EduNinja

C2.1 Introduction to algebra

A letter can represent any number allowed by a situation. An algebraic expression then describes one calculation that works for every permitted value of that letter.

Notation	Meaning	Example in context
$x$	one variable quantity	number of pens
$60x$	60 multiplied by $x$	cost in cents of $x$ pens at 60 cents each
$60x+29y$	sum of two variable costs	cost of $x$ pens and $y$ rulers
$d/n$	one total shared among $n$ equal parts	cost of one bag when $n$ bags cost $d$ dollars

Identify what each letter counts or measures, attach its rate or multiplier, then combine parts using the action in the context. If a team earns 3 points per win and 1 per draw, $w$ wins and $d$ draws give $3w+d$ points.

Order matters when quantities enter or leave. If a train starts with $p$ passengers, then $x$ get off and $y$ get on, the new number is $p-x+y$ . The expression remains general because it works for any valid $p$ , $x$ and $y$ .

A formula names the quantity produced by an expression. A car hire charge of 56 dollars per day plus a fixed 436 dollars can be written $C=56d+436$ , where $C$ is the total cost for $d$ days.

Adjacent symbols indicate multiplication: $60x$ means $60\times x$ , not the two-digit number '60x'. A letter does not always mean one fixed unknown; it may vary across cases. Keep units consistent before forming a general expression.

Substitution replaces each letter with its given numerical value while preserving the original operations, brackets and powers.

Step	Action	Check
1	write the expression or formula unchanged	every symbol is present
2	replace each letter with its value in brackets	negative values stay grouped
3	evaluate powers, then multiplication/division, then addition/subtraction	follow operation order
4	attach the required unit and apply requested rounding	interpret the result

For $s=ut+\dfrac12at^2$ with $u=5.2$ , $t=7$ and $a=1.6$ , substitute first: $s=5.2(7)+\dfrac12(1.6)(7)^2=36.4+39.2=75.6$ . The square applies to the substituted value of $t$ .

Brackets protect signs. For $8x-3y$ with $x=5$ and $y=-2$ , write $8(5)-3(-2)=40+6=46$ . Without brackets, the second negative can be lost.

If a formula needs an intermediate quantity, calculate it explicitly. For $M=2.5AT$ , a rectangle 1.9 m by 0.6 m has $A=1.9\times0.6=1.14$ m²; with $T=8$ , $M=2.5(1.14)(8)=22.8$ kg.

Substitution evaluates a given expression; it does not change the formula or solve for a different letter. In $3t^2$ , square only $t$ before multiplying by 3, and for $t=-4$ use $3(-4)^2$ , not $3(-4^2)$ .

C2.2 Algebraic manipulation

Like terms have exactly the same variable part, including the same powers. They represent the same kind of quantity, so only their numerical coefficients can be combined.

Pair	Like terms?	Reason
$3a$ and $-5a$	yes	both have variable part $a$
$4y^2$ and $-y^2$	yes	both have variable part $y^2$
$2a$ and $2b$	no	different variables
$x$ and $x^2$	no	different powers

Group each set of like terms, add or subtract their coefficients, and keep the common variable part unchanged. For example, $2a+3b+5a-9b=(2+5)a+(3-9)b=7a-6b$ .

Treat different powers as different groups. Thus $4y^2+3y-y^2+2y=(4-1)y^2+(3+2)y=3y^2+5y$ . Neither $y^2$ nor $y$ can be absorbed into the other.

A sign belongs to the term immediately after it. In $a-2b-3a+7b$ , collect $a-3a=-2a$ and $-2b+7b=5b$ , giving $-2a+5b$ .

Do not combine unlike terms by adding their coefficients: $3a+4b$ is already simplified. Collecting terms changes coefficients, not variables or powers; for instance, $3x+7x-4x=6x$ , not $6x^2$ .

Expansion uses the distributive law to replace a product containing brackets with an equivalent sum or difference. Every term in one factor must multiply every term in the other factor.

Form	Products to write	Then simplify
$k(a+b)$	$ka+kb$	collect if possible
$k(a-b)$	$ka-kb$	keep the second sign
$(a+b)(c+d)$	$ac+ad+bc+bd$	collect like middle terms

Multiply the outside term by each term inside. The syllabus example becomes $3x(2x-4y)=3x\cdot2x+3x\cdot(-4y)=6x^2-12xy$ . The variable products must also be multiplied.

For $(2x+1)(x-4)$ , write all four products: $2x^2-8x+x-4$ . The two middle terms are like terms, so the simplified result is $2x^2-7x-4$ .

A minus before a bracket changes the sign of every term in that bracket. For $4(x-5)-(3-2x)$ , expansion gives $4x-20-3+2x=6x-23$ .

Do not stop after multiplying only the first terms, and do not drop a negative product. Expansion rewrites an expression; it does not solve an equation. A quick check is that substituting the same value of $x$ before and after expansion gives the same result.

Factorising by a common factor reverses expansion: write an expression as a product by taking outside the greatest factor shared by every term.

Part of the common factor	How to choose it	Example
number	greatest common divisor of all coefficients	for 9 and 15, choose 3
variable	include only variables present in every term	$x$ is common to $x^2$ and $xy$
power	use the smallest shared power	$a^2$ and $ab$ share $a$

Find the greatest common factor, divide every term by it to form the bracket, then multiply back to check. For $9x^2+15xy$ , the greatest common factor is $3x$ , so $9x^2+15xy=3x(3x+5y)$ .

Factorisation is complete only when the bracket has no further common factor. For $18x^2-12x$ , $2x(9x-6)$ is only partial; extracting $6x$ gives the fully factorised form $6x(3x-2)$ .

Preserve the operation signs when dividing each term. For $5x-20x^2$ , extracting $5x$ gives $5x(1-4x)$ because $5x\div5x=1$ and $-20x^2\div5x=-4x$ .

Only extract a factor shared by every term. Do not use quadratic-factor methods here: this Core objective is common-factor extraction. Expansion of the final product must reproduce the original expression exactly.

C2.4 Indices II

In $a^n$ , $a$ is the base and $n$ is the index. The index describes repeated multiplication when it is positive, and the same pattern extends consistently to zero and negative indices.

Index type	Meaning	Example
positive	multiply the base by itself	$3^4=3 imes3 imes3 imes3=81$
zero	the value is 1 for a non-zero base	$7x^0=7 imes1=7$
negative	take the reciprocal of the matching positive power	$x^{-3}=1/x^3$ for $x
e0$

Each step down in the index divides by the base: $5^3=125$ , $5^2=25$ , $5^1=5$ , $5^0=1$ , $5^{-1}=1/5$ . This pattern explains both the zero-index and negative-index meanings.

The index applies only to its stated base. In $(-2)^4$ , the base is $-2$ , so the value is 16. In $-2^4$ , the power applies to 2 first, so the value is $-16$ .

A negative index does not make the value negative: $y^{-2}=1/y^2$ . It moves a non-zero factor across a fraction bar and changes the sign of the index.

$a^0=1$ and $a^{-n}=1/a^n$ require $a
e0 $; division by zero is undefined. Do not multiply the base by the index:$ 4^3 $means$ 4 imes4 imes4 $, not$ 4 imes3$.

Index laws preserve the meaning of repeated multiplication. First identify the operation and confirm that the bases match; then change the indices using the corresponding rule.

Structure	Rule	Example
same base, multiply	$a^m a^n=a^{m+n}$	$x^3x^5=x^8$
same base, divide	$a^m/a^n=a^{m-n}$	$y^3/y^5=y^{-2}=1/y^2$
power of a power	$(a^m)^n=a^{mn}$	$(w^5)^4=w^{20}$
power of a product	$(ab)^n=a^nb^n$	$(3x^2y^4)^3=27x^6y^{12}$

Treat numerical coefficients separately from variable powers. For $6x^7y^4 imes5x^{-5}y$ , multiply coefficients and add indices for each matching base: $30x^{7+(-5)}y^{4+1}=30x^2y^5$ .

For $12a^5\div3a^{-2}$ , divide coefficients and subtract the denominator index: $4a^{5-(-2)}=4a^7$ . Subtracting a negative index increases the result's index.

To find an unknown index, write both sides with the same base and equate indices. Since $2^x=32=2^5$ , $x=5$ . Likewise, $9^{12}/9^w=9^4$ gives $12-w=4$ , so $w=8$ ; logarithms are not required.

Add or subtract indices only when multiplying or dividing powers with the same base. There is no rule that turns $a^m+a^n$ into $a^{m+n}$ , and $(a^m)^n$ uses multiplication of indices, not addition.

C2.5 Equations

Algebra translates quantities and relationships into symbols. Decide what each letter represents, express each quantity once, and use an equals sign only when two complete quantities have the same value.

Algebraic statement	Purpose	Example
expression	represents a quantity; no equals sign	2 more than $n$ : $n+2$
equation	states that two quantities are equal	$3n-5=22$
formula	defines one quantity using others	$C=56d+436$
simultaneous equations	two conditions on the same two unknowns	$3a+4c=38.5$ , $6a+5c=65$

Translate operations in their stated order. 'Three less than twice $x$ ' is $2x-3$ , while 'twice three less than $x$ ' is $2(x-3)$ . Brackets preserve which operation happens first.

Build a total by adding every contribution once. If Rovers score $x$ , United score $x+8$ , and City score $2x-3$ , a total of 117 gives $x+(x+8)+(2x-3)=117$ .

Use the same letters with the same meanings in both conditions. If a minibus costs $h$ dollars per hire-hour and $p$ dollars per passenger, the two trips give $3h+10p=61$ and $5h+8p=80$ .

Do not insert an equals sign into a requested expression, and do not solve unless asked. Keep units consistent: $3a+4c=38.5$ uses dollar prices, whereas cent prices would require 3850 on the right.

An equation is balanced because both sides have equal value. Any operation applied to one side must be applied to the other side; use inverse operations until the unknown is isolated.

Step	Action	Example for $12x-3=4x+21$
1	remove brackets or fractions if useful	none here
2	collect unknown terms on one side	$8x-3=21$
3	collect constants on the other side	$8x=24$
4	divide by the coefficient	$x=3$
5	substitute back to check	both sides equal 33

Either divide by an outside factor first or expand correctly. For $5(2x+4)=85$ , divide by 5 to get $2x+4=17$ , then $2x=13$ and $x=6.5$ .

Clear fractions by multiplying every term by a common denominator. For $3w/16-1=1/2$ , multiply all terms by 16: $3w-16=8$ , so $3w=24$ and $w=8$ .

When the unknown appears on both sides, move all unknown terms together before dividing. From $5x-6=x+3$ , subtract $x$ and add 6 to obtain $4x=9$ , so $x=2.25$ .

A term does not simply 'move and change sign'; the sign changes because the same inverse operation is performed on both sides. Never divide by the coefficient until every remaining term on that side belongs to the unknown term.

A simultaneous solution is one pair of values that makes both equations true at the same time. Elimination combines equivalent equations so that one unknown cancels.

Step	Action
1	align like terms and equals signs
2	multiply one or both equations so one pair of coefficients is equal or opposite
3	add or subtract the whole equations to eliminate that variable
4	solve the resulting one-variable equation
5	substitute back to find the second value and check both equations

Solve $5x-2y=44$ and $2x+3y=10$ . Multiply the first equation by 3 and the second by 2: $15x-6y=132$ and $4x+6y=20$ . Add to get $19x=152$ , so $x=8$ . Then $2(8)+3y=10$ , giving $y=-2$ .

Choose addition when coefficients are opposite and subtraction when they are equal. If neither is ready, multiply complete equations first; every term, including the right-hand side, must be multiplied.

In a context, define both unknowns and form two equations before eliminating. For adult price $x$ and child price $y$ , the conditions $2x+3y=15$ and $3x+5y=23.5$ use the same meanings and lead to one common price pair.

Finding values that satisfy only one equation is not enough. Substitute the final pair into both original equations. Do not add or subtract selected terms from an equation while leaving its other terms unchanged.

The subject of a formula is the single letter isolated on one side. To change the subject, preserve equality while undoing the operations around the required letter in reverse order.

Operation around the new subject	Apply to both sides
add $c$	subtract $c$
subtract $c$	add $c$
multiply by $m$	divide by $m$
divide by $m$	multiply by $m$

Make $x$ the subject of $y=mx+c$ . First subtract $c$ : $y-c=mx$ . Then divide by $m$ : $x=(y-c)/m$ , where $m
e0$. The numerator brackets show that the entire difference is divided.

Make $m$ the subject of $q=m/7+r$ . Subtract $r$ to get $q-r=m/7$ , then multiply by 7: $m=7(q-r)$ . Multiplying out to $7q-7r$ is equivalent.

For $5w-3y+7=0$ , isolate the term containing $w$ : $5w=3y-7$ . Divide by 5 to obtain $w=(3y-7)/5$ .

This Core objective uses formulas where the new subject appears once and is not under a power or root. Do not divide only one term of a numerator: $(y-c)/m$ means both $y$ and $-c$ are divided by $m$ .

C2.6 Inequalities

An inequality describes a set of possible values rather than one value. Its symbol tells both the boundary and whether the boundary value itself belongs to the set.

Inequality	Read as	Endpoint on a number line	Direction from the endpoint
$x<a$	$x$ is less than $a$	open circle	left
$x>a$	$x$ is greater than $a$	open circle	right
$x\le a$	$x$ is at most $a$	closed circle	left
$x\ge a$	$x$ is at least $a$	closed circle	right

An open circle means the endpoint is excluded, matching $<$ or $>$ . A closed circle means the endpoint is included, matching $\le$ or $\ge$ . The line or ray shows all the other permitted values.

A bounded interval combines two conditions. The syllabus example $-3\le x<1$ means values from $-3$ up to but not including 1: use a closed endpoint at $-3$ , an open endpoint at 1, and join the points between them.

When reading a number line, inspect each endpoint before writing the compound inequality. A closed point at $-5$ and open point at 1 with the interval between them gives $-5\le x<1$ ; keep the smaller boundary on the left.

If only integer solutions are requested, list the integers inside the interval. For $-3\le x<3$ , they are $-3,-2,-1,0,1,2$ : include $-3$ because of $\le$ , but exclude 3 because of $<$ .

Do not decide endpoint inclusion from the direction of the line: inclusion depends only on whether the circle is closed. This Core objective represents and interprets given inequalities; solving inequalities algebraically is outside this card.

C2.7 Sequences

To continue a sequence, identify the operation that connects consecutive terms and check that it works across every given step before applying it again.

Given terms	Changes	Continuation
$7,13,19,25$	add 6 each time	$31$
$3,9,27,81$	multiply by 3 each time	$243$
$27,26,23,18,11$	subtract $1,3,5,7$	subtract 9, then 11: $2,-9$
$18,21,26,33,42$	add $3,5,7,9$	add 11: $53$

Write the first differences beneath the terms. If they are constant, repeat that difference. If they form a simple sequence themselves, continue the difference pattern first and then use the new difference on the last term.

Differences are not always the controlling pattern. For $18,20,24,32,48$ , the added amounts double: $+2,+4,+8,+16$ , so add 32 next to obtain 80.

Use at least three transitions to test a proposed rule. A rule that fits only the final pair may not explain the whole sequence; after predicting a term, verify that its change continues the same pattern.

Continue only the stated or strongly supported pattern. A short finite list can fit many invented rules, so prefer the simplest consistent rule signalled by all given terms; finding a general nth term is handled in the later card.

A term-to-term rule tells how to obtain the next term from the current one. A pattern table may contain several linked sequences, so compare both changes within each row and relationships between rows.

Evidence	Likely rule to test
constant first differences	add or subtract a fixed amount
constant ratio	multiply or divide by a fixed amount
differences follow a pattern	continue the difference sequence
two rows change together	test a sum, difference, multiple or square relationship

For $30,26,22,18$ , each term is 4 less than the previous one, so the term-to-term rule is 'subtract 4'. State the operation and amount, not merely the next term.

A rule can be reversed. If 'multiply by 3, then subtract 1' produces the fourth term 68, undo subtraction first and multiplication second: $(68+1)/3=23$ , so the third term is 23.

In a pattern table, track each row separately before comparing rows. If small-square counts are $2,4,6,8$ while dot counts are $6,9,12,15$ , their term-to-term rules are add 2 and add 3; the rows share the same diagram number but not the same change.

A term-to-term rule uses the preceding term; an nth-term rule uses the position number directly. Do not call 'add 6' the nth term, and do not assume two linked rows must have identical differences.

An nth-term formula gives a term directly from its position $n$ . Constant first, second or third differences reveal whether a sequence is linear, quadratic or cubic.

Sequence type	Constant layer	Leading coefficient	Starting model
linear	first difference $d$	$d$	$dn+b$
quadratic	second difference $D_2$	$a=D_2/2$	$an^2+bn+c$
cubic	third difference $D_3$	$a=D_3/6$	$an^3+bn^2+cn+d$

For $12,19,26,33,40$ , the first difference is 7. Start with $7n$ ; its first term is 7, which is 5 too small, so the nth term is $7n+5$ .

For $2,5,10,17$ , first differences are $3,5,7$ and the second difference is 2, so the leading term is $n^2$ . Since $n^2$ gives $1,4,9,16$ , add 1 to every term: the nth term is $n^2+1$ .

For $3,17,55,129,251$ , the constant third difference is 12, so the leading term is $(12/6)n^3=2n^3$ . Comparing with $2,16,54,128,250$ shows a constant remainder 1, giving $2n^3+1$ .

To generate terms, substitute $n=1,2,3,\ldots$ . To find a position, set the formula equal to the target and solve for a positive integer $n$ . For $7n+5=40$ , $n=5$ , so 40 is the fifth term.

Differences diagnose the degree; do not use the linear shortcut when first differences change. Sequence positions begin at $n=1$ unless stated otherwise, and a non-integer or non-positive solution for $n$ is not a valid term position.

C2.9 Graphs in practical situations

A practical graph connects two measured quantities. Read the axis labels, units and scale first; then interpret coordinates and gradients in the situation rather than as isolated numbers.

Feature on a distance–time graph	Meaning
rising segment	distance from the start increases
horizontal segment	distance is unchanged, so the traveller is stationary
falling segment	the traveller moves back towards the start
steeper segment	greater distance change per unit time, so greater speed
intersection of two journeys	same distance from the reference point at the same time

For a straight segment, $ext{gradient}=\Delta ext{distance}/\Delta ext{time}$ , so its units are a rate such as km/h. A rise of 6 km in 40 minutes is $6/(40/60)=9$ km/h.

To read a value, start at the known quantity on one axis, move to the graph, then move parallel to the other axis to read the corresponding quantity. Interpolate between scale marks rather than rounding too early.

A conversion graph maps one unit or currency to another. If the line shows 1 dollar corresponds to 110 yen, $100 corresponds to 11000 yen. To convert in the reverse direction, start from the yen axis and read back to dollars.

When comparing two cost or journey graphs, the lower cost is the lower vertical value at the same horizontal input. At an intersection the values are equal; which graph is cheaper can change after the crossing.

A horizontal distance–time segment means stopped, not travelling at constant speed. A falling segment can still represent positive speed towards the start; the sign records direction of distance change, while speed is the magnitude of the gradient.

A practical graph is built from ordered data pairs. Choose labelled axes and a scale that uses the grid well, plot each pair accurately, and join points in the way the context supports.

Step	Drawing decision
1	put the independent quantity, usually time, on the horizontal axis
2	label both axes with quantities and units
3	choose simple uniform scales covering all values
4	calculate any missing endpoint values or times
5	plot coordinates accurately and join required straight journey segments
6	check every endpoint against the narrative

Starting at 15:00 and distance 0, walking for 20 minutes at 4.5 km/h covers $4.5 imes20/60=1.5$ km, so draw from $(15{:}00,0)$ to $(15{:}20,1.5)$ . Running a further 6 km in 40 minutes ends at $(16{:}00,7.5)$ .

A stop is a horizontal segment from the arrival time to the departure time. A return to the starting point is a segment ending at distance 0; calculate its endpoint time from $ext{time}= ext{distance}/ ext{speed}$ .

For a proportional conversion, include the origin. If 1 litre is 0.22 gallons, useful pairs include $(0,0)$ and $(100,22)$ ; plot them and draw one straight line through both.

When data are supplied in a table, keep coordinate order consistent and use the graph type requested. Do not force a line through the origin unless the context states zero of one quantity corresponds to zero of the other.

On a distance–time graph, plot cumulative distance, not speed. Constant speed appears as a straight sloping segment; a faster stage is represented by a steeper gradient rather than a larger labelled speed value on the vertical axis.

C2.10 Graphs of functions

A table of values turns a function rule into coordinates. Substitute each chosen $x$ -value to calculate $y$ , plot each $(x,y)$ pair on uniform scales, then join the points in the way the function requires.

Function form	Graph to recognise	Construction boundary
$y=ax+b$	a straight line	two accurate points determine the line; extra points check it
$y=\pm x^2+ax+b$	a smooth parabola	$+x^2$ opens upwards and $-x^2$ opens downwards
$y=\dfrac{a}{x}$ , $x\ne0$	two separate reciprocal branches	calculate values on both sides of zero and never join across $x=0$

Use this order: (1) copy the required $x$ -values; (2) substitute with brackets, especially for negative $x$ ; (3) check the table for arithmetic errors; (4) label axes and choose scales that cover every coordinate; (5) plot small accurate crosses; (6) draw a ruled straight line or a single smooth curve through the plotted pattern.

For $y=x^2+2x-4$ , the values at $x=-2,-1,0,1$ are $-4,-5,-4,-1$ . For instance, at $x=-2$ , $y=(-2)^2+2(-2)-4=-4$ . These coordinates form part of a smooth upward-opening parabola.

Interpret a graph by reading coordinates and visible features. The $y$ -intercept is where $x=0$ ; an $x$ -intercept is where $y=0$ . Read values from the stated scale and interpolate carefully between grid lines.

Do not connect reciprocal branches through $x=0$ : $a/x$ is undefined there. Do not replace a smooth quadratic or reciprocal curve with straight segments between plotted points.

A graphical solution is an $x$ -value where the required graphs have the same $y$ -value. The equation decides which intersection to read.

Equation	What to find on the graph
$f(x)=0$	where $y=f(x)$ crosses or touches the $x$ -axis
$f(x)=k$	intersections of $y=f(x)$ with the horizontal line $y=k$
$f(x)=g(x)$	intersections of the two graphs $y=f(x)$ and $y=g(x)$

Draw any additional line requested, identify every relevant intersection, then project vertically to the $x$ -axis. Record all $x$ -values allowed by the displayed domain. If coordinates are requested, read both $x$ and $y$ from each intersection.

To solve $x^2+2x-4=2x+2$ graphically, use the parabola $y=x^2+2x-4$ and the line $y=2x+2$ . Their intersections have equal $y$ -values, so the corresponding $x$ -coordinates are the solutions of the equation.

Graphical answers are usually approximate. Use the finest grid interval to estimate between scale marks, give sensible precision, and check that each reported point lies on both graphs. The number of intersections is the number of graphical solutions visible in the stated domain.

A root of $f(x)=0$ is an $x$ -coordinate, not the full coordinate pair. For an intersection equation, do not read where either graph crosses an axis unless that point is also an intersection of the required graphs.

C2.11 Sketching curves

A sketch shows the correct graph family and its defining features without requiring a full table of values or exact plotting. Mark roots and intercepts clearly, then draw the shape consistently with them.

Feature	Linear graph	Quadratic graph
overall shape	one straight line	one smooth parabola
roots	where the line crosses the $x$ -axis	where the curve crosses or touches the $x$ -axis
symmetry	not required	a vertical line; paired points at the same height lie equally far from it
direction	rises or falls at a constant rate	opens upwards or downwards

For $y=-2x+4$ , the $y$ -intercept is $(0,4)$ and the root is $(2,0)$ . Mark these points and join them with one straight falling line. A linear sketch must not bend.

For $y=(x+1)(x-5)$ , the roots are $x=-1$ and $x=5$ . Their midpoint is $(-1+5)/2=2$ , so the line of symmetry is $x=2$ . Since the coefficient of $x^2$ is positive, draw a smooth upward-opening curve through both roots, mirrored about $x=2$ . The $y$ -intercept is $(0,-5)$ , which helps place the curve.

To interpret a quadratic graph, read each root as an $x$ -value where $y=0$ . Read its symmetry line as $x=k$ : points with equal $y$ -values occur at equal horizontal distances on either side of $x=k$ . If the curve only touches the $x$ -axis, that root lies on the symmetry line.

Do not calculate or label an exact turning point unless the question supplies enough information and asks for it; C2.11 requires roots and symmetry, not turning-point knowledge. A sketch still needs the correct intercepts, symmetry and opening direction.

2. Algebra and graphs

Exam analysis

Most tested topics

C2.1 Introduction to algebra

Use letters to describe general quantities

Substitute values into expressions and formulas

C2.2 Algebraic manipulation

Collect like terms without changing their variable part

Expand every product across algebraic brackets

Factorise fully by extracting the greatest common factor

C2.4 Indices II

Interpret positive, zero and negative indices

Choose and apply the index laws

C2.5 Equations

Translate situations into expressions, equations and formulas

Solve linear equations by preserving balance

Solve simultaneous linear equations by eliminating one unknown

Change the subject using inverse operations

C2.6 Inequalities

Represent and interpret inequality intervals

C2.7 Sequences

Continue sequences from their changing pattern

Recognise term-to-term rules and linked sequence patterns

Find and use linear, quadratic and cubic nth terms

C2.9 Graphs in practical situations

Interpret travel and conversion graphs in context

Draw practical graphs from data and journey stages

C2.10 Graphs of functions

Construct and recognise linear, quadratic and reciprocal graphs

Solve equations by reading roots and intersections

C2.11 Sketching curves

Sketch linear and quadratic graphs from their key features