A Algebra · equations and reasoning — notes on one page

Every Edexcel GCSE Maths 1MA1 specification point in A, read straight through: 8 points with definitions, explanations, mark-scheme wording and worked examples. Practice questions are in the printable question pack.

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.

How this checking works

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

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 $2n$ for an even integer or $2n+1$ 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.Let the first integer be $n$ , so the next is $n+1$ .
2.Their sum is $n+(n+1)=2n+1$ .
3. $2n$ is even for every integer $n$ , so $2n+1$ is odd.

Answer: The sum has form $2n+1$ , 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.

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)

A function maps each allowed input to one output. To evaluate $f(a)$ , substitute the complete input $a$ 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 $fg(x)=f(g(x))$ , $g$ 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 $x$ .

Worked example

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

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

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

Common mistakes

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

Exam tip. Put the complete input in brackets everywhere $x$ appears before simplifying.

A8 · Work with coordinates in all four quadrants

A coordinate $(x,y)$ gives horizontal position first and vertical position second. Positive $x$ is right, negative $x$ is left, positive $y$ is up and negative $y$ 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 $x$ -coordinates and separately averaging the two $y$ -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)

in each.

Worked example

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

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

Answer: $(-1,-1)$ .

Common mistakes

Don't write the vertical coordinate first and swap $(x,y)$ .
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.

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

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

A straight line

y=mx+c

crosses the vertical axis at

(0,c)

and has constant gradient.

Worked example

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

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

Answer: $y=-x+3$ .

Common mistakes

Don't subtract coordinates in different orders in the gradient numerator and denominator.
Don't use the $x$ -intercept as $c$ in $y=mx+c$ .
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 $c$ .

A10 · Identify and interpret gradients and intercepts of linear functions graphically and algebraically

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 $y$ -intercept is the output when $x=0$ ; the $x$ -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 $V=120-6t$ , where $V$ is litres and $t$ is minutes. Interpret the gradient and intercept.

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

Answer: The tank starts with $120$ litres and loses $6$ 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.

A11 · Identify and interpret roots, intercepts, turning points of quadratic functions graphically; deduce roots algebraically and turning points by completing the square

A root of a quadratic is an $x$ -coordinate where its graph meets the $x$ -axis, so $y=0$ . The $y$ -intercept is found by setting $x=0$ .
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 $a(x-h)^2+k$ reveals turning point $(h,k)$ .
Examiners expect coordinates for points, equations for axes, and algebraic working when roots are to be deduced.

A quadratic's roots are its

x

-axis intersections, and its turning point lies on the axis of symmetry.

Worked example

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

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

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

Common mistakes

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

Exam tip. State roots as $x$ -values, intercepts as coordinates, and the symmetry line as an equation.

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

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 $y=\dfrac{1}{x}$ has two reciprocal branches with asymptotes $x=0$ and $y=0$ .
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/x

has branches in quadrants I and III and approaches, but never reaches, both axes.

Worked example

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

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

Answer: Asymptotes $x=0$ and $y=0$ ; 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.

A13 · Sketch translations and reflections of a given function [Higher only]

For $y=f(x)+a$ , translate the graph vertically by vector $\begin{pmatrix}0\\a\end{pmatrix}$ . For $y=f(x-a)$ , translate it horizontally by $\begin{pmatrix}a\\0\end{pmatrix}$ ; the sign inside the function appears opposite to the movement.
The graph $y=-f(x)$ is the reflection of $y=f(x)$ in the $x$ -axis, while $y=f(-x)$ is its reflection in the $y$ -axis.
Track distinctive points, including intercepts and turning points, and preserve the graph's exact shape and scale.
A point $(p,q)$ 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 $(p,q)$ lies on $y=f(x)$ . Find its image on $y=f(x-3)-2$ .

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

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

Common mistakes

Don't move $f(x-3)$ three units left instead of right.
Don't reflect $y=f(-x)$ in the $x$ -axis rather than the $y$ -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.