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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.

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A6

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
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Set state:

The Secure button is a self-rating. Evidence-secure needs the latest Tier 2/3 attempt correct, plus 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 make this mistake: Higher tier: checks several numerical cases and calls the pattern a proof.
  • Don't make this mistake: Higher tier: finishes with algebra but does not state why its form proves the claim.

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

2 marks
ORIGINAL

Show that 4(n+2)3(n1)4(n+2)-3(n-1) is equivalent to n+11n+11.

Tier 2 · Standard

3 marks
ORIGINAL

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

Tier 3 · Hard

4 marks
ORIGINAL

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

A7

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
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Set state:

The Secure button is a self-rating. Evidence-secure needs the latest Tier 2/3 attempt correct, plus 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)=3x4f(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)=6a4f(2a)=6a-4.

Answer: f(6)=14f(6)=14 and f(2a)=6a4f(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 make this mistake: Higher tier: reads f1(x)f^{-1}(x) as 1f(x)\dfrac{1}{f(x)}.

Exam tip

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

Tier 1 · Easy

1 mark
ORIGINAL

Given f(x)=3x4f(x)=3x-4, work out f(6)f(6).

Tier 2 · Standard

3 marks
ORIGINAL

Given f(x)=5x+2f(x)=5x+2, find f1(x)f^{-1}(x) and work out f1(27)f^{-1}(27).

Tier 3 · Hard

4 marks
ORIGINAL

Let f(x)=2x+3f(x)=2x+3 and g(x)=x21g(x)=x^2-1. Solve fg(x)=19fg(x)=19, where fg(x)=f(g(x))fg(x)=f(g(x)).

A8

Work with coordinates in all four quadrants

Notes
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Set state:

The Secure button is a self-rating. Evidence-secure needs the latest Tier 2/3 attempt correct, plus 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.
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 swaps (x,y)(x,y).
  • Don't lose a negative sign when averaging coordinates.
  • Don't find half the coordinate differences but forgets 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

1 mark
ORIGINAL

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

Tier 2 · Standard

2 marks
ORIGINAL

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

Tier 3 · Hard

3 marks
ORIGINAL

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.

A9

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
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Set state:

The Secure button is a self-rating. Evidence-secure needs the latest Tier 2/3 attempt correct, plus 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=y2y1x2x1m=\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.
Worked example

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

  1. 1.m=154(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 make this mistake: Higher tier: changes only the sign of a gradient to make a perpendicular line.

Exam tip

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

Tier 1 · Easy

1 mark
ORIGINAL

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

Tier 2 · Standard

3 marks
ORIGINAL

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.

Tier 3 · Hard

4 marks
ORIGINAL

Find the equation of the line through (3,5)(3,-5) that is perpendicular to 2x3y=62x-3y=6. Give your answer in the form y=mx+cy=mx+c.

A10

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

Notes
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Set state:

The Secure button is a self-rating. Evidence-secure needs the latest Tier 2/3 attempt correct, plus 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.
Worked example

Water volume follows V=1206tV=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

2 marks
ORIGINAL

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

Tier 2 · Standard

3 marks
ORIGINAL

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

Tier 3 · Hard

4 marks
ORIGINAL

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.

A11

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

Notes
Unseen
Set state:

The Secure button is a self-rating. Evidence-secure needs the latest Tier 2/3 attempt correct, plus 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(xh)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.
Worked example

Find the roots and yy-intercept of y=x25x+6y=x^2-5x+6.

  1. 1.Factorise: x25x+6=(x2)(x3)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 make this mistake: Higher tier: reads (xh)2+k(x-h)^2+k as having turning point (h,k)(-h,k).

Exam tip

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

Tier 1 · Easy

2 marks
ORIGINAL

Find the roots and the yy-intercept of y=x25x+6y=x^2-5x+6.

Tier 2 · Standard

3 marks
ORIGINAL

For y=(x4)29y=(x-4)^2-9, state the turning point and axis of symmetry, and find the roots.

Tier 3 · Hard

4 marks
ORIGINAL

Complete the square for y=2x2+8x10y=2x^2+8x-10. Hence state the turning point and find the roots.

A12

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
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Set state:

The Secure button is a self-rating. Evidence-secure needs the latest Tier 2/3 attempt correct, plus 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.
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 sketche a cubic as a parabola rather than an S-shaped curve.
  • Don't make this mistake: Higher tier: treats exponential growth as a straight line.

Exam tip

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

Tier 1 · Easy

1 mark
ORIGINAL

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.

Tier 2 · Standard

3 marks
ORIGINAL

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

Tier 3 · Hard

4 marks
ORIGINAL

For y=cosxy=\cos x on 180x360-180^\circ\leq x\leq360^\circ, list the xx-intercepts and the coordinates of every maximum and minimum needed for an accurate sketch.

A13

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

Notes
Unseen
Set state:

The Secure button is a self-rating. Evidence-secure needs the latest Tier 2/3 attempt correct, plus 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(xa)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.
Worked example

The point (p,q)(p,q) lies on y=f(x)y=f(x). Find its image on y=f(x3)2y=f(x-3)-2.

  1. 1.f(x3)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,q2)(p+3,q-2).

Answer: The image is (p+3,q2)(p+3,q-2), under translation by (32)\begin{pmatrix}3\\-2\end{pmatrix}.

Common mistakes

  • Don't move f(x3)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 changes the graph's shape.

Exam tip

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

Tier 1 · Easy

2 marks
ORIGINAL

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

Tier 2 · Standard

3 marks
ORIGINAL

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.

Tier 3 · Hard

4 marks
ORIGINAL

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

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