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Edexcel GCSE Maths revision notes

Algebra · equations and reasoning

Section A
8 specification points

Notes and three levels of exam-style practice for each registered specification point in this section.

Checked against Edexcel 1MA1 section A

Checked against Edexcel 1MA1 section A. Review basis: the qualification registry sourced from the Pearson Edexcel Level 1/Level 2 GCSE (9-1) in Mathematics (1MA1) specification; registry verification recorded 9 July 2026.

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In the exam: Formula sheet provided · Paper 1 non-calculator

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Know the difference between an equation and an identity; argue mathematically to show algebraic expressions are equivalent, and use algebra to support and construct arguments and proofs

Notes
Worked answers & exam appearances →
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • An equation is satisfied by particular values, whereas an identity states that two expressions are equivalent for every permitted value.
  • To show expressions are equivalent, expand or factorise one side until it matches the other.
  • Foundation tier: use algebra to support and construct an argument by defining quantities with variables, translating the claim and linking the resulting algebra back to it.
  • Higher tier: extend this to a proof by using a general form, such as 2n2n for an even integer or 2n+12n+1 for an odd integer, and reasoning that covers every permitted case.
  • Checking examples alone is not a proof.
Worked example

Higher tier: prove algebraically that the sum of two consecutive integers is odd.

  1. 1.Let the first integer be nn, so the next is n+1n+1.
  2. 2.Their sum is n+(n+1)=2n+1n+(n+1)=2n+1.
  3. 3.2n2n is even for every integer nn, so 2n+12n+1 is odd.

Answer: The sum has form 2n+12n+1, so it is odd.

Common mistakes

  • Don't treat expressions that agree for one value as equivalent without simplifying them generally.
  • Don't check several numerical cases and call the pattern a proof (Higher tier).
  • Don't finish with algebra and fail to state why its form proves the claim (Higher tier).

Exam tip

Foundation tier: show each algebraic step and link the result to the argument. Higher tier: a “prove” question needs a general variable-based argument, not examples.

Tier 1 · Easy

ORIGINAL

1

Show that 4(n+2)−3(n−1)4(n+2)-3(n-1) is equivalent to n+11n+11.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1

Prove algebraically that the sum of two consecutive integers is odd.

(3)

(Total for Question 1 is 3 marks)

Tier 3 · Hard

ORIGINAL

1

An odd number is written as 2n+12n+1. Demonstrate algebraically that squaring it leaves remainder 11 after division by 88.

(4)

(Total for Question 1 is 4 marks)

Your progress and exam materials

Interpret simple expressions as functions with inputs and outputs; interpret the reverse as the 'inverse function' and two successive functions as a 'composite function' (formal notation expected)

Notes
Worked answers & exam appearances →
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • A function maps each allowed input to one output. To evaluate f(a)f(a), substitute the complete input aa into every occurrence of the variable and then simplify.
  • A function can be represented by a rule, mapping diagram or input-output table.
  • Higher tier: an inverse function reverses a one-to-one function, while a composite applies two functions successively; in fg(x)=f(g(x))fg(x)=f(g(x)), gg acts first.
  • Formal function notation is expected for those extensions.
  • Examiners require careful brackets when the input is an expression, because the whole input replaces xx.
Worked example

Given f(x)=3x−4f(x)=3x-4, work out f(6)f(6) and f(2a)f(2a).

  1. 1.Substitute 66: f(6)=3(6)−4=14f(6)=3(6)-4=14.
  2. 2.Substitute the complete input 2a2a: f(2a)=3(2a)−4f(2a)=3(2a)-4.
  3. 3.Simplify to f(2a)=6a−4f(2a)=6a-4.

Answer: f(6)=14f(6)=14 and f(2a)=6a−4f(2a)=6a-4.

Common mistakes

  • Don't treat f(x)f(x) as f×xf\times x instead of function notation.
  • Don't substitute only part of an expression supplied as the input.
  • Don't read f−1(x)f^{-1}(x) as 1f(x)\dfrac{1}{f(x)} (Higher tier).

Exam tip

Put the complete input in brackets everywhere xx appears before simplifying.

Tier 1 · Easy

ORIGINAL

1

Given f(x)=3x−4f(x)=3x-4, work out f(6)f(6).

(1)

(Total for Question 1 is 1 mark)

Tier 2 · Standard

ORIGINAL

1

Given f(x)=5x+2f(x)=5x+2, find f−1(x)f^{-1}(x) and work out f−1(27)f^{-1}(27).

(3)

(Total for Question 1 is 3 marks)

Tier 3 · Hard

ORIGINAL

1

Let f(x)=2x+3f(x)=2x+3 and g(x)=x2−1g(x)=x^2-1. Solve fg(x)=19fg(x)=19, where fg(x)=f(g(x))fg(x)=f(g(x)).

(4)

(Total for Question 1 is 4 marks)

Work with coordinates in all four quadrants

Notes
Worked answers & exam appearances →
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • A coordinate (x,y)(x,y) gives horizontal position first and vertical position second. Positive xx is right, negative xx is left, positive yy is up and negative yy is down.
  • The signs identify the quadrant, numbered anticlockwise from the top right.
  • Find a displacement by subtracting starting coordinates from ending coordinates.
  • Find a midpoint by averaging the two xx-coordinates and separately averaging the two yy-coordinates.
  • Examiners expect coordinate order and negative signs to be preserved, with the two component calculations shown clearly.
The four coordinate quadrants and the signs of (x,y)(x,y) in each.
Worked example

Find the midpoint of the segment joining (−5,2)(-5,2) and (3,−4)(3,-4).

  1. 1.Average the xx-coordinates: −5+32=−1\dfrac{-5+3}{2}=-1.
  2. 2.Average the yy-coordinates: 2+(−4)2=−1\dfrac{2+(-4)}{2}=-1.
  3. 3.Write the coordinates in (x,y)(x,y) order.

Answer: (−1,−1)(-1,-1).

Common mistakes

  • Don't write the vertical coordinate first and swap (x,y)(x,y).
  • Don't lose a negative sign when averaging coordinates.
  • Don't find half the coordinate differences but forget to add them to an endpoint.

Exam tip

Write the midpoint as two separate averages before combining them into one ordered pair.

Tier 1 · Easy

ORIGINAL

1

State the quadrant containing the point (−4,3)(-4,3).

(1)

(Total for Question 1 is 1 mark)

Tier 3 · Hard

ORIGINAL

1

The point PP divides the line segment from A(−6,5)A(-6,5) to B(4,−5)B(4,-5) in the ratio AP:PB=3:2AP:PB=3:2. Find the coordinates of PP.

(3)

(Total for Question 1 is 3 marks)

Plot graphs of straight-line equations; use y = mx + c to identify parallel and perpendicular lines; find the equation of a line through two given points, or one point with a given gradient

Notes
Worked answers & exam appearances →
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • A non-vertical straight line has equation y=mx+cy=mx+c, where mm is the gradient and (0,c)(0,c) is the yy-intercept. Plot a line by calculating at least two accurate coordinate pairs and joining them carefully with a ruler.
  • From two points, find m=y2−y1x2−x1m=\dfrac{y_2-y_1}{x_2-x_1}, then substitute either point to find cc.
  • Parallel lines have equal gradients.
  • Higher tier: perpendicular non-vertical lines have gradients whose product is −1-1.
  • Examiners expect the gradient calculation, substitution for the intercept and a final equation in a requested form.
A straight line y=mx+cy=mx+c crosses the vertical axis at (0,c)(0,c) and has constant gradient.
Worked example

Find the equation of the line through (−2,5)(-2,5) and (4,−1)(4,-1).

  1. 1.m=−1−54−(−2)=−66=−1m=\dfrac{-1-5}{4-(-2)}=\dfrac{-6}{6}=-1.
  2. 2.Use y=−x+cy=-x+c and substitute (−2,5)(-2,5): 5=2+c5=2+c.
  3. 3.c=3c=3, so the equation is y=−x+3y=-x+3.

Answer: y=−x+3y=-x+3.

Common mistakes

  • Don't subtract coordinates in different orders in the gradient numerator and denominator.
  • Don't use the xx-intercept as cc in y=mx+cy=mx+c.
  • Don't change only the sign of a gradient to make a perpendicular line (Higher tier).

Exam tip

For a line through two points, show the gradient first and then substitute one point to find cc.

Tier 1 · Easy

ORIGINAL

1

Write the equation of the line with gradient 44 and yy-intercept −7-7.

(1)

(Total for Question 1 is 1 mark)

Tier 2 · Standard

ORIGINAL

1

Line AA has equation 4y=6x+74y=6x+7. Line BB passes through the points (−2,4)(-2,4) and (2,10)(2,10). Show that the two lines are parallel.

(3)

(Total for Question 1 is 3 marks)

Tier 3 · Hard

ORIGINAL

1

Find an equation of the line through (−6,5)(-6,5) that is perpendicular to 7x+4y=137x+4y=13. Give your answer in the form y=mx+cy=mx+c.

(4)

(Total for Question 1 is 4 marks)

Identify and interpret gradients and intercepts of linear functions graphically and algebraically

Notes
Worked answers & exam appearances →
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • The gradient of a linear function is the change in the vertical quantity per unit change in the horizontal quantity. Its sign shows whether the line rises or falls.
  • The yy-intercept is the output when x=0x=0; the xx-intercept is where the output is zero.
  • Read axis scales and units before calculating or interpreting either feature.
  • In a context, a gradient is a rate and an intercept is often an initial value or fixed charge.
  • Examiners require a value, its unit and a sentence explaining what it represents.
The intercept is the output at zero; the gradient is rise divided by run.
Worked example

Water volume follows V=120−6tV=120-6t, where VV is litres and tt is minutes. Interpret the gradient and intercept.

  1. 1.The coefficient of tt is −6-6, so volume changes by −6-6 litres per minute.
  2. 2.Therefore the volume decreases by 66 litres each minute.
  3. 3.When t=0t=0, V=120V=120, so the intercept is the initial volume.

Answer: The tank starts with 120120 litres and loses 66 litres per minute.

Common mistakes

  • Don't calculate run divided by rise for the gradient.
  • Don't read the wrong intercept because the axes have been confused.
  • Don't give a contextual gradient as a bare number without units or meaning.

Exam tip

For “interpret”, state what happens per horizontal-axis unit and what the intercept means at zero.

Tier 1 · Easy

ORIGINAL

1

State the gradient and yy-intercept of y=−3x+8y=-3x+8.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1

A straight line crosses the axes at (0,12)(0,12) and (6,0)(6,0). Find its gradient and both intercepts.

(3)

(Total for Question 1 is 3 marks)

Tier 3 · Hard

ORIGINAL

1

A straight-line graph of water volume VV litres against time tt minutes passes through (0,120)(0,120) and (8,72)(8,72). Find and interpret its gradient and VV-intercept, then write VV in terms of tt.

(4)

(Total for Question 1 is 4 marks)

Identify and interpret roots, intercepts, turning points of quadratic functions graphically; deduce roots algebraically and turning points by completing the square

Notes
Worked answers & exam appearances →
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • A root of a quadratic is an xx-coordinate where its graph meets the xx-axis, so y=0y=0. The yy-intercept is found by setting x=0x=0.
  • A turning point is the maximum or minimum point, and its vertical line is the axis of symmetry.
  • Factorising can reveal roots algebraically and the graph confirms their positions.
  • Higher tier: completing the square into a(x−h)2+ka(x-h)^2+k reveals turning point (h,k)(h,k).
  • Examiners expect coordinates for points, equations for axes, and algebraic working when roots are to be deduced.
A quadratic's roots are its xx-axis intersections, and its turning point lies on the axis of symmetry.
Worked example

Find the roots and yy-intercept of y=x2−5x+6y=x^2-5x+6.

  1. 1.Factorise: x2−5x+6=(x−2)(x−3)x^2-5x+6=(x-2)(x-3).
  2. 2.Set y=0y=0: each factor can be zero, giving x=2x=2 or x=3x=3.
  3. 3.Set x=0x=0: y=6y=6, so the yy-intercept is (0,6)(0,6).

Answer: Roots x=2x=2 and x=3x=3; yy-intercept (0,6)(0,6).

Common mistakes

  • Don't report roots as yy-values instead of xx-coordinates.
  • Don't find the yy-intercept by setting y=0y=0.
  • Don't read (x−h)2+k(x-h)^2+k as having turning point (−h,k)(-h,k) (Higher tier).

Exam tip

State roots as xx-values, intercepts as coordinates, and the symmetry line as an equation.

Tier 2 · Standard

ORIGINAL

1

For y=(x−4)2−9y=(x-4)^2-9, state the turning point and axis of symmetry, and find the roots.

(3)

(Total for Question 1 is 3 marks)

Tier 3 · Hard

ORIGINAL

1

Complete the square for y=2x2+8x−10y=2x^2+8x-10. Hence state the turning point and find the roots.

(4)

(Total for Question 1 is 4 marks)

Recognise, sketch and interpret graphs of linear, quadratic and simple cubic functions, the reciprocal y = 1/x (x ≠ 0), exponential y = k^x (k > 0), and y = sin x, cos x, tan x for angles of any size

Notes
Worked answers & exam appearances →
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Recognise graph families from their defining shapes and features: linear graphs have constant gradient, quadratics are symmetric parabolas, simple cubics have an S-shape, and y=1xy=\dfrac{1}{x} has two reciprocal branches with asymptotes x=0x=0 and y=0y=0.
  • Sketch by marking intercepts, roots, turning points, asymptotes and representative values.
  • Higher tier: also recognise exponentials and sine, cosine and tangent graphs for angles of any size, using their periods and standard values.
  • Examiners expect a sketch to show correct shape and position, not merely a collection of plotted points.
The reciprocal graph y=1/xy=1/x has branches in quadrants I and III and approaches, but never reaches, both axes.
Worked example

For y=6xy=\dfrac{6}{x}, state both asymptotes and the quadrants containing its branches.

  1. 1.x=0x=0 is excluded, so the vertical asymptote is x=0x=0.
  2. 2.As ∣x∣|x| grows, 6x\dfrac{6}{x} approaches 00, so the horizontal asymptote is y=0y=0.
  3. 3.xx and yy have the same sign, so the branches lie in quadrants I and III.

Answer: Asymptotes x=0x=0 and y=0y=0; branches in quadrants I and III.

Common mistakes

  • Don't draw a reciprocal branch touching or crossing an axis.
  • Don't sketch a cubic as a parabola rather than an S-shaped curve.
  • Don't treat exponential growth as a straight line (Higher tier).

Exam tip

Before sketching, list the intercepts, turning points and asymptotes that fix the graph's shape.

Tier 1 · Easy

ORIGINAL

1

A graph passes through (0,1)(0,1) and its yy-value doubles whenever xx increases by 11. Name the function y=2xy=2^x as linear, quadratic, cubic, reciprocal or exponential.

(1)

(Total for Question 1 is 1 mark)

Evidence from answers you checked

Checked automatically against the model answer once you submit.

Tier 2 · Standard

ORIGINAL

1

For the graph y=6xy=\frac6x, state both asymptotes and the two quadrants containing its branches.

(3)

(Total for Question 1 is 3 marks)

Tier 3 · Hard

ORIGINAL

1

For y=cos⁡xy=\cos x on −180∘≤x≤360∘-180^\circ\leq x\leq360^\circ, list the xx-intercepts and the coordinates of every maximum and minimum needed for an accurate sketch.

(4)

(Total for Question 1 is 4 marks)

Sketch translations and reflections of a given function [Higher only]

Notes
Worked answers & exam appearances →
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • For y=f(x)+ay=f(x)+a, translate the graph vertically by vector (0a)\begin{pmatrix}0\\a\end{pmatrix}. For y=f(x−a)y=f(x-a), translate it horizontally by (a0)\begin{pmatrix}a\\0\end{pmatrix}; the sign inside the function appears opposite to the movement.
  • The graph y=−f(x)y=-f(x) is the reflection of y=f(x)y=f(x) in the xx-axis, while y=f(−x)y=f(-x) is its reflection in the yy-axis.
  • Track distinctive points, including intercepts and turning points, and preserve the graph's exact shape and scale.
  • A point (p,q)(p,q) provides a reliable coordinate check after transforming.
  • Examiners expect a fully described transformation, including the correct axis or translation vector.
A translation moves every point by the same vector while preserving the graph's shape.
Worked example

The point (p,q)(p,q) lies on y=f(x)y=f(x). Find its image on y=f(x−3)−2y=f(x-3)-2.

  1. 1.f(x−3)f(x-3) translates the graph 33 units right.
  2. 2.Subtracting 22 outside the function translates it 22 units down.
  3. 3.Therefore (p,q)(p,q) maps to (p+3,q−2)(p+3,q-2).

Answer: The image is (p+3,q−2)(p+3,q-2), under translation by (3−2)\begin{pmatrix}3\\-2\end{pmatrix}.

Common mistakes

  • Don't move f(x−3)f(x-3) three units left instead of right.
  • Don't reflect y=f(−x)y=f(-x) in the xx-axis rather than the yy-axis.
  • Don't move only selected points and change the graph's shape.

Exam tip

For a translation, state the vector; for a reflection, name the mirror axis.

Tier 1 · Easy

ORIGINAL

1

Describe fully the transformation from y=f(x)y=f(x) to y=f(x)+4y=f(x)+4.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1

The point (3,−2)(3,-2) lies on y=f(x)y=f(x). Find the corresponding point on y=f(−x)y=f(-x) and name the transformation.

(3)

(Total for Question 1 is 3 marks)

Tier 3 · Hard

ORIGINAL

1

The point (−1,5)(-1,5) lies on y=f(x)y=f(x). Find the corresponding point on y=f(2−x)−3y=f(2-x)-3, and describe the reflection and translations that produce the new graph.

(4)

(Total for Question 1 is 4 marks)

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