IB MYP Mathematics: Algebra, equations and sequences (MYP Mathematics)

This guide covers the Algebra rows of the numerical and abstract reasoning branch in the IB Mathematics guide for the MYP (for use from September 2020 and January 2021, first assessment 2022). The framework has three columns: MYP 1-3, MYP 4-5 standard and MYP 4-5 extended, plus a list of enrichment ideas. Items from the extended column and the enrichment list are labelled below. MYP has no tiers. For the other branches see the MYP Mathematics hub, the printable checklist and the syllabus guide. Pair it with the revision notes, the practice set and the free diagnostics.

What this unit covers

Topic	What you should be able to do	Stage
Sequences	Find and verify rules; justify and prove general rules	MYP 1-3 (find, verify); MYP 4-5 (justify, prove)
Equations	Form, rearrange and solve; simultaneous equations by algebra and graphs	MYP 1-3; MYP 4-5
Expressions	Substitute, expand, factorize, including quadratics; spot equivalent expressions	MYP 1-3; MYP 4-5
Inequalities	Solve, including compound and double	MYP 4-5
Quadratic equations	Solve by factorization, the formula and graphs	MYP 4-5
Formulae	Use and rearrange	MYP 1-3; MYP 4-5
Algorithms	Read and build flowcharts	MYP 1-3
Sequence notation	Arithmetic and geometric formulae	Extended
Enrichment	Exponential equations, sums, sigma notation, convergence	Enrichment

The guide ties this to solving versus satisfying an equation and equivalent expressions. The related concepts it names for the branch include equivalence, generalization, simplification and validity. Criterion B (investigating patterns) asks you to describe general rules and, at the top levels, to prove them, or verify and justify them.

Sequences: find, verify, justify, prove

Find a rule by looking at differences. A constant first difference means a linear rule. Constant second differences mean a quadratic rule whose n² coefficient is half the second difference.

Sequence 13, 19, 25, 31, 37. The first difference is 6 each time, so the rule is 6n + c. At n = 1 the term is 13, so c = 7 and the term is 6n + 7.

Verify by testing terms the rule was not built from. The rule 6n + 7 gives 13, 19, 25, 31, 37 for n = 1 to 5. Continue the list by adding 6 up to position 8 and you reach 55; the rule gives 6(8) + 7 = 55. Verifying shows the rule fits; it does not show it always will.

Justify means giving a reason: the difference is always 6 because each step adds 6, and the rule adds 6 per position.

Prove means an argument that works for every n. Claim: the sum of two consecutive terms of 6n + 7 is always a multiple of 4.

(6n + 7) + (6(n + 1) + 7) = 6n + 7 + 6n + 13 = 12n + 20
12n + 20 = 4(3n + 5), and 3n + 5 is a whole number.

Quadratic example: 5, 9, 15, 23, 33. First differences: 4, 6, 8, 10. Second differences: all 2, so the rule contains n². Subtract n² (1, 4, 9, 16, 25) and the leftovers are 4, 5, 6, 7, 8, which is n + 3. The rule is n² + n + 3. To justify it, the gap between terms n and n + 1 is ((n + 1)² + (n + 1) + 3) − (n² + n + 3) = 2n + 2, which is 4, 6, 8, 10 for n = 1 to 4.

Forming and solving equations

Define the unknown, build the equation from the words, solve, then check against the story.

A plumber charges a £28 call-out fee plus £16.50 per hour. The bill is £110.50. Let h be the number of hours.

28 + 16.5h = 110.5
16.5h = 82.5
h = 5

The job took 5 hours. When the unknown is on both sides, collect terms: 8x + 3 = 3x + 38 gives 5x = 35, so x = 7. With fractions, multiply by the common denominator: (x + 5)/3 = (2x − 1)/5 becomes 5x + 25 = 6x − 3, so x = 28.

Solving versus satisfying

A value satisfies an equation if substituting it makes both sides equal. Solving means finding every value that does this. For 8x + 3 = 3x + 38, x = 7 satisfies it (59 = 59) and nothing else does. For x² = 16, both 4 and −4 satisfy it, so a solution of only 4 is incomplete.

Equivalent expressions

Two expressions are equivalent if they have the same value for every input. Equations hold for particular values, so check them by substitution. Equivalence needs algebra: 4(x + 2) and 4x + 8 are equivalent because expanding turns one into the other. Testing one value is not enough. (x + 3)² and x² + 9 both equal 9 when x = 0, but at x = 1 they give 16 and 10, so they are not equivalent. Use the symbol ≡ for “is identical to”.

Simultaneous equations

Two equations in x and y have a solution that satisfies both.

Elimination: 4x + 3y = 24 and 3x − 2y = 1. Scale the first equation by 2 and the second by 3 so the y terms match:

8x + 6y = 48
9x − 6y = 3
Add: 17x = 51, so x = 3.
Substitute: 12 + 3y = 24, so y = 4.
Check in the second: 9 − 8 = 1.

Substitution: y = x − 4 and 3x + 2y = 7. Replace y: 3x + 2x − 8 = 7, so x = 3 and y = −1.

Graphically: draw both lines; the coordinates of the intersection are the solution. For y = x/2 + 1 and x + y = 7 the lines cross at (4, 3): check 4/2 + 1 = 3 and 4 + 3 = 7. Parallel lines never meet, so there is no solution.

Expressions: substitute, expand, factorize

Substitute with brackets around negatives. For a = −4 and b = 5, 2a² − ab = 2(16) − (−4)(5) = 32 + 20 = 52.

Expand every term in the first bracket times every term in the second:

(3x − 2)(x + 6) = 3x² + 18x − 2x − 12 = 3x² + 16x − 12.

Factorize by taking out the highest common factor first: 8p²q − 12pq² = 4pq(2p − 3q). Then look for a difference of two squares: 25y² − 36 = (5y − 6)(5y + 6).

Quadratic expressions: for x² − 3x − 28 look for two numbers with product −28 and sum −3. They are −7 and 4, so (x − 7)(x + 4). When the x² term has a coefficient, split the middle term: 6x² + 7x − 5 needs two numbers with product 6 × (−5) = −30 and sum 7, which are 10 and −3:

6x² + 10x − 3x − 5 = 2x(3x + 5) − (3x + 5) = (2x − 1)(3x + 5).

Expanding the answer is always the fastest check.

Inequalities

Solve like an equation, with one change: multiplying or dividing by a negative reverses the sign.

9 − 4x ≥ 21 gives −4x ≥ 12, so x ≤ −3.

Compound inequalities join two conditions with “and” or “or”. For “and” the answer is the overlap: 2x + 3 < 15 and x − 1 ≥ −4 give x < 6 and x ≥ −3, so −3 ≤ x < 6. For “or” the answer is both pieces: 3x + 1 ≤ −8 or 2 − x < −1 give x ≤ −3 or x > 3.

Double inequalities are solved in one go by doing the same to all three parts: −5 ≤ 2x − 3 < 9 gives −2 ≤ 2x < 12, so −1 ≤ x < 6. On a number line use an open circle for < and > and a filled circle for ≤ and ≥. The integers that satisfy it are −1, 0, 1, 2, 3, 4, 5, which is 7 values.

Quadratic equations

Rearrange to one side equal to zero first. Then pick a method.

Factorization: x² − 3x − 28 = 0 gives (x − 7)(x + 4) = 0, so x = 7 or x = −4. Do not divide both sides by x in x² = 7x; write x² − 7x = 0, so x(x − 7) = 0 and x = 0 or 7.

The quadratic formula: for ax² + bx + c = 0, x = (−b ± √(b² − 4ac)) / (2a). For 3x² + 4x − 5 = 0, a = 3, b = 4, c = −5:

x = (−4 ± √(16 + 60)) / 6 = (−4 ± √76) / 6, so x = 0.786 or x = −2.12 to 3 s.f.

If the number under the root is negative, the graph never reaches the x-axis and there is no real solution.

Graphically: draw y = ax² + bx + c and read off the x-values at its crossings of the x-axis; those are the solutions of ax² + bx + c = 0. The graph is symmetrical about the vertical line through the midpoint of the two roots.

In context: a picture frame is 4 cm longer than it is wide and its area is 96 cm². With width w, w(w + 4) = 96, so w² + 4w − 96 = 0 and (w + 12)(w − 8) = 0. Width cannot be negative, so w = 8 cm and the length is 12 cm.

Using and rearranging formulae

Substitute known values, keeping units. To rearrange, undo operations in reverse order, treating both sides equally.

E = ½mv²: multiply by 2, divide by m, then take the square root, so v = √(2E/m). With E = 50 and m = 4, v = √25 = 5.

I = Prt: dividing by Pr gives t = I/(Pr).

If the new subject appears twice, gather those terms on one side and factorize. Make x the subject of y = (5x − 1)/(x + 2):

y(x + 2) = 5x − 1
yx + 2y = 5x − 1
yx − 5x = −1 − 2y
x(y − 5) = −(1 + 2y)
x = (1 + 2y)/(5 − y)

Check with x = 3: y = 14/5 = 2.8, and the formula gives (1 + 5.6)/(5 − 2.8) = 3.

Flowcharts and simple algorithms

A flowchart uses ovals for start and stop, parallelograms for input and output, rectangles for steps and diamonds for yes/no decisions. Arrows show the order. A trace table records each variable after every pass.

Algorithm: set p = 3 and t = 0. Repeat: add p to t, then double p. Stop when t is above 60, and output t.

Pass	t	p
start	0	3
1	3	6
2	9	12
3	21	24
4	45	48
5	93	96

The output is 93. The algorithm is adding the terms of the sequence 3, 6, 12, 24, 48 until the total passes 60.

Extended: arithmetic and geometric sequences

For terms a₁, a₂, a₃, … (the guide’s notation list writes terms this way) an arithmetic sequence adds a common difference d, so aₙ = a₁ + (n − 1)d. A geometric sequence multiplies by a common ratio r, so aₙ = a₁ × rⁿ⁻¹.

Arithmetic: 50, 44, 38, … has a₁ = 50 and d = −6, so aₙ = 56 − 6n and a₁₅ = −34.

Geometric: 5, 15, 45, … has a₁ = 5 and r = 3, so aₙ = 5 × 3ⁿ⁻¹ and a₆ = 1215. The progression shows growth by tripling; the arithmetic one shows steady decline.

Enrichment

Exponential equations: write both sides with the same base. 16ˣ = 8 gives 2⁴ˣ = 2³, so 4x = 3 and x = 3/4. And 2 × 3ⁿ⁻¹ = 54 gives 3ⁿ⁻¹ = 27 = 3³, so n = 4.

Sums: for an arithmetic sequence, Sₙ = (n/2)(2a₁ + (n − 1)d). For 50, 44, 38, … S₁₂ = 6(100 − 66) = 204. For a geometric sequence, Sₙ = a₁(rⁿ − 1)/(r − 1). For 5, 15, 45, … S₆ = 5 × 728 / 2 = 1820.

Sigma notation writes a sum compactly. Σ from k = 1 to 6 of (3k − 2) = 1 + 4 + 7 + 10 + 13 + 16 = 51.

Convergence and divergence: 12, 6, 3, … halves each time, so its running total approaches 24 and the series converges. A sequence with r = 3 gets ever larger, so its total diverges. Convergence needs a ratio between −1 and 1.

Common errors

• Writing x = 4 for x² = 16 and forgetting −4.
• Keeping the original sign after dividing an inequality by a negative number.
• Expanding (x + 3)² as x² + 9.
• Dividing by x in a quadratic and losing a root.
• Rejecting a verified rule because one term fails, or accepting it because three terms fit. Verify beyond the data and prove with algebra.
• Writing a power as 2^n instead of 2ⁿ; the guide’s notation list asks for mathematical notation, not calculator notation.

Official syllabus

Mathematics guide for the Middle Years Programme (for use from September 2020/January 2021), International Baccalaureate Organization, first assessment 2022, numerical and abstract reasoning, Algebra rows.