Maths · Ireland · NCCA Primary Mathematics Curriculum (2023) and Junior Cycle Mathematics specification (2018)

Maths in Ireland, Fourth Class to Third Year, and where our lectures fit

Children in the Republic of Ireland follow the Primary Mathematics Curriculum (2023) through eight years of primary school, reaching Fourth to Sixth Class at about 9–12. Post-primary starts at about 12 with the three-year Junior Cycle, a common Mathematics specification ending in a State Examinations Commission (SEC) paper at Higher or Ordinary level in June of Third Year (about age 15). After an optional Transition Year, students take the two-year Leaving Certificate at Higher, Ordinary or Foundation level; this is where calculus starts.

  • 9–15ages covered
  • 6areas of maths
  • 12lectures built
  • 52practice problems

School years and ages

Ireland

There is no single national cut-off date. Children must be in school by age 6 and most start junior infants at 4–5, so ages in a class vary by about a year. The school year runs from late August or September to June. After Third Year comes Transition Year (optional in most schools, about 15–16), then Fifth and Sixth Year for the Leaving Certificate (about 16–18).

  1. Primary (Stage 3)Fourth ClassAge 9–10Mandatory standardised test in maths and English reading (May/June); results go to parents and aggregates to the Department
  2. Primary (Stage 4)Fifth ClassAge 10–11No national test
  3. Primary (Stage 4)Sixth ClassAge 11–12Mandatory standardised test in maths and English reading; end of primary
  4. Post-primary: Junior CycleFirst YearAge 12–13No national test; common specification; all students in mixed or school-set classes
  5. Post-primary: Junior CycleSecond YearAge 13–14Classroom-Based Assessment 1 (Mathematical investigation) normally completed at end of year
  6. Post-primary: Junior CycleThird YearAge 14–15CBA 2 (Statistical investigation) in first term; SEC final exam in June at Higher or Ordinary level; Junior Cycle Profile of Achievement (JCPA)

What this page describes One national curriculum for the whole state, with no regional variation. Primary: the Primary Mathematics Curriculum (NCCA, 2023). Teachers became familiar with it in 2023/24, and all classes from junior infants to sixth class began implementing it from September 2024 (Circular 0039/2023). Its learning outcomes cover two-year stages (Stage 3 = Third and Fourth Class, Stage 4 = Fifth and Sixth Class), so the split between individual classes below is indicative. Post-primary: Junior Cycle Mathematics specification (2018; first examined 2021). Its learning outcomes cover the whole three years and schools choose the order, so the split between First Year and Second–Third Year is typical practice, not prescribed. The specification is common to all students, and the extra Higher-level outcomes are printed in bold. Leaving Certificate: the syllabus for examination from 2015. A revised Leaving Certificate Mathematics specification is due in schools from September 2027 (NCCA Tranche 3).

What is taught, by area

Select a cell

Three primary columns are used because the primary curriculum has no tracks. First Year is one common column. Second and Third Year are split by the exam level students work towards: the Ordinary column holds the outcomes common to all students, and the Higher column adds the outcomes the specification prints in bold for Higher level. The exam level is not fixed until Third Year (and can be changed up to the exam), although many schools set classes by level from Second Year. The bottom rows place each Ludumina lecture at the stage it teaches.

What pupils learn in each area of maths, by stage. Select a cell for detail.
4th ClassAge 9–105th ClassAge 10–116th ClassAge 11–121st YearAge 12–132nd–3rd Year OrdinaryAge 13–152nd–3rd Year HigherAge 13–15
Number
Algebra
Ratio and proportion
Geometry and measures
Statistics and probability
Calculus
Algebra lecturesM01.0M01.1M01.5M01.2M01.4
Geometry lecturesM02.1M02.0M02.2M02.3M02.6
  • Not taught
  • Informal
  • Introduced
  • Core
  • Advanced
  • Ludumina lecture

Taught later here · Leaving Certificate · Fifth–Sixth Year, age 16–18 M01.3M01.6

Select any cell for the full list of what is taught

The exam at 16

Junior Cycle Mathematics final examination (SEC), reported on the Junior Cycle Profile of Achievement (JCPA)
  • Set and marked by the State Examinations Commission (SEC) at two levels, Higher and Ordinary, from one common specification. The final written exam takes 2 hours in June of Third Year: the 2025 Higher level paper was one 270-mark booklet of 13 compulsory questions.
  • Calculators are allowed (candidates write the make and model on the paper). The SEC supplies the Formulae and Tables booklet.
  • Grade descriptors from 2025: Distinction 85–100%, Higher Merit 70–84%, Merit 55–69%, Achieved 40–54%, Partially Achieved 20–39%, Not Graded 0–19%. Before 2025 the bands were Distinction 90–100 and Higher Merit 75–89.
  • The specification includes two Classroom-Based Assessments marked by the teacher: CBA 1, a mathematical investigation (end of Second Year), and CBA 2, a statistical investigation (Third Year). It also includes an SEC-marked Assessment Task worth 10%. In 2025 and 2026 each student needs only one CBA per subject, the Assessment Task is not examined, and the grade is based on the exam paper alone.
  • Leaving Certificate (after Fifth and Sixth Year): Mathematics at Higher, Ordinary and Foundation level. Higher and Ordinary have two papers (Paper 1 and Paper 2), each with Section A (concepts and skills) and Section B (contexts and applications). Foundation has one paper.

Compared with England

  • Junior Cycle is one common specification with the exam level (Higher or Ordinary) chosen late, unlike England's GCSE tiers. The exam is taken at about 15, a year earlier than GCSE.
  • Most algebraic-fraction work that GCSE Higher expects by 16 is Leaving Certificate content here (about 16–18). Only adding a/(bx + c) is in Junior Cycle Higher.
  • Geometry is axiomatic: students learn numbered axioms and theorems from Geometry for Post-Primary School Mathematics and cite them as reasons. Circle theorems are limited to the angle at the centre and its corollaries.
  • Sets and Venn-diagram set operations, and financial maths (VAT, income tax, net pay, compound interest), are explicit Junior Cycle outcomes.
  • Calculus is not in Junior Cycle. It starts in the Leaving Certificate (Strand 5.2): differentiation at Ordinary level, with first principles and integration at Higher level. A revised Leaving Certificate specification is due from September 2027.

How every lecture runs

Same shape in both courses
  1. HookA real problem on the set: a pony pen, a pediment, a column too tall to measure.
  2. ExplanationFour to seven short narrated lines, captioned, stepped with Next and Back.
  3. DemonstrationThe lecturer solves a live problem on the board. Every move is checked by the maths engine.
  4. ExerciseThe learner solves the same problem, with move-by-move guidance.
  5. Your problemA new problem of the same kind, with one hint.
  6. Three storiesReal-life word problems. Choose the fact (geometry), choose the equation, then solve. One hint each.
  7. Report cardWhat went well and what to practise, for the learner and the adult.
  8. RewardA spacewalk or a flight: earned by finishing all three stories.

The lecturer never marks an answer by eye. Each move on the board is checked by the maths engine, so a learner cannot reach the answer by a wrong step. Lectures are in English.

M01 · Algebra

Math Orbital Lab · with René Descartes

Seven lectures that take a learner from the words of algebra to solving equations with algebraic fractions: the manipulation skills that the hardest exam papers at 16 use in almost every question.

Set on a glass lecture deck on a space station in orbit. The reward for finishing a lecture is a three-minute spacewalk in Descartes' spacesuit.

  • 7lectures built
  • 32practice problems
  • 21real-life stories

Course outcome: by the end, a learner can

  • Tell a term from a factor, and an expression from an equation, an identity and a formula
  • Expand brackets, collect like terms, and cancel factors (never terms)
  • Solve linear equations containing fractions by doing the same to both sides
  • Factorise quadratics, including the difference of two squares and ax² + bx + c
  • Simplify, multiply, divide, add and subtract algebraic fractions, noting excluded values
  • Solve quadratic equations by factorising and reject answers that make no sense in context
  • Solve equations with algebraic fractions and check every answer against the excluded values
  • Turn a real-life story into an equation and solve it

M01.0 · Lecture 0 · René Descartes

Terms, factors and the equals sign

  • Stage1st Year
  • Typical age12–13 · First Year
  • LevelCommon (all students)
  • Exam levelAchieved/Merit on Ordinary level
  • CurriculumJC AF.2a–c, AF.3a.I, AF.3b.I, U.3 (letter-symbols start at primary PMC Stage 4)
  • StatusBuilt

Outcome

Terms are added, factors are multiplied. Expressions are simplified, equations are solved. Cancel factors, never terms.

Formal expressions, expanding one bracket and equality as a relationship are core Junior Cycle work, usually in First Year; primary gives only informal letter-symbol work.

Needs first Letters standing for numbers (end of primary)

What happens in this lecture

Hook

Look at this pen for a pony. Its sides are written with x. To find the length of fence, we must build an expression and make it simpler.

But first, the words of algebra. Every other lecture uses them.

Explanation

  1. Algebra has its own words. A term is a piece that is added or taken away. In three x plus six, the terms are three x and six.
  2. A factor is a piece that is multiplied. In three x, the factors are three and x. In three times x plus two, the factors are three and the bracket.
  3. An expression is terms put together, like three x plus six. It has no equals sign. You can simplify an expression, but you cannot solve it.
  4. An equation has an equals sign, like three x plus six equals twelve. It is true only for some values of x, and solving finds them.
  5. An identity is true for every value of x, like two times x plus three, equals two x plus six. A formula links quantities, like area equals length times width.
  6. And the golden rule for fractions: you may cancel factors, never terms. Two x plus six, over two, is not x plus six. The two divides every term, so it is x plus three.

Demonstration

  1. Let us simplify an expression: three times x plus two, plus two x. On the left is the question, and it stays the same. I make the copy after the equals sign simpler.
  2. The three is a factor of the whole bracket, so it multiplies every term inside: three x, and three times two.
  3. Three times two is six.
  4. Three x and two x are like terms: both are a number times x. Together they make five x. Six has no x, so it stays on its own.
  5. Five x plus six. The like terms are joined and the bracket is gone, so it is as simple as it gets. It is an identity: true for every x.

Exercise

Your turn. Tap the three to multiply out the bracket. Then join the like terms.

Well done. Three times x plus two, plus two x, is five x plus six.

Practice problems for M01.0
#ProblemEquationAnswer
1Your problemSimplify it yourself

Now one on your own. Two times x plus four, plus three x. Make it as simple as you can. You have one hint.

2(x + 4) + 3xIt is five x plus eight.
2Real-life storyPony pen

A rectangular pen is x + 3 metres long and 2 metres wide. Find an expression for the length of fence all the way round.

2(x + 3) + 4The fence is two x plus ten metres long.
3Real-life storySharing

Two friends share 2x + 6 sweets equally. How many does each friend get?

(2x + 6)/2Each friend gets x plus three sweets. The two divides both terms: two x over two is x, and six over two is three.
4Real-life storyFamily ages

Sam is x years old. His sister is 3 years older. Their dad is 4 times Sam's age. Find their total age.

x + x + 3 + 4xTogether they are six x plus three years old.

M01.1 · Lecture 1 · René Descartes

Rearrange a fractional formula

  • Stage1st Year
  • Typical age12–14 · First–Second Year
  • LevelCommon (Ordinary and Higher)
  • Exam levelMerit on Ordinary level
  • CurriculumJC AF.4a (linear equations with coefficients in ℚ)
  • StatusBuilt

Outcome

Undo the fraction: multiply both sides by 3, take away the number next to x, and x stands alone.

Linear equations with fractional coefficients are common-level content, usually met in First Year and revisited in Second Year.

Needs first M01.0; inverse operations

What happens in this lecture

Hook

Every day we meet problems like this one. A taxi, a phone bill, a plant that grows. To solve them, we use equations.

But first, you need to know how to solve the equation. Let me show you how.

Explanation

  1. Today I am going to show you how to solve an equation with fractions. First I will explain. Then I will do it in front of you. Then you will do it on your own.
  2. Here is our equation: x divided by three, plus two, equals four.
  3. The equals sign means both sides are the same. If we change one side, we must change the other side in the same way.
  4. x over three means x is cut into three equal parts, and we have one part. Then two is added to it.
  5. First we get rid of the fraction. Multiply everything on both sides by three. Then x over three becomes just x.
  6. Next we get rid of the number next to x. Take it away from both sides. Now x is on its own.
  7. So the plan is: clear the fraction, then clear the extra number. Watch me do it once. Then you try.

Demonstration

  1. Let us start. x over three, plus two, equals four.
  2. I pick up a three. Every part of both sides gets multiplied by it.
  3. Three on the left side.
  4. And three on the right side. Both sides are three times bigger, so they are still equal.
  5. The three multiplies each part inside the bracket.
  6. Three divided by three is one. So x over three, times three, is just x.
  7. Three times two is six.
  8. Three times four is twelve. No more fraction. Now it says x plus six equals twelve.
  9. Now I take six away from both sides.
  10. Six take away six is zero. It is gone.
  11. Twelve take away six is six. So x is six.
  12. Let us check. Six divided by three is two. Two plus two is four. It works. Try moving the numbers yourself. If you get stuck, ask me a question.

Exercise

Your turn. Take the three from the tray and multiply both sides. Then take away the six. I will watch.

Well done. You found that x is six. Great work.

Practice problems for M01.1
#ProblemEquationAnswer
1Your problemSolve it yourself

Now a new equation, all on your own. x over four, plus three, equals seven. It works just like mine. You have one hint, so keep it for when you are really stuck.

x/4 + 3 = 7x is sixteen. Sixteen divided by four is four, and four plus three is seven. It checks out.
2Real-life storyTaxi fare

A taxi charges £3 to start, plus £1 for every 2 miles. Your ride cost £8. How many miles did you travel? Let x be the number of miles.

x/2 + 3 = 8Ten miles. Ten over two is five, and five plus three is eight pounds. You used the same method on a real problem. That is what the exam calls problem solving.
3Real-life storyPhone bill

A phone plan costs £5 a month, plus £1 for every 3 GB of data. This month's bill was £9. How many GB did you use? Let x be the number of GB.

x/3 + 5 = 9Twelve gigabytes. Twelve over three is four, and four plus five is nine pounds.
4Real-life storyGrowing plant

A plant is 4 cm tall. It grows 1 cm every 5 days. Now it is 7 cm tall. How many days have passed? Let x be the number of days.

x/5 + 4 = 7Fifteen days. Fifteen over five is three, and three plus four is seven centimetres.

M01.2 · Lecture 2 · René Descartes

Simplify an algebraic fraction by factorising

  • Stage2nd–3rd Year Higher
  • Typical age13–15 · Second–Third Year
  • LevelOrdinary for x² + bx + c and x² − a²; Higher for ax² + bx + c
  • Exam levelHigher level, Merit/Higher Merit
  • CurriculumJC AF.3c, AF.3d.IV–V (bold parts Higher only)
  • StatusBuilt

Outcome

Factorise the top of the fraction, cancel the bracket it shares with the bottom, and note which x is not allowed.

Factorising monic quadratics, difference of two squares and dividing a quadratic by a linear factor are common content. Non-monic factorising is Higher only, and excluded values are not named in the specification.

Needs first Factorising x² + bx + c

What happens in this lecture

Hook

Look at this problem. A rectangle's area and width are written with x. To find its length, we must divide one expression by another, and then make the answer simpler.

But first, let me show you how to simplify a fraction with letters in it.

Explanation

  1. This fraction has letters in it. On top: x squared plus five x plus six. On the bottom: x plus two. We will make it simpler.
  2. A fraction gets simpler when the same thing is on the top and on the bottom. But we cannot cross out a piece of a sum. First we turn the top into brackets.
  3. To do that, find two numbers that multiply to make six and add to make five. They are two and three. So the top is x plus two, times x plus three.
  4. Now x plus two is on the top and on the bottom. Anything divided by itself is one. So they cancel, and x plus three is left.
  5. One warning. If x was minus two, the bottom would be zero. You cannot divide by zero. So we say x is not allowed to be minus two.
  6. So: make brackets, cancel the match, and note the value that is not allowed. Watch me, then you try.

Demonstration

  1. Here is the fraction, written twice. On the left is the question, and it never changes. After the equals sign is a copy. I make the copy simpler, one step at a time.
  2. Now the brackets go back into the fraction, in place of the top.
  3. x plus two is on the top and the bottom. They cancel. And I note that x cannot be minus two.
  4. What is left is x plus three. So the fraction is equal to x plus three, for every x except minus two. If you get stuck, ask me.

Exercise

Your turn. Tap the top of the fraction to make brackets. Then drag the matching bracket onto the one below the line.

Well done. You made brackets and cancelled the match. The fraction is x plus three, and x may not be minus two.

Practice problems for M01.2
#ProblemEquationAnswer
1Your problemSimplify it yourself

Now one on your own. x squared plus seven x plus twelve, over x plus three. Make it as simple as you can. You have one hint.

(x^2 + 7x + 12)/(x + 3)It is x plus four, and x may not be minus three. You found the brackets and cancelled the match.
2Your problemTwo squares

A new pattern. x squared minus nine, over x plus three. There is no middle term. x squared is x times x, and nine is three times three. One square minus another square factorises as x minus three, times x plus three. You have one hint.

(x^2 - 9)/(x + 3)It is x minus three, and x may not be minus three. Remember: a squared minus b squared is a minus b, times a plus b.
3ChallengeTop-grade challenge

Now a top-grade one. Two x squared plus seven x plus three, over two x plus one. There is a two in front of x squared. Multiply that two by the last number, three: six. Find two numbers that multiply to six and add to seven. One and six. Then the top is two x plus one, times x plus three. You have one hint.

(2x^2 + 7x + 3)/(2x + 1)It is x plus three, and x may not be minus a half. That was a top-grade question.
4Real-life storyRectangle

A rectangle has area x^2 + 5x + 4 square metres. Its width is x + 1 metres. Find an expression for its length.

(x^2 + 5x + 4)/(x + 1)The length is x plus four metres. And x can never be minus one, because a width of zero makes no rectangle.
5Real-life storySharing sweets

A jar holds x^2 + 6x + 8 sweets. They are shared equally between x + 2 children. How many does each child get?

(x^2 + 6x + 8)/(x + 2)Each child gets x plus four sweets.
6Real-life storyAverage speed

A car travels x^2 + 3x + 2 kilometres in x + 1 hours. Find an expression for its average speed.

(x^2 + 3x + 2)/(x + 1)The speed is x plus two kilometres per hour.

M01.3 · Lecture 3 · René Descartes

Multiply and divide algebraic fractions

  • StageLeaving Certificate · Fifth–Sixth Year, age 16–18
  • Typical age16–18 · Fifth–Sixth Year (Leaving Certificate)
  • LevelLeaving Certificate Higher level
  • Exam levelLeaving Certificate Higher level
  • CurriculumLC Mathematics syllabus (2015) Strand 4.1: arithmetic operations on rational algebraic expressions (Higher level)
  • StatusBuilt

Outcome

Multiply fractions top by top and bottom by bottom, divide by flipping, then cancel what matches.

Not in the Junior Cycle specification. Multiplying and dividing algebraic fractions first appears as a Leaving Certificate Higher-level outcome, later than in England.

Needs first M01.2

What happens in this lecture

Hook

Look at this garden. Its length and its width are fractions with x in them. To find its area, we must multiply two fractions.

But first, let me show you how to multiply fractions with letters.

Explanation

  1. Today we multiply two fractions with letters. x plus one over two x, times four x over x plus one.
  2. Multiplying fractions is easy. Top times top, and bottom times bottom. You get one fraction.
  3. Then we make it simpler. Anything that is on both the top and the bottom cancels. Each cancelled bracket gives a value that is not allowed.
  4. Dividing has one extra step. To divide by a fraction, turn it upside down and multiply. Dividing by a half is the same as multiplying by two.
  5. So: multiply tops and bottoms, cancel the matches, note the values that are not allowed. Watch.

Demonstration

  1. Here are the two fractions, multiplied together. On the left is the question, and it stays as it is. I make the copy after the equals sign simpler.
  2. Top times top, bottom times bottom. Now it is one fraction.
  3. There is an x on the top and an x on the bottom. They cancel. So x cannot be zero.
  4. And x plus one is on the top and the bottom. They cancel too. So x cannot be minus one.
  5. Four over two is two. So the whole thing is just two.
  6. So the whole product is just two, for every x except zero and minus one. If you get stuck, ask me.

Exercise

Your turn. Tap the fractions to multiply them. Then drag each matching piece from the top onto the same piece below the line.

Well done. You multiplied and cancelled. It is just two, and x may not be zero or minus one.

Practice problems for M01.3
#ProblemEquationAnswer
1Your problemSimplify it yourself

Now one on your own. x plus two over three x, times six x over x plus two. Multiply and cancel until it is as simple as it can be. You have one hint.

(x + 2)/(3x) × (6x)/(x + 2)It is just two, and x may not be zero or minus two.
2Real-life storyGarden

A garden is (x + 3)/2 metres long and 4/(x + 3) metres wide. Find its area.

(x + 3)/2 × 4/(x + 3)The area is two square metres, whatever x is, as long as x is not minus three.
3Real-life storyRibbon

A ribbon is (x + 1)/x metres long. It is cut into pieces (x + 1)/(3x) metres long. How many pieces are there?

((x + 1)/x)/((x + 1)/(3x))Three pieces. Dividing by a fraction is the same as multiplying by it flipped over.
4Real-life storyMoon rover

A moon rover moves at 2x/(x + 4) metres per second for (x + 4)/x seconds. How far does it travel?

2x/(x + 4) × (x + 4)/xTwo metres. Distance is speed times time, and everything else cancels.

M01.4 · Lecture 4 · René Descartes

Add and subtract algebraic fractions

  • Stage2nd–3rd Year Higher
  • Typical age14–15 · Third Year
  • LevelHigher level only
  • Exam levelHigher level, Higher Merit
  • CurriculumJC AF.3a.III (expressions of the form a/(bx + c), bold = Higher only)
  • StatusBuilt

Outcome

Give both fractions the same bottom, add the tops, then tidy the brackets.

Adding algebraic fractions with linear denominators is Higher only at Junior Cycle. At Leaving Certificate Ordinary level it returns as Strand 4.1.

Needs first M01.2; expanding brackets

What happens in this lecture

Hook

Here is a journey in two parts. Each part takes a fraction of an hour with x in it. To find the total time, we must add two fractions.

But first, let me show you how to add fractions with letters.

Explanation

  1. Today we add two fractions with letters. Two over x plus one, plus three over x minus two.
  2. You can only add fractions when the bottoms are the same. Halves add to halves. Thirds add to thirds. Here the bottoms are different, so first we make them the same.
  3. Each fraction is missing the other one's bottom. So we multiply the top and the bottom of each fraction by the bracket it is missing. The value stays the same.
  4. Now both bottoms are x plus one, times x minus two. So we add the tops. Then we open the brackets and collect the like terms.
  5. Taking away works the same way. Just keep the minus sign on the top. Watch me add these two.

Demonstration

  1. Two over x plus one, plus three over x minus two. The question stays on the left. I work on the copy after the equals sign.
  2. I give each fraction the other one's bottom, on the top and the bottom. Now the bottoms match.
  3. The bottoms match, so I add the tops over the same bottom.
  4. Now I open each bracket on the top.
  5. Two x and three x make five x. Minus four and three make minus one.
  6. Five x minus one, over x plus one times x minus two. One fraction, and it cannot be made simpler. If you get stuck, ask me.

Exercise

Your turn. Drag one fraction onto the other to make the bottoms match. Do it again to add them. Tap the brackets on top to open them. Then join the like terms.

Well done. Match the bottoms, add, open the brackets, collect. You added fractions with letters.

Practice problems for M01.4
#ProblemEquationAnswer
1Your problemSimplify it yourself

Now one on your own. One over x plus one, plus two over x plus three. Make one fraction. You have one hint.

1/(x + 1) + 2/(x + 3)Three x plus five, over x plus one times x plus three.
2Real-life storyTwo-part trip

You walk 2 km at x + 1 km per hour, then cycle 3 km at x + 3 km per hour. Find the total time.

2/(x + 1) + 3/(x + 3)The total time is five x plus nine, over x plus one times x plus three hours.
3Real-life storyTwo taps

Two taps fill a tank. One fills 1/(x + 1) of it each minute, the other 1/(x + 3). How much do they fill together each minute?

1/(x + 1) + 1/(x + 3)Together they fill two x plus four, over x plus one times x plus three, of the tank every minute.
4Real-life storyCutting a plank

A plank is 5/(x + 1) metres long. You cut off 2/(x + 3) metres. How much is left?

5/(x + 1) - 2/(x + 3)Three x plus thirteen, over x plus one times x plus three metres are left.

M01.5 · Lecture 5 · René Descartes

Solve quadratic equations

  • Stage2nd–3rd Year Ordinary
  • Typical age13–15 · Second–Third Year
  • LevelOrdinary (integer coefficients and solutions); Higher adds ℚ coefficients, real roots and forming equations from roots
  • Exam levelOrdinary level Merit; Higher level Merit
  • CurriculumJC AF.4b, AF.5 (Higher)
  • StatusBuilt

Outcome

Get zero on one side, factorise into two brackets, set each bracket to zero, and check each answer.

Solving factorisable quadratics is common content, matching England's Higher tier timing.

Needs first M01.1; factorising from M01.2

What happens in this lecture

Hook

Here is a rectangle. We know its area, and its sides are written with x. That gives an equation with x squared in it.

But first, let me show you how to solve an equation with x squared in it.

Explanation

  1. An equation with x squared in it is called a quadratic equation. It usually has two answers.
  2. Here is the key idea. If two numbers multiply to make zero, one of them must be zero. Three times zero is zero. Zero times five is zero.
  3. So first, get zero on one side. Then factorise the other side into two brackets, just like in the simplifying lecture.
  4. Now either bracket could be the zero. Set each bracket equal to zero and solve it. Each one gives an answer.
  5. Always check each answer in the equation. And in a real problem, ask if the answer makes sense. A length cannot be negative.

Demonstration

  1. Here is x squared plus five x plus six, equals zero. Zero is already on one side.
  2. Two and three multiply to six and add to five. So it is x plus two, times x plus three, equals zero.
  3. The two brackets multiply to make zero, so one of them is zero. First, x plus two equals zero.
  4. Take two from both sides, and x is on its own.
  5. So x is minus two. Keep that answer. Now the other bracket: x plus three equals zero.
  6. Take three from both sides.
  7. So x is minus two, or minus three. Check minus two: four, minus ten, plus six, is zero.

Exercise

Your turn. Factorise, set each bracket equal to zero, and solve each one.

Well done. x is minus two or minus three. Two answers, one from each bracket.

Practice problems for M01.5
#ProblemEquationAnswer
1Your problemSolve it yourself

Now one on your own. x squared plus x equals six. Zero is not on one side yet, so start there. You have one hint.

x^2 + x = 6x is two or minus three. Check two: four plus two is six.
2Real-life storyRectangle

A rectangle is x + 3 cm long and x cm wide. Its area is 40 square centimetres. Find x.

x(x + 3) = 40x is five. The other answer, minus eight, cannot be a width, so we reject it. The rectangle is five by eight centimetres.
3Real-life storyNumber puzzle

I think of a number. I square it and add the number itself. I get 12. What could the number be?

x^2 + x = 12The number is three or minus four. Both work: nine plus three is twelve, and sixteen minus four is twelve.
4Real-life storyPatio

A square patio has sides of x metres. One side is made 2 m longer, and the new area is 15 square metres. Find x.

x(x + 2) = 15x is three. Minus five cannot be a length, so we reject it. The patio is three by five metres.

M01.6 · Lecture 6 · René Descartes

Solve the equation and check the excluded values

  • StageLeaving Certificate · Fifth–Sixth Year, age 16–18
  • Typical age16–18 · Fifth–Sixth Year (Leaving Certificate)
  • LevelLeaving Certificate Ordinary and Higher
  • Exam levelLeaving Certificate Ordinary/Higher level
  • CurriculumLC Mathematics syllabus (2015) Strand 4.2: equations f(x) = g(x) with f(x) = a/(bx + c) ± p/(qx + r)
  • StatusBuilt

Outcome

Clear the bottoms by multiplying both sides, solve for x, and check it is not an excluded value.

Not a Junior Cycle outcome. Equations with algebraic fractions are Leaving Certificate work, a year or two later than England's GCSE Higher.

Needs first M01.1, M01.4, M01.5

What happens in this lecture

Hook

A boat goes upstream and downstream in the same time. That gives us an equation with fractions in it, and we must find x.

But first, let me show you how to solve an equation with fractions in it.

Explanation

  1. Our last lesson. Three over x plus one, equals one over x minus one. This time we find x.
  2. The bottoms are brackets with x in them. To clear a bracket from the bottom, multiply both sides by it. It cancels on one side and appears on the other.
  3. Every time a bracket cancels, note the value that is not allowed. x plus one is zero when x is minus one. x minus one is zero when x is one.
  4. When both bottoms are gone, a normal equation is left. Open the brackets, collect the x terms, and divide.
  5. At the end, compare your answer with the values that are not allowed. If they are the same, the answer is wrong. Watch.

Demonstration

  1. Three over x plus one, equals one over x minus one.
  2. I multiply both sides by x plus one.
  3. On the left, the bracket cancels with the bottom. So x cannot be minus one. Three is left.
  4. Now I multiply both sides by x minus one.
  5. It cancels on the right. So x cannot be one. No more fractions.
  6. Open the bracket. Three x minus three equals x plus one.
  7. Take x away from both sides, and collect.
  8. Add three to both sides, and collect again.
  9. Two x equals four. Divide both sides by two.
  10. x is two. Now check. Two is not minus one, and two is not one. So the answer is good. Try moving the numbers yourself. If you get stuck, ask me a question.

Exercise

Your turn. Drag a bottom bracket to the Multiply spot. Cancel it with its match. Do the same for the other one. Then finish the equation and check the values that are not allowed.

Well done. You solved it, and x is allowed: it does not make any bottom zero. Now try one on your own.

Practice problems for M01.6
#ProblemEquationAnswer
1Your problemSolve it yourself

Now one on your own. Two over x plus one, equals one over x minus two. Find x, and check it is allowed. You have one hint.

2/(x + 1) = 1/(x - 2)x is five. It is not minus one and not two, so it is allowed.
2ChallengeTop-grade challenge

Now a top-grade one. Six over x equals x minus one. Clear the fraction first. This time you get x squared, so use what you learned about quadratic equations. You have one hint.

6/x = x - 1x is three or minus two. Neither one makes the bottom zero, so both are allowed.
3Your problemAlways check

Here is a tricky one. x over x minus two, equals two over x minus two. Solve it, then check your answer against the values that are not allowed. You have one hint.

x/(x - 2) = 2/(x - 2)Look closely: x is two, but two is not allowed, because it makes the bottom zero. So this equation has no solution at all. That is why we always check.
4Real-life storyRiver boat

A boat goes 3 km downstream at x + 1 km/h. In the same time it goes 2 km upstream at x - 1 km/h. Find x.

3/(x + 1) = 2/(x - 1)x is five kilometres per hour. Three over six is half an hour, and two over four is half an hour. It checks out.
5Real-life storyTwo cars

Car A goes 90 km in the same time car B goes 60 km. Car A is 10 km/h faster. Let x be car B's speed.

90/(x + 10) = 60/xCar B goes at twenty kilometres per hour, and car A at thirty. Both take three hours.
6Real-life storyWalker and runner

A walker covers 12 km in the same time a runner covers 18 km. The runner is 3 km/h faster. Let x be the walker's speed.

12/x = 18/(x + 3)The walker goes at six kilometres per hour, the runner at nine. Both take two hours.

Algebra taught by 16 here that this course does not cover yet

So a parent or teacher knows what to find elsewhere.

  • Sets and Venn diagrams with set notation: ∪, ∩, complement, set difference, cardinal number (N.5)
  • Simultaneous linear equations in two variables (AF.4c)
  • Linear inequalities with solution sets on the number line for ℕ, ℤ and ℝ (AF.4d)
  • Functions and graphs: domain, range, f(x) notation, graphical solutions of f(x) = g(x) (AF.7)
  • Changing the subject of a formula and forming quadratics from roots (AF.6, AF.5, Higher)
  • Linear, quadratic and exponential patterns and their general terms (AF.1)
  • Co-ordinate geometry of the line: slope, midpoint, distance, y = mx + c (GT.5)
  • Dividing quadratic and cubic expressions by linear expressions (AF.3c)

M02 · Geometry

Parthenon Geometry Lab · with Pythagoras

The most-examined geometry of the age-16 exams, taught on the temple itself. Every question ends as an equation, and the learner must first choose the geometric fact that justifies it, because exams award marks for stated reasons.

Set on the Parthenon on the Acropolis, 432 BC. The reward for finishing a lecture is a three-minute flight over the Acropolis.

  • 5lectures built
  • 20practice problems
  • 15real-life stories

Course outcome: by the end, a learner can

  • Use angle facts on lines, round points, in triangles and with parallel lines, giving a reason each time
  • Find interior and exterior angles of any polygon with (n − 2) × 180° and the 360° rule
  • Find any side of a right-angled triangle with Pythagoras' theorem
  • Choose sine, cosine or tangent (SOH CAH TOA) and use the exact values without a calculator
  • Find missing lengths in similar shapes from the scale factor
  • Choose the fact, then the equation, then solve it: the full reasoning chain

M02.0 · Lecture 0 · Pythagoras

Angle facts and reasons

  • Stage1st Year
  • Typical age12–13 · First Year
  • LevelCommon (all students)
  • Exam levelAchieved/Merit on Ordinary level
  • CurriculumJC GT.3b: axioms 1–5, theorems 1–6 (Geometry for Post-Primary School Mathematics); primary PMC Stage 4 angle problems
  • StatusBuilt

Outcome

Angles on a line make 180°, round a point 360°, in a triangle 180°. Parallel lines: alternate and corresponding angles are equal. Give a reason for every angle.

Angle facts are stated as numbered theorems (vertically opposite, isosceles, alternate, triangle sum, corresponding, exterior angle), and students give theorem numbers as reasons.

Needs first Measuring angles; angles in a triangle (primary)

What happens in this lecture

Hook

Look up at the end of the roof. This triangle is called the pediment. Its top angle is one hundred and fifty-four degrees. How big are the two angles at the bottom?

But first, the angle facts. Every geometry question uses them.

Explanation

  1. Angles on a straight line add up to one hundred and eighty degrees. Angles all the way round a point add up to three hundred and sixty.
  2. The three angles inside a triangle add up to one hundred and eighty degrees. In an isosceles triangle, two sides are equal, and so are the two angles at their feet.
  3. When a line crosses two parallel lines, alternate angles are equal: they make a Z shape. Corresponding angles are equal too: they make an F shape.
  4. Co-interior angles sit between the parallel lines, on the same side. They make a C shape, and they add up to one hundred and eighty degrees.
  5. In the exam, every angle you find needs a reason. Say which fact you used, not just the number.

Demonstration

  1. Here is the pediment. The two bottom angles are equal, so call each one x. The angles in a triangle add up to one hundred and eighty, so two x plus one hundred and fifty-four equals one hundred and eighty.
  2. So each bottom angle is thirteen degrees. The reason: angles in a triangle add up to one hundred and eighty, and the base angles of an isosceles triangle are equal.

Exercise

Your turn. Take one hundred and fifty-four away from both sides, then share both sides by two.

Well done. x is thirteen degrees. Check: thirteen, plus thirteen, plus one hundred and fifty-four, is one hundred and eighty.

Practice problems for M02.0
#ProblemEquationAnswer
1Your problemSolve it yourself

Now one on your own. A line crosses two parallel lines. The alternate angles are two x plus ten, and seventy. Alternate angles are equal, so two x plus ten equals seventy. You have one hint.

2x + 10 = 70x is thirty. Two times thirty, plus ten, is seventy, so the alternate angles match.
2Real-life storyThe pediment

The pediment at the end of the roof is an isosceles triangle. Its top angle is 154°. Find each base angle, x.

Fact: Angles in a triangle add up to 180
2x + 154 = 180Each base angle is thirteen degrees. Thirteen, plus thirteen, plus one hundred and fifty-four, is one hundred and eighty.
3Real-life storyTram lines

A road crosses two parallel tram lines. One angle is 3x − 20°, and the corresponding angle is 2x + 15°. Find x.

Fact: Corresponding angles are equal
3x - 20 = 2x + 15x is thirty-five. Both angles are eighty-five degrees: three times thirty-five minus twenty, and two times thirty-five plus fifteen.
4Real-life storyThe ladder

A ladder stands on flat ground. The angles on each side of it are x and 3x + 20°. Find x.

Fact: Angles on a straight line add up to 180
x + 3x + 20 = 180x is forty. The angles are forty and one hundred and forty degrees, and together they make a straight line.

M02.1 · Lecture 1 · Pythagoras

Angles in polygons

  • Stage6th Class
  • Typical age11–13 · Sixth Class to First Year
  • LevelPrimary Stage 4; common at Junior Cycle
  • CurriculumPMC Stage 4 Shape: investigate and construct angles in the context of shape; Geometry for Post-Primary School Mathematics definitions 28 and 33 (polygon, regular polygon)
  • StatusBuilt

Outcome

The angles inside a polygon add up to (n − 2) × 180°. The outside angles always add up to 360°. In a regular polygon every angle is the same.

The (n − 2) × 180° rule is not a named Junior Cycle learning outcome. It is usually treated as an application of Theorem 4 (angle sum of a triangle).

Needs first M02.0

What happens in this lecture

Hook

Look at this column. Its surface is cut into twenty grooves called flutes, so its cross-section is a regular polygon with twenty sides. How big is each corner?

But first, let me show you the angles of any polygon.

Explanation

  1. A polygon is a flat shape with straight sides. A pentagon has five sides, a hexagon six, and an octagon eight.
  2. Split a polygon into triangles from one corner. There are always two fewer triangles than sides. So the angles inside add up to n minus two, times one hundred and eighty.
  3. Walk round the outside of any polygon, and you turn all the way round once. So the outside angles always add up to three hundred and sixty.
  4. In a regular polygon, all the sides and all the angles are equal. Share the total between the angles.

Demonstration

  1. Here is a regular octagon, with eight equal angles. Eight sides make six triangles, and six times one hundred and eighty is one thousand and eighty. So eight x equals one thousand and eighty.
  2. Each angle is one hundred and thirty-five degrees.

Exercise

Your turn. Share both sides by eight.

Well done. Each angle of a regular octagon is one hundred and thirty-five degrees.

Practice problems for M02.1
#ProblemEquationAnswer
1Your problemSolve it yourself

Now one on your own. A pentagon has angles x, two x, one hundred, one hundred and ten, and one hundred and twenty. A pentagon makes three triangles, so its angles add up to five hundred and forty. You have one hint.

x + 2x + 100 + 110 + 120 = 540x is seventy, so the two unknown angles are seventy and one hundred and forty. Together with the others they make five hundred and forty.
2Real-life storyThe column

A Doric column has 20 flutes, so its cross-section is a regular polygon with 20 sides. Find each inside angle, x.

Fact: Inside angles add up to (n - 2) × 180
20x = 3240Each angle is one hundred and sixty-two degrees. Check with the outside angle: three hundred and sixty over twenty is eighteen, and one hundred and eighty minus eighteen is one hundred and sixty-two.
3Real-life storyHoneycomb

Bees build their cells as regular hexagons. Find each inside angle, x. Why do three cells fit round a point?

Fact: Inside angles add up to (n - 2) × 180
6x = 720Each angle is one hundred and twenty degrees. Three of them make three hundred and sixty, a full turn, so hexagons fit together with no gaps.
4Real-life storyThe market tent

A market tent has a floor shaped like a regular polygon. Each outside angle is 24°. How many sides, x, does it have?

Fact: Outside angles add up to 360
24x = 360Fifteen sides. Fifteen turns of twenty-four degrees make three hundred and sixty, one full turn.

M02.2 · Lecture 2 · Pythagoras

Pythagoras' theorem

  • Stage2nd–3rd Year Ordinary
  • Typical age13–15 · Second–Third Year
  • LevelCommon (Ordinary and Higher)
  • Exam levelOrdinary level Merit
  • CurriculumJC GT.3b: Theorem 14 (Pythagoras) and Theorem 15 (converse)
  • StatusBuilt

Outcome

In a right-angled triangle, the square on the longest side equals the squares on the other two added: a² + b² = c². Take the square root to find the length.

Pythagoras is common content, taught at about the same age as in England.

Needs first Squares and square roots

What happens in this lecture

Hook

The builders of this temple needed perfect right angles at its corners. They used a rope with knots in it, making a triangle with sides three, four and five. Why does that work?

The answer carries my name. Let me show you my theorem.

Explanation

  1. A right-angled triangle has one square corner. The longest side is opposite the right angle. It is called the hypotenuse.
  2. Draw a square on each side. The square on the hypotenuse is exactly as big as the other two squares together. We write a squared plus b squared equals c squared.
  3. Three, four, five: nine plus sixteen is twenty-five. That is why the builders' rope always makes a right angle.
  4. To find the hypotenuse, square the two short sides, add them, and take the square root. To find a short side, take away instead of adding.

Demonstration

  1. Here is a right-angled triangle. The short sides are six and eight, and the hypotenuse is x. So x squared equals six squared plus eight squared.
  2. The hypotenuse is ten. Check: thirty-six plus sixty-four is one hundred, and ten squared is one hundred.

Exercise

Your turn. Work out the squares, add them, then take the square root of both sides.

Well done. The hypotenuse is ten.

Practice problems for M02.2
#ProblemEquationAnswer
1Your problemSolve it yourself

Now one on your own, and this time a short side is missing. The hypotenuse is thirteen, one side is five, and the other is x. So x squared plus five squared equals thirteen squared. You have one hint.

x^2 + 5^2 = 13^2x is twelve. Check: one hundred and forty-four plus twenty-five is one hundred and sixty-nine, thirteen squared.
2Real-life storyThe builders' rope

The builders pegged out two sides of a corner, 3 m and 4 m long, at a right angle. How long is the rope across, x?

Fact: Pythagoras: a^2 + b^2 = c^2
x^2 = 3^2 + 4^2The rope is five metres. Three, four, five: that is how the builders made perfect right angles.
3Real-life storyThe ladder

A 10 m ladder leans against a column. Its foot is 6 m from the base of the column. How high up the column does it reach, x?

Fact: Pythagoras: a^2 + b^2 = c^2
x^2 + 6^2 = 10^2The ladder reaches eight metres up. Check: sixty-four plus thirty-six is one hundred, ten squared.
4Real-life storyThe map

On a map, the temple is at (1, 2) and the theatre at (7, 10), in hundreds of metres. How far apart are they in a straight line, x?

Fact: Pythagoras: a^2 + b^2 = c^2
x^2 = 6^2 + 8^2They are ten, that is one thousand metres, apart in a straight line.

M02.3 · Lecture 3 · Pythagoras

Trigonometry in right-angled triangles

  • Stage2nd–3rd Year Ordinary
  • Typical age13–15 · Second–Third Year
  • LevelOrdinary (angles in whole degrees); Higher adds angles in decimal form
  • Exam levelOrdinary level Merit; Higher level Merit
  • CurriculumJC GT.4
  • StatusBuilt

Outcome

Name the sides from the angle: opposite, adjacent, hypotenuse. SOH CAH TOA picks the ratio. Learn the exact values: sin 30° = ½, cos 60° = ½, tan 45° = 1.

SOH CAH TOA is common content. Exact surd values of the ratios are listed only at Leaving Certificate Higher level (Strand 2.3).

Needs first M02.2; M01.1

What happens in this lecture

Hook

How tall is a column? You cannot climb it with a tape measure. But you can stand back, measure the angle up to its top, and let a triangle do the rest.

First, the three ratios of a right-angled triangle.

Explanation

  1. Pick an angle in a right-angled triangle. The side across from it is the opposite. The side next to it is the adjacent. The longest side is the hypotenuse.
  2. Sine is opposite over hypotenuse. Cosine is adjacent over hypotenuse. Tangent is opposite over adjacent. Remember it as SOH, CAH, TOA.
  3. Some values are exact, and you need them without a calculator. Sine of thirty degrees is one half. Cosine of sixty degrees is one half. Tangent of forty-five degrees is one.
  4. Write the ratio as an equation, put in the exact value, and solve it, just like an equation with a fraction.

Demonstration

  1. The hypotenuse is twelve, the angle is thirty degrees, and we want the opposite side, x. Opposite and hypotenuse: that is sine. Sine of thirty is one half, so x over twelve equals one half.
  2. So x is six. Half the hypotenuse, because sine of thirty is one half.

Exercise

Your turn. Multiply both sides by twelve to clear the fraction.

Well done. The opposite side is six.

Practice problems for M02.3
#ProblemEquationAnswer
1Your problemSolve it yourself

Now one on your own. The angle is sixty degrees, the adjacent side is five, and the hypotenuse is x. Adjacent over hypotenuse is cosine, and cosine of sixty is one half. So one half equals five over x. You have one hint.

1/2 = 5/xx is ten. The adjacent side is half the hypotenuse, because cosine of sixty is one half.
2Real-life storyThe column

You stand 10 m from a column. The angle up to its top is 45°. How tall is the column, x?

Fact: tan = opposite / adjacent
x/10 = 1The column is ten metres tall. At forty-five degrees the height and the distance are equal. The real columns are about ten and a half metres.
3Real-life storyThe kite

A kite flies on a 40 m string. The string makes an angle of 30° with the ground. How high is the kite, x?

Fact: sin = opposite / hypotenuse
x/40 = 1/2The kite is twenty metres high: half the string, because sine of thirty is one half.
4Real-life storyThe ramp

A plank leans on a step at 60° to the ground. Its foot is 2 m from the point under its top. How long is the plank, x?

Fact: cos = adjacent / hypotenuse
1/2 = 2/xThe plank is four metres long. The ground is half the plank, because cosine of sixty is one half.

M02.6 · Lecture 6 · Pythagoras

Similar shapes and scale factors

  • Stage2nd–3rd Year Ordinary
  • Typical age13–15 · Second–Third Year
  • LevelCommon (Ordinary and Higher)
  • Exam levelOrdinary level Merit
  • CurriculumJC GT.3b: Axiom 4 (congruent triangles SAS/ASA/SSS), Theorem 13 (similar triangles); GT.2a scaled diagrams
  • StatusBuilt

Outcome

Similar shapes have equal angles and sides in the same ratio. Match the sides, write the ratio as an equation, and solve it. Areas scale by k², volumes by k³.

Congruence and similar triangles are common content. Enlargement with its effect on area (k²) is Leaving Certificate Strand 2.4, and the k³ rule for volume is not named.

Needs first Ratio; M01.6

What happens in this lecture

Hook

Long ago, Thales measured a pyramid without climbing it. He used its shadow. We can do the same with a column of this temple.

First, what it means for two shapes to be similar.

Explanation

  1. Two shapes are similar when one is an enlargement of the other. The angles stay the same, and every side is multiplied by the same scale factor.
  2. Match the sides that sit in the same place. The ratio of matching sides is the same all the way round.
  3. Write one ratio equal to the other, as an equation with fractions. Then solve it.
  4. Careful with areas and volumes. If the lengths are multiplied by k, areas are multiplied by k squared, and volumes by k cubed.

Demonstration

  1. Here are two similar triangles. The small one has sides four and ten. The big one has matching sides six and x. So x over ten equals six over four.
  2. So x is fifteen. The scale factor is one and a half, and ten times one and a half is fifteen.

Exercise

Your turn. Clear the fractions, then find x.

Well done. x is fifteen.

Practice problems for M02.6
#ProblemEquationAnswer
1Your problemSolve it yourself

Now one on your own. Two similar triangles: the small one has sides six and nine, and the big one has matching sides eight and x. So x over nine equals eight over six. You have one hint.

x/9 = 8/6x is twelve. The scale factor is eight over six, which is four thirds, and nine times four thirds is twelve.
2Real-life storyThales' shadow

A 2 m stick casts a 3 m shadow. At the same moment a column casts a 15 m shadow. How tall is the column, x?

Fact: The two triangles are similar
x/15 = 2/3The column is ten metres tall. The real columns are about ten and a half metres.
3Real-life storyThe model temple

A museum model of the Parthenon is built to a scale of 1 : 50. The real front is 31 m wide. How wide is the model's front, x?

Fact: The model and the temple are similar
x/31 = 1/50The model's front is thirty-one fiftieths of a metre, sixty-two centimetres. Areas would shrink by fifty squared, two thousand five hundred.
4Real-life storyThe photo

A photo is 6 cm wide and 10 cm tall. It is enlarged to 15 cm wide. How tall is the enlargement, x?

Fact: The two photos are similar
x/10 = 15/6The enlargement is twenty-five centimetres tall. The scale factor is two and a half.

Planned: the rest of the geometry course

Designed, not yet built.

  • M02.4Area, perimeter and circlesTrapezium, compound shapes, arcs and sectors, answers in terms of π
  • M02.5Volume and surface areaPrisms, cylinder, cone, pyramid, sphere, frustum
  • M02.7Sine rule and cosine ruleAny triangle; choosing the rule; the ambiguous case
  • M02.8½ab sin C and 3D trigonometryArea from two sides and the included angle; line–plane angles
  • M02.9Circle theoremsThe eight theorems; angle chasing with reasons; proof
  • M02.10VectorsAB = b − a, midpoints, ratio on a line, proving lines parallel
  • M02.11TransformationsReflect, rotate, translate, enlarge; 'describe fully'
  • M02.12Constructions, loci and bearingsBisectors, loci regions, three-figure bearings
  • M02.13Exam problems and geometric proofMixed multi-step questions, 'prove' and 'show that'

Suggested order by year

A guide, not a rule
  1. Sixth Class – First Year · age 11–13

    Letter-symbols, angle theorems 1–6 and polygon angles bridge primary Stage 4 and early Junior Cycle work.

  2. First–Second Year · age 12–14

    Linear equations with fractions, Pythagoras and similar/congruent triangles are common-level outcomes, usually met before Third Year.

  3. Second–Third Year · age 13–15

    Quadratics, factorising and right-angled trigonometry are central to both exam levels; the non-monic cases are Higher level.

  4. Third Year Higher, then Leaving Certificate · age 14–18

    Adding algebraic fractions is Junior Cycle Higher only; multiplying, dividing and solving with algebraic fractions come at Leaving Certificate.