers.

Show solution
  1. 1Let the smaller integer be \(n\), so the next consecutive integer is \((n+1)\).
  2. 2Product \(=72\): \(n(n+1)=72\)
  3. 3Expand and set to zero: \(n^2+n-72=0\)
  4. 4Factorise: \((n+9)(n-8)=0\Rightarrow n=-9\) or \(n=8\)
  5. 5The problem asks for positive integers, so reject \(n=-9\): \(\boxed{8\text{ and }9}\) (check: \(8\times9=72\)✓)
Always check for sense
A quadratic equation usually gives two solutions, but in a real-world context (lengths, ages, quantities) a negative answer is often impossible — reject it and state only the value that makes sense.
Quick Check

Solve for x and show graphically: \(2x-4<6\)

Two numbers have a sum of 15 and a difference of 3 (let \(x\) be the larger number: \(x+y=15,\ x-y=3\)). What is the larger number, \(x\)?

Grade 10 Mastery Sprint

Do all four without notes. Then check and correct in a different colour.

01 Factorise: \(3x^2+7x+2\)
Answer: \((3x+1)(x+2)\)
02 Simplify: \(\dfrac{x^2-9}{x^2-3x}\)
Answer: \(\dfrac{x+3}{x},\ x\ne0,3\)
03 Solve: \(2^{x+1}=16\)
Answer: \(16=2^4\Rightarrow x+1=4\Rightarrow x=3\)
04 Solve: \(-2x+5\le11\)
Answer: \(-2x\le6\Rightarrow x\ge-3\) (flip the sign).

Exam Strategy

Checklist before you submit
  • Factorised fully? Check for a common factor first, every time.
  • Quadratic equation: is it set to zero before factorising?
  • Exponential equation: are both sides written with the same base before equating exponents?
  • Inequality: did you flip the sign when multiplying/dividing by a negative?
  • Word problem: does your final answer actually make sense in context?
Exam mantra
Factorise first. Set to zero. Check your answer makes sense. Almost every algebra method on this page starts with getting the expression into its simplest, fully factorised form.
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What CAPS Expects You to Know

The Grade 10 Algebra & Equations knowledge statement this page is built from.

  1. 1

    Understand that real numbers can be rational or irrational; establish between which two integers a given simple surd lies; round real numbers to an appropriate degree of accuracy.

  2. 2

    Multiply a binomial by a trinomial; factorise expressions including trinomials, grouping in pairs, and the sum and difference of two cubes.

  3. 3

    Simplify algebraic fractions using factorisation, limited to denominators involving the sum or difference of two cubes.

  4. 4

    Revise the laws of exponents learnt in Grade 9, extended to include \(x^0=1\) and negative exponents; use the laws to simplify expressions and solve equations.

  5. 5

    Revise the solution of linear equations; solve quadratic equations by factorisation; solve simultaneous linear equations in two unknowns.

  6. 6

    Solve word problems involving linear, quadratic, or simultaneous linear equations; solve literal equations (changing the subject of a formula).

  7. 7

    Solve linear inequalities and show the solution graphically, using interval notation.

How to Use This Lesson

A few practical notes before you start.

  • Always check for a common factor before applying any other factorising method.
  • Memorise the sum/difference of two cubes formulas — there's no shortcut derivation at this level.
  • Never divide an equation by a variable — move everything to one side and factorise instead, or you'll silently lose a solution.
  • For inequalities, remember the one rule that's different from equations: flip the sign when multiplying or dividing by a negative.
  • Try each worked example yourself before pressing “Show solution.”

Learn It in Short Videos

Four free, independent videos — not made by Equation Station SA.

Factorising Trinomials

Splitting the middle term when the leading coefficient isn't 1.

Khan Academy · Factoring trinomials with a non-1 leading coefficient by grouping

Sum & Difference of Cubes

Applying the two new cube-factorisation formulas.

Khan Academy · Factoring sum of cubes

Zero & Negative Exponents

Why any nonzero base to the power 0 is 1, and what a negative exponent means.

Khan Academy · Zero, negative, and fractional exponents

Quadratic Equations

Solving by factorisation, step by step.

Khan Academy · Solving quadratics by factoring

Algebra Study Path

Learn → practise → visualise → stretch.

Step 1 · Learn

Siyavula: Grade 10 Mathematics

Learn the CAPS sequence step by step.

Open Siyavula
Step 2 · Practise

Khan Academy: Algebra Basics

Practise expressions, equations and factorising.

Start Algebra Basics
Step 3 · Visualise

GeoGebra Graphing Calculator

Make equations visible on a graph.

Open GeoGebra
Step 4 · Extend

DBE CAPS Learner Books

Use the official Grade 10 learner book for extra CAPS examples.

Open DBE books

Common Exam Mistakes

Avoid these errors. They cost marks every year.

Cancelling terms, not factors

\(\dfrac{x+5}{x}\) does not simplify by cancelling the \(x\)'s — only common factors of the whole numerator and denominator can be cancelled.

Dividing an equation by the variable

Dividing \(x^2=5x\) by \(x\) silently discards the solution \(x=0\). Move everything to one side and factorise instead.

Forgetting to flip an inequality sign

Only multiplying or dividing by a negative number flips the sign — it never happens for addition, subtraction, or positive multiplication/division.

Mixing up the two cube formulas' signs

The first bracket matches the original sign; the second bracket's middle term always has the opposite sign, and its outer terms are always positive.

Equating exponents with different bases

You can only equate exponents once both sides share the exact same base — rewrite every term as a power of one common base first.

Not rejecting an impossible answer

In a word problem about lengths or quantities, a negative solution from a quadratic is usually impossible — state only the value that makes sense.

This Topic in Other Grades

Algebra & Equations builds across all three FET grades — preview the other levels here.

Grade 11

Rational exponents, surd equations, completing the square, the quadratic formula, quadratic inequalities, and nature of roots

Study this

Grade 12

Combining every method into full NSC-style riders

Study this

Practise This Topic

You've done the notes above — now practise and test yourself.

BANK
Mastery Bank
Grade 10 Algebra & Equations Mastery Bank

Exam-style Grade 10 questions arranged by level, combining original practice with clearly identified paper-and-memo matches.

Step 1 • Practise • Cognitive Levels
Open Mastery Bank
TEST
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Algebra & Equations Grade 10 Test Your Knowledge

Auto-marked quiz with instant feedback, explanations and a complete answer review.

Step 2 • Test Yourself • Auto-Marked + Review
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Student Questions

Frequently Asked Questions

Straight answers to common Grade 10 CAPS questions about algebra and equations.

What is Grade 10 Algebra & Equations about?

Grade 10 Algebra covers real numbers and simple surds, factorising (revision plus new types: trinomials, grouping in pairs, and the sum/difference of two cubes), simplifying algebraic fractions using factorisation, the laws of exponents (including zero and negative exponents), solving exponential equations, revising linear equations, solving quadratic equations by factorisation, simultaneous linear equations, literal equations (changing the subject of a formula), and linear inequalities with interval notation and a graphical solution.

Do I need to memorise the sum and difference of two cubes formulas?

Yes — \(a^3+b^3=(a+b)(a^2-ab+b^2)\) and \(a^3-b^3=(a-b)(a^2+ab+b^2)\) are new Grade 10 formulas with no shortcut derivation expected at this level. Practise spotting a sum or difference of two perfect cubes before applying the formula.

What is a literal equation?

A literal equation has more than one letter (variable), and you are asked to make one of them the subject — for example, solving \(V=\pi r^2h\) for \(r\). The algebra steps are identical to a normal equation; only the goal (isolate a chosen letter) is different.

Why does the inequality sign flip in some problems?

Multiplying or dividing both sides of an inequality by a negative number flips the direction of the inequality sign. It never happens when you multiply or divide by a positive number.

Which resource should I open first for this topic?

Everything you need to learn the topic is on this page already. Once you've been through the notes above, work through the Mastery Bank, then finish with the Test Your Knowledge quiz as a self-check.

What mistakes should I avoid?

Common mistakes include forgetting to check for a common factor before using a formula, forgetting that a base must match before comparing exponents in an equation, losing a solution by dividing an equation by a variable instead of factorising, and forgetting to flip an inequality sign when multiplying or dividing by a negative number.