Mini-Lesson

Algebra & Graphs

11 missions. Guess, learn, then check. Your place is saved.

Mission 1A square grows 3 cm each way. Is its new area just $x^{2} + 9$ x² + 9?
Mission 2Is there one temperature that reads the same in °C and °F?
Mission 3When does the pricier phone plan become the cheaper one?
Mission 4A thrown ball is 10 m up at two different times. When, and why?
Mission 53 coffees + 2 cakes = £13.50; 2 coffees + 3 cakes = £13. A cake?
Mission 6Four graphs, four equations: can you match them without plotting?
Mission 7A machine doubles your number, then adds 3. How do you undo it?
Mission 8Chessboard: 1 grain, then double each square. How many on square 64?
Mission 9Two runners' distance–time graphs cross. What does the crossing mean?
Mission 10Cycle out at 10 km/h, back at 30 km/h. Is your average 20 km/h?
Mission 1140 m of fence and a wall: which rectangle encloses the most area?

The specification points and sources are listed under the lesson: see Sources.

Mission 1

A square grows 3 cm each way. Is its new area just $x^{2} + 9$ x² + 9?

Make a guess: A square grows 3 cm each way. Is its new area just $x^{2} + 9$ x² + 9?

Algebra · expressions

Terms, expressions and like terms

An expression is a collection of terms such as $5 a$ 5a, $- 9 b$ −9b or $x^{2}$ x². The number in front of a letter is its coefficient.
Algebra shorthand: $ab$ ab means $a \times b$ a × b, $3 y$ 3y means $y + y + y$ y + y + y, $a^{2} b$ a²b means $a \times a \times b$ a × a × b, $\frac{a}{b}$ a/b means $a \div b$ a ÷ b.
Like terms have exactly the same letters and powers. Collect them by adding their coefficients: $2 a + 3 b + 5 a - 9 b = 7 a - 6 b$ 2a + 3b + 5a − 9b = 7a − 6b.
$x^{2}$ x² and $x$ x are not like terms: $3 x^{2} + 2 x$ 3x² + 2x cannot be simplified.
To expand a single bracket, multiply every term inside by the term outside: $3 x (2 x - 4 y) = 6 x^{2} - 12 x y$ 3x(2x − 4y) = 6x² − 12xy and $- 4 (x - 3) = - 4 x + 12$ −4(x − 3) = −4x + 12.

Quick check

?Expand and simplify

3 (2 x - 1) - 2 (x - 4)

.

Algebra · expressions

Expanding two brackets

Multiply every term in the first bracket by every term in the second. A grid keeps track: four boxes, four products.
$(x + 5) (x + 2)$ (x + 5)(x + 2): the boxes are $x^{2}$ x², $2 x$ 2x, $5 x$ 5x and $10$ 10. Collect the like terms: $x^{2} + 7 x + 10$ x² + 7x + 10.
The grid is also an area picture: a rectangle $(x + 5)$ (x + 5) by $(x + 2)$ (x + 2) splits into four smaller rectangles.
Extended: for three brackets, expand two of them first, then multiply the result by the third: $(x + 1) (x - 1) (2 x + 1) = (x^{2} - 1) (2 x + 1) = 2 x^{3} + x^{2} - 2 x - 1$ (x + 1)(x − 1)(2x + 1) = (x² − 1)(2x + 1) = 2x³ + x² − 2x − 1.

Worked example Algebra · expressions

Expanding and simplifying two brackets

Question

Expand and simplify

(2 x + 1) (x - 4)

.

1
Multiply the first terms: $2 x \times x = 2 x^{2}$ 2x × x = 2x²
2
Multiply the outer and inner pairs: $2 x \times (- 4) = - 8 x$ 2x × (−4) = −8x and $1 \times x = x$ 1 × x = x
3
Multiply the last terms: $1 \times (- 4) = - 4$ 1 × (−4) = −4
4
Collect the like terms: $- 8 x + x = - 7 x$ −8x + x = −7x

Answer

2 x^{2} - 7 x - 4

⚠️ Watch out

the sign belongs to the term after it. The

- 4

in the second bracket makes two of the four products negative.

Calculate

?Expand

(x - 6)^{2}

and write it in the form

x^{2} + b x + c

. Enter the value of

b

(exact).

Algebra · expressions

Factorising: expanding in reverse

Factorise means write as a product. Take out the highest common factor of every term: $9 x^{2} + 15 x y = 3 x (3 x + 5 y)$ 9x² + 15xy = 3x(3x + 5y). "Factorise fully" means nothing more can come out.
A quadratic $x^{2} + b x + c$ x² + bx + c factorises to $(x + p) (x + q)$ (x + p)(x + q) where $p$ p and $q$ q multiply to $c$ c and add to $b$ b: $x^{2} + 7 x + 10 = (x + 5) (x + 2)$ x² + 7x + 10 = (x + 5)(x + 2).
Difference of two squares: $x^{2} - a^{2} = (x - a) (x + a)$ x² − a² = (x − a)(x + a), so $x^{2} - 16 = (x - 4) (x + 4)$ x² − 16 = (x − 4)(x + 4).
Check any factorisation by expanding it again.
Extended: factorising quadratics is Extended content. You also need grouping, $x y + 3 x + 2 y + 6 = x (y + 3) + 2 (y + 3) = (x + 2) (y + 3)$ xy + 3x + 2y + 6 = x(y + 3) + 2(y + 3) = (x + 2)(y + 3), the perfect square $a^{2} + 2 ab + b^{2} = (a + b)^{2}$ a² + 2ab + b² = (a + b)², and cubics such as $2 x^{3} + 5 x^{2} - 3 x = x (2 x - 1) (x + 3)$ 2x³ + 5x² − 3x = x(2x − 1)(x + 3) (take out $x$ x first).

Quick check

?Factorise

x^{2} + 2 x - 24

.

Worked example Algebra · expressions

Factorising $a x^{2} + b x + c$ ax² + bx + c (Extended)

Question

Factorise

6 x^{2} - 5 x - 6

.

1
Multiply $a$ a by $c$ c: $6 \times (- 6) = - 36$ 6 × (−6) = −36
2
Find two numbers that multiply to $- 36$ −36 and add to $- 5$ −5: $- 9$ −9 and $4$ 4
3
Split the middle term with them: $6 x^{2} - 9 x + 4 x - 6$ 6x² − 9x + 4x − 6
4
Factorise each pair: $3 x (2 x - 3) + 2 (2 x - 3)$ 3x(2x − 3) + 2(2x − 3)
5
Take out the common bracket: $(2 x - 3) (3 x + 2)$ (2x − 3)(3x + 2)

Answer

(2 x - 3) (3 x + 2)

⚠️ Watch out

check by expanding.

(2 x - 3) (3 x + 2) = 6 x^{2} + 4 x - 9 x - 6 = 6 x^{2} - 5 x - 6

.

Calculate

?A square photo of side

x

cm sits in a square frame of side

(x + 4)

cm. The visible frame (the outer square minus the photo) has an area of 200 cm². Form an expression for the frame's area, simplify it, and find

x

(exact).

Mission check

?A square photo of side

x

cm is made 3 cm wider and 3 cm taller. Which expression gives its new area in cm²?

Mission 2

Is there one temperature that reads the same in °C and °F?

Make a guess: Is there one temperature that reads the same in °C and °F?

Algebra · formulae

Expressions, equations, formulae

An expression has no equals sign: $3 x + 5$ 3x + 5.
An equation is true only for particular values of the letter: $3 x + 5 = 20$ 3x + 5 = 20 is true only when $x = 5$ x = 5.
A formula links two or more quantities, so you can work one out from the others: $v = u + a t$ v = u + at.

Sort it

?Sort each line into the right group.

Tap a statement, then tap the group it belongs in. Then press Check.

Algebra · formulae

Substituting into formulae

Substitute means replace each letter by its value, then follow the order of operations: brackets, powers, then $\times$ × and $\div$ ÷, then $+$ + and $-$ −.
Put negative numbers in brackets: if $x = - 3$ x = −3, then $x^{2} = (- 3)^{2} = 9$ x² = (−3)² = 9 and $2 x^{2} = 2 \times 9 = 18$ 2x² = 2 × 9 = 18.
Watch the minus: $5 - 2 x = 5 - 2 \times (- 3) = 5 + 6 = 11$ 5 − 2x = 5 − 2 × (−3) = 5 + 6 = 11.
A formula in words works the same way: write it with letters first, then substitute.

Worked example Algebra · formulae

Substituting into a kinematics formula

Question

Use

s = u t + \frac{1}{2} a t^{2}

to find

s

when

u = 12

,

t = 3

and

a = - 4

.

1
Substitute, keeping the negative in brackets: $s = 12 \times 3 + \frac{1}{2} \times (- 4) \times 3^{2}$ s = 12 × 3 + 1/2 × (−4) × 3²
2
Work out the power first: $3^{2} = 9$ 3² = 9
3
Work out each product: $12 \times 3 = 36$ 12 × 3 = 36 and $\frac{1}{2} \times (- 4) \times 9 = - 18$ 1/2 × (−4) × 9 = −18
4
Add: $36 + (- 18) = 18$ 36 + (−18) = 18

Answer

s = 18

⚠️ Watch out

only

t

is squared. Squaring

a t

gives

\frac{1}{2} \times 144 = 72

and the wrong answer 108.

Calculate

?The formula

F = 1.8 C + 32

converts a temperature

C

in °C to

F

in °F. Find

F

when

C = - 25

(exact).

Algebra · formulae

Changing the subject

The subject is the letter on its own on one side, like $v$ v in $v = u + a t$ v = u + at.
To change the subject, undo the operations on the letter you want, in reverse order, doing the same to both sides.
$v = u + a t$ v = u + at: subtract $u$ u, then divide by $a$ a, so $t = \frac{v - u}{a}$ t = (v − u)/a.
$y = 3 x - 2$ y = 3x − 2: add 2, then divide by 3, so $x = \frac{y + 2}{3}$ x = (y + 2)/3.
Extended: if the subject is squared, undo the square last with a square root: $A = π r^{2}$ A = π r² gives $r = \frac{A}{π}$ r = √(A/π).

Worked example Algebra · formulae

Rearranging the temperature formula

Question

Make

C

the subject of

F = \frac{9 C}{5} + 32

.

1
Subtract 32 from both sides: $F - 32 = \frac{9 C}{5}$ F − 32 = 9C/5
2
Multiply both sides by 5: $5 (F - 32) = 9 C$ 5(F − 32) = 9C
3
Divide both sides by 9: $C = \frac{5 ( F - 32 )}{9}$ C = (5(F − 32))/9
4
Check with a known value: $F = 212$ F = 212 gives $C = \frac{5 \times 180}{9} = 100$ C = (5 × 180)/9 = 100

Answer

C = \frac{5 ( F - 32 )}{9}

⚠️ Watch out

the bracket matters.

\frac{5 F - 32}{9}

subtracts 32 after multiplying by 5, which is a different formula.

Worked example Algebra · formulae

When the subject appears twice (Extended)

Question

Make

x

the subject of

a x + 5 = 3 x + b

.

1
Get every $x$ x term on one side, everything else on the other: $a x - 3 x = b - 5$ ax − 3x = b − 5
2
Factorise out $x$ x: $x (a - 3) = b - 5$ x(a − 3) = b − 5
3
Divide by the bracket: $x = \frac{b - 5}{a - 3}$ x = (b − 5)/(a − 3)
4
Check with numbers, e.g. $a = 7$ a = 7, $b = 13$ b = 13: $x = 2$ x = 2, and $7 \times 2 + 5 = 19 = 3 \times 2 + 13$ 7 × 2 + 5 = 19 = 3 × 2 + 13

Answer

x = \frac{b - 5}{a - 3}

⚠️ Watch out

you can't divide by

a

and by 3 separately. Collect the

x

terms, then factorise: that is the only way to get one

x

.

Calculate

?A taxi charges a fixed £3.50 plus £1.80 for every mile. Write a formula for the cost

C

(in £) of a journey of

m

miles. A journey costs £26. How many miles was it? Give your answer as a decimal.

miles

Mission check

?Is there one temperature that reads the same in °C and °F? Use

F = 1.8 C + 32

.

Mission 3

When does the pricier phone plan become the cheaper one?

Make a guess: When does the pricier phone plan become the cheaper one?

Algebra · linear equations and inequalities

Solving linear equations

Keep the equation balanced: whatever you do to one side, do to the other.
Brackets: expand them first. $7 (x + 3) = 5 x - 8$ 7(x + 3) = 5x − 8 becomes $7 x + 21 = 5 x - 8$ 7x + 21 = 5x − 8.
Unknown on both sides: subtract the smaller $x$ x term from both sides so $x$ x appears once: $2 x + 21 = - 8$ 2x + 21 = −8, so $x = - 14.5$ x = −14.5.
Fractions: multiply every term by the denominator: $\frac{4 x + 5}{2} = 3$ (4x + 5)/2 = 3 gives $4 x + 5 = 6$ 4x + 5 = 6, so $x = 0.25$ x = 0.25.
Extended: equations with fractions are Extended content.

Worked example Algebra · linear equations and inequalities

Brackets and the unknown on both sides

Question

Solve

5 (x - 1) = 4 - x

.

1
Expand the bracket: $5 x - 5 = 4 - x$ 5x − 5 = 4 − x
2
Add $x$ x to both sides: $6 x - 5 = 4$ 6x − 5 = 4
3
Add 5 to both sides: $6 x = 9$ 6x = 9
4
Divide both sides by 6: $x = 1.5$ x = 1.5

Answer

x = 1.5

⚠️ Watch out

check by substituting.

5 (1.5 - 1) = 2.5

and

4 - 1.5 = 2.5

, so both sides agree.

Algebra · linear equations and inequalities

Forming an equation from words

Choose a letter for the unknown and say what it stands for: "let $m$ m be the number of months".
Write each quantity as an expression in that letter. A fixed charge plus a rate per unit is "fixed + rate × units".
When two quantities are equal, write expression = expression and solve.
Finish by answering the question asked, in context and with units.

Worked example Algebra · linear equations and inequalities

When do two costs match?

Question

Gym A charges £60 to join plus £15 a month. Gym B has no joining fee and charges £27 a month. After how many months is the total cost the same?

1
Let $m$ m be the number of months. Gym A costs $60 + 15 m$ 60 + 15m and gym B costs $27 m$ 27m.
2
Set the costs equal: $60 + 15 m = 27 m$ 60 + 15m = 27m
3
Subtract $15 m$ 15m from both sides: $60 = 12 m$ 60 = 12m
4
Divide by 12: $m = 5$ m = 5

Answer

after 5 months (both cost £135). After that, gym A is cheaper, because it adds less each month.

⚠️ Watch out

the question asks for months, so "

m = 5

" alone is not the full answer. Say what it means.

Calculate

?A rectangle has length

(3 x + 2)

cm and width

(x - 1)

cm. Its perimeter is 50 cm. Form and solve an equation, then find the length of the rectangle (exact).

cm

Algebra · linear equations and inequalities

Linear inequalities

$<$ < less than, $>$ > greater than, $\leq$ ≤ at most, $\geq$ ≥ at least.
Number line: open circle for $<$ < or $>$ > (end value not included), closed circle for $\leq$ ≤ or $\geq$ ≥.
Solve an inequality like an equation: $3 x - 2 < 10$ 3x − 2 < 10 gives $3 x < 12$ 3x < 12, so $x < 4$ x < 4.
If you multiply or divide by a negative number, reverse the inequality sign: $- 2 x < 6$ −2x < 6 gives $x > - 3$ x > −3.
Extended: solving them is Extended; Core shows and reads them on a number line.

Calculate

?Solve

5 - 2 x < 13

. What is the smallest integer

x

that satisfies it? (exact)

Algebra · linear equations and inequalities

Regions on a graph (Extended)

An inequality in $x$ x and $y$ y, such as $x + y \leq 5$ x + y ≤ 5, describes a region of the grid.
Draw the boundary line $x + y = 5$ x + y = 5: solid for $\leq$ ≤ or $\geq$ ≥, dashed for $<$ < or $>$ >.
Test a point not on the line, such as $(0, 0)$ (0, 0): $0 + 0 \leq 5$ 0 + 0 ≤ 5 is true, so the region is on the side containing $(0, 0)$ (0, 0).
Several inequalities together give the region where all are true.
Cambridge: shade the unwanted regions unless told otherwise.

Sort it

?The region R is defined by

x \geq 0

,

y \geq 1

and

x + y \leq 5

. Sort each point.

Tap a statement, then tap the group it belongs in. Then press Check.

Mission check

?Plan A costs £12 a month plus £0.08 a minute. Plan B costs £20 a month plus £0.03 a minute. When is Plan B cheaper?

Mission 4

A thrown ball is 10 m up at two different times. When, and why?

Make a guess: A thrown ball is 10 m up at two different times. When, and why?

Algebra · quadratic equations

Solving by factorising

Make one side 0: write the equation as $a x^{2} + b x + c = 0$ ax² + bx + c = 0.
Factorise, then use: if $A B = 0$ AB = 0, then $A = 0$ A = 0 or $B = 0$ B = 0.
Each bracket gives one root, so there are usually two solutions.
The roots are where the graph of $y = a x^{2} + b x + c$ y = ax² + bx + c crosses the $x$ x-axis; a graph gives approximate solutions.
Cambridge: solving quadratic equations is Extended content.

Worked example Algebra · quadratic equations

Solving $x^{2} + 2 x - 15 = 0$ x² + 2x − 15 = 0 (Extended)

Question

Solve

x^{2} + 2 x - 15 = 0

.

1
Find two numbers that multiply to $- 15$ −15 and add to $2$ 2: they are $5$ 5 and $- 3$ −3
2
Factorise: $(x + 5) (x - 3) = 0$ (x + 5)(x − 3) = 0
3
Set each bracket equal to zero: $x + 5 = 0$ x + 5 = 0 or $x - 3 = 0$ x − 3 = 0

Answer

x = - 5

or

x = 3

⚠️ Watch out

x + 5 = 0

gives

x = - 5

, not

5

. Each root has the opposite sign to the number in its bracket.

Quick check

?Solve

x^{2} - x - 12 = 0

.

Calculate

?A rectangle is

x

cm wide and

(x + 3)

cm long. Its area is 40 cm², which gives the equation

x^{2} + 3 x - 40 = 0

. Find the perimeter of the rectangle.

cm

Algebra · quadratic equations

The quadratic formula (Extended)

For $a x^{2} + b x + c = 0$ ax² + bx + c = 0:

x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

Use it when the quadratic does not factorise, or when you need decimals or surds.
Rearrange to $\dots = 0$ … = 0 first, and take each of $a$ a, $b$ b, $c$ c with its sign.
If $b^{2} - 4 a c$ b² − 4ac is negative there are no solutions: the graph does not cross the $x$ x-axis.
The formula is given on the formula sheet, but practise using it until it is quick and accurate.

Worked example Algebra · quadratic equations

Using the formula (Extended)

Question

Solve

2 x^{2} - 3 x - 4 = 0

. Give your answers to 2 decimal places.

1
Identify the coefficients: $a = 2$ a = 2, $b = - 3$ b = −3, $c = - 4$ c = −4
2
Work out $b^{2} - 4 a c = (- 3)^{2} - 4 \times 2 \times (- 4) = 9 + 32 = 41$ b² − 4ac = (−3)² − 4 × 2 × (−4) = 9 + 32 = 41
3
Substitute: $x = \frac{3 \pm 41}{4}$ x = (3 ± √41)/4
4
Evaluate each sign: $\frac{3 + 6.403 \dots}{4}$ (3 + 6.403…)/4 and $\frac{3 - 6.403 \dots}{4}$ (3 − 6.403…)/4

Answer

x = 2.35

or

x = - 0.85

⚠️ Watch out

- b

is

+ 3

here because

b = - 3

; and

(- 3)^{2} = 9

, not

- 9

.

Calculate

?(Extended) Solve

2 x^{2} + 7 x - 3 = 0

. Give the positive solution to 3 significant figures.

Algebra · quadratic equations

Completing the square (Extended)

Halve the coefficient of $x$ x: $x^{2} + b x + c = (x + \frac{b}{2})^{2} - (\frac{b}{2})^{2} + c$ x² + bx + c = (x + b/2)² − (b/2)² + c
Solve $x^{2} - 8 x + 13 = 0$ x² − 8x + 13 = 0: $(x - 4)^{2} - 16 + 13 = 0$ (x − 4)² − 16 + 13 = 0, so $(x - 4)^{2} = 3$ (x − 4)² = 3 and $x = 4 \pm 3$ x = 4 ± √3.
Keep the ± when you square-root. Leave surds exact unless told to round.
$y = (x + p)^{2} + q$ y = (x + p)² + q has its turning point at $(- p, q)$ (−p, q), so $y = (x - 4)^{2} - 3$ y = (x − 4)² − 3 has minimum point $(4, - 3)$ (4, −3).
Choosing a method: factorise if you can; use the formula for decimals; complete the square for exact surds or a turning point.

Quick check

?Solve

x^{2} + 10 x + 7 = 0

by completing the square.

Mission check

?A ball's height after

t

seconds is

h = 15 t - 5 t^{2}

metres. It is 10 m up when

5 t^{2} - 15 t + 10 = 0

. When is the ball 10 m up?

Mission 5

3 coffees + 2 cakes = £13.50; 2 coffees + 3 cakes = £13. A cake?

Make a guess: 3 coffees + 2 cakes = £13.50; 2 coffees + 3 cakes = £13. A cake?

Algebra · simultaneous equations

Two unknowns, two equations

One equation in $x$ x and $y$ y has endless solutions. Simultaneous equations are two equations that must both be true at once.
The solution is the pair $x$ x, $y$ y that fits both.
Graphically, each linear equation is a line; the solution is where the lines intersect. $y = 2 x - 1$ y = 2x − 1 and $x + y = 5$ x + y = 5 meet at $(2, 3)$ (2, 3).
Always check your pair in both original equations.

Worked example Algebra · simultaneous equations

Solving by elimination

Question

Solve

3 x + 2 y = 16

and

5 x - 3 y = 14

.

1
Match the $y$ y coefficients: multiply the first by 3 and the second by 2: $9 x + 6 y = 48$ 9x + 6y = 48 and $10 x - 6 y = 28$ 10x − 6y = 28
2
The signs of $6 y$ 6y differ, so add: $19 x = 76$ 19x = 76
3
Divide: $x = 4$ x = 4
4
Substitute into $3 x + 2 y = 16$ 3x + 2y = 16: $12 + 2 y = 16$ 12 + 2y = 16, so $y = 2$ y = 2
5
Check in the other equation: $5 \times 4 - 3 \times 2 = 14$ 5 × 4 − 3 × 2 = 14 ✓

Answer

x = 4

,

y = 2

⚠️ Watch out

add when the matched terms have different signs and subtract when they have the same sign. Multiply every term, including the number on the right.

Quick check

?Solve

5 x + 2 y = 20

and

3 x - 2 y = 4

.

Algebra · simultaneous equations

Solving by substitution

Use substitution when one equation already says $y = \dots$ y = … or $x = \dots$ x = ….
Replace that letter in the other equation, using brackets: $y = 2 x - 1$ y = 2x − 1 into $x + y = 5$ x + y = 5 gives $x + (2 x - 1) = 5$ x + (2x − 1) = 5.
Solve for one letter: $3 x = 6$ 3x = 6, so $x = 2$ x = 2; then $y = 2 \times 2 - 1 = 3$ y = 2 × 2 − 1 = 3.
Elimination and substitution give the same answer; pick whichever is quicker.

Calculate

?Solve the simultaneous equations

y = 2 x - 3

and

3 x + 2 y = 22

. Give the value of

y

.

Calculate

?At a cinema, 3 adult tickets and 2 child tickets cost £34. One adult ticket and 4 child tickets cost £23. Find the cost of one child ticket. Give your answer as a decimal.

£

Algebra · simultaneous equations

One linear, one quadratic (Extended)

Rearrange the linear equation to $y = \dots$ y = … (or $x = \dots$ x = …) and substitute it into the quadratic one.
You get a quadratic equation in one letter: make it $= 0$ = 0 and solve.
Find the partner value for each root using the linear equation. Give the answers as pairs.
A line can meet a curve at two points, one point (it touches) or no points.
The curve might be a circle such as $x^{2} + y^{2} = 25$ x² + y² = 25: the same method works.

Worked example Algebra · simultaneous equations

A line meeting a curve (Extended)

Question

Solve

y = 2 x + 1

and

y = x^{2} - 2

.

1
Set the two expressions for $y$ y equal: $x^{2} - 2 = 2 x + 1$ x² − 2 = 2x + 1
2
Make one side zero: $x^{2} - 2 x - 3 = 0$ x² − 2x − 3 = 0
3
Factorise: $(x - 3) (x + 1) = 0$ (x − 3)(x + 1) = 0, so $x = 3$ x = 3 or $x = - 1$ x = −1
4
Use $y = 2 x + 1$ y = 2x + 1 for each: $y = 7$ y = 7 or $y = - 1$ y = −1

Answer

x = 3, y = 7

and

x = - 1, y = - 1

⚠️ Watch out

two

x

values need two matching

y

values. Stopping at

x

loses marks.

Quick check

?Extended. Solve

y = x + 4

and

y = x^{2} - 2 x

.

Mission check

?3 coffees and 2 cakes cost £13.50. 2 coffees and 3 cakes cost £13.00. How much is one cake?

Mission 6

Four graphs, four equations: can you match them without plotting?

Make a guess: Four graphs, four equations: can you match them without plotting?

Algebra · graphs of functions

Lines and parabolas

A linear graph $y = m x + c$ y=mx+c is a straight line: gradient $m$ m, crossing the $y$ y-axis at $(0, c)$ (0, c).
A quadratic graph $y = a x^{2} + b x + c$ y=ax²+bx+c is a parabola: U-shaped if $a > 0$ a>0, ∩-shaped if $a < 0$ a<0.
The roots are where the curve crosses the $x$ x-axis ( $y = 0$ y=0). The $y$ y-intercept is at $(0, c)$ (0, c).
The turning point lies on the line of symmetry, halfway between the roots.
Core: you need roots and symmetry; turning points are Extended.

Worked example Algebra · graphs of functions

Reading a quadratic graph

Question

The graph of

y = x^{2} - 2 x - 3

is drawn. Find its roots, its turning point, and use it to solve

x^{2} - 2 x - 3 = 5

.

1
Read where the curve crosses the $x$ x-axis: $x = - 1$ x = −1 and $x = 3$ x = 3
2
The line of symmetry is halfway between: $x = \frac{- 1 + 3}{2} = 1$ x = (−1+3)/2 = 1
3
Substitute $x = 1$ x = 1: $y = 1 - 2 - 3 = - 4$ y = 1 − 2 − 3 = −4, so the turning point is $(1, - 4)$ (1, −4)
4
Draw the line $y = 5$ y = 5 and read where it meets the curve: $x = - 2$ x = −2 and $x = 4$ x = 4

Answer

roots

x = - 1

and

x = 3

; turning point

(1, - 4)

; solutions

x = - 2

or

x = 4

⚠️ Watch out

to solve

x^{2} - 2 x - 3 = 5

you read off at

y = 5

, not at

y = 0

.

Calculate

?The graph of

y = x^{2} + 2 x - 8

is drawn for

- 6 \leq x \leq 4

. Use the graph to solve

x^{2} + 2 x - 8 = 7

. Enter the positive solution.

Algebra · graphs of functions

Cubics and reciprocals

A cubic $y = a x^{3} + \dots$ y=ax³+… is S-shaped: rising left to right if $a > 0$ a>0, falling if $a < 0$ a<0. $y = x^{3} - 4 x$ y = x³ − 4x has roots $- 2$ −2, 0 and 2.
The reciprocal graph $y = \frac{a}{x}$ y=a/x has two branches that never touch the axes (the asymptotes); $x \neq = 0$ x ≠ 0.
To match an equation to a graph: check the highest power of $x$ x and its sign, then the intercepts.
Core: draw and recognise $y = \frac{a}{x}$ y=a/x from a table of values. Cubic graphs, and sketching reciprocal graphs, are Extended.

Sort it

?Sort each equation by the shape of its graph.

Tap a statement, then tap the group it belongs in. Then press Check.

Calculate

?Exam-style. The graphs of

y = \frac{6}{x}

and

y = x + 1

are drawn on the same axes. Use the graphs to find the

x

-coordinate of the intersection point in the first quadrant.

Algebra · graphs of functions

Exponential graphs (Extended)

An exponential graph $y = k^{x}$ y = k^x ( $k > 0$ k > 0) always passes through $(0, 1)$ (0, 1), because $k^{0} = 1$ k⁰ = 1.
If $k > 1$ k > 1 it shows growth: $y = 2^{x}$ y = 2^x goes 1, 2, 4, 8 as $x$ x goes 0, 1, 2, 3.
If $0 < k < 1$ 0 < k < 1 it shows decay: $y = (\frac{1}{2})^{x}$ y = (1/2)^x falls towards 0.
It never touches the $x$ x-axis: $y = 0$ y=0 is an asymptote, and $y$ y is always positive.

Algebra · graphs of functions

Trigonometric graphs (Extended)

$y = sin x$ y = sin x and $y = cos x$ y = cos x are waves between $- 1$ −1 and 1 that repeat every $36 0^{\circ}$ 360°.
$sin x$ sin x starts at 0 and peaks at $x = 9 0^{\circ}$ x = 90°; $cos x$ cos x starts at 1 and is $- 1$ −1 at $x = 18 0^{\circ}$ x = 180°.
$y = tan x$ y = tan x repeats every $18 0^{\circ}$ 180° and has asymptotes at $x = 9 0^{\circ}$ x = 90°, $27 0^{\circ}$ 270°, ….
Symmetry gives more solutions: $sin x = sin (18 0^{\circ} - x)$ sin x = sin (180° − x).
You need these graphs for $0^{\circ} \leq x \leq 36 0^{\circ}$ 0° ≤ x ≤ 360°.

Quick check

?You know that

sin 3 0^{\circ} = 0.5

. Use the graph of

y = sin x

to find the other angle between

0^{\circ}

and

36 0^{\circ}

with

sin x = 0.5

.

Mission check

?Four graphs, four equations. Graph P has two separate branches and never meets either axis. Which equation matches it, and why?

Mission 7

A machine doubles your number, then adds 3. How do you undo it?

Make a guess: A machine doubles your number, then adds 3. How do you undo it?

Algebra · functions

Function notation (Extended)

A function turns each input into one output. $f (x) = 3 x - 5$ f(x) = 3x − 5 means "multiply by 3, then subtract 5".
$f (4)$ f(4) means the output when the input is 4: $f (4) = 3 \times 4 - 5 = 7$ f(4) = 3 × 4 − 5 = 7.
Solving $f (x) = 13$ f(x) = 13 means finding the input: $3 x - 5 = 13$ 3x − 5 = 13, so $x = 6$ x = 6.
Composite function: $f g (x)$ fg(x) means $f (g (x))$ f(g(x)). Do $g$ g first, then $f$ f.

Worked example Algebra · functions

Composite functions (Extended)

Question

f (x) = 2 x + 3

and

g (x) = x^{2}

. Find

f g (3)

,

g f (3)

and an expression for

f g (x)

.

1
For $f g (3)$ fg(3), apply $g$ g first: $g (3) = 9$ g(3) = 9
2
Then apply $f$ f: $f (9) = 2 \times 9 + 3 = 21$ f(9) = 2 × 9 + 3 = 21
3
For $g f (3)$ gf(3), apply $f$ f first: $f (3) = 9$ f(3) = 9, then $g (9) = 81$ g(9) = 81
4
For $f g (x)$ fg(x), put $g (x)$ g(x) into $f$ f: $f (x^{2}) = 2 x^{2} + 3$ f(x²) = 2x² + 3

Answer

f g (3) = 21

,

g f (3) = 81

,

f g (x) = 2 x^{2} + 3

⚠️ Watch out

f g (x)

is not

f (x) \times g (x)

, and the order matters:

f g (3) \neq = g f (3)

.

Calculate

?

f (x) = 5 - x

and

g (x) = x^{2}

. Work out

f g (- 3)

.

Algebra · functions

Inverse functions (Extended)

The inverse function $f^{- 1}$ f⁻¹ undoes $f$ f: if $f (5) = 13$ f(5) = 13 then $f^{- 1} (13) = 5$ f⁻¹(13) = 5.
Undo the steps in reverse order, like taking off shoes before socks.
Algebra method: write $y = f (x)$ y = f(x), rearrange to make $x$ x the subject, then write it in terms of $x$ x.

y = 2 x + 3 \Rightarrow x = \frac{y - 3}{2} \Rightarrow f^{- 1} (x) = \frac{x - 3}{2}

Check: $f^{- 1} (f (x)) = x$ f⁻¹(f(x)) = x.

Quick check

?

f (x) = \frac{x + 4}{5}

. Which is

f^{- 1} (x)

?

Calculate

?Exam-style.

f (x) = 2 x - 1

and

g (x) = x^{2} + 1

. Solve

g f (x) = 26

. Enter the positive solution.

Algebra · functions

Domain and range (Extended)

The domain is the set of inputs allowed; the range is the set of outputs.
Exclude any input that would mean dividing by zero: $f (x) = \frac{1}{x - 2}$ f(x) = 1/(x − 2) needs $x \neq = 2$ x ≠ 2.
Exclude any input that would mean the square root of a negative: $g (x) = x - 3$ g(x) = √(x − 3) needs $x \geq 3$ x ≥ 3.
$h (x) = x^{2} + 1$ h(x) = x² + 1 for all $x$ x has range $h (x) \geq 1$ h(x) ≥ 1, because $x^{2}$ x² is never negative.

Mission check

?A machine doubles your number, then adds 3. Which machine undoes it?

Mission 8

Chessboard: 1 grain, then double each square. How many on square 64?

Make a guess: Chessboard: 1 grain, then double each square. How many on square 64?

Algebra · sequences

Two kinds of rule

A term-to-term rule gets each term from the one before: 5, 9, 13, 17, … is "add 4".
A position-to-term rule (the nth term) gets any term from its position $n$ n: here $4 n + 1$ 4n + 1.
Substitute $n$ n to find a term: the 10th term is $4 \times 10 + 1 = 41$ 4 × 10 + 1 = 41.
The chessboard has term-to-term rule "double".

Worked example Algebra · sequences

Finding the nth term of a linear sequence

Question

Find the nth term of 7, 10, 13, 16, … and its 100th term. Is 100 a term?

1
Find the common difference: $10 - 7 = 3$ 10 − 7 = 3, so the nth term starts $3 n$ 3n
2
Compare with $3 n$ 3n (3, 6, 9, 12): each term is 4 more, so add 4
3
Write the rule: nth term $= 3 n + 4$ = 3n + 4
4
Substitute $n = 100$ n = 100: $3 \times 100 + 4 = 304$ 3 × 100 + 4 = 304
5
Solve $3 n + 4 = 100$ 3n + 4 = 100: $n = 32$ n = 32, a whole number, so 100 is the 32nd term

Answer

nth term

3 n + 4

; 100th term 304; 100 is a term (the 32nd).

⚠️ Watch out

the nth term is not "

n + 3

". The difference 3 multiplies

n

.

Calculate

?The sequence 40, 37, 34, 31, … is linear. Find its 20th term.

Calculate

?Exam-style. Pattern

n

is a row of

n

hexagons made from matchsticks. Pattern 1 uses 6 sticks, pattern 2 uses 11 and pattern 3 uses 16. Which pattern uses exactly 101 matchsticks?

Algebra · sequences

Special sequences

Square numbers $n^{2}$ n²: 1, 4, 9, 16, … and cube numbers $n^{3}$ n³: 1, 8, 27, 64, ….
Triangular numbers: 1, 3, 6, 10, 15, … (add 2, then 3, then 4, …).
A geometric sequence multiplies by the same number $r$ r each time: 3, 6, 12, 24 ( $r = 2$ r = 2). The chessboard 1, 2, 4, 8, … is geometric: square $n$ n holds $2^{n - 1}$ 2ⁿ⁻¹.
A Fibonacci-type sequence adds the two previous terms: 2, 5, 7, 12, 19, ….
Extended: geometric nth term $a r^{n - 1}$ arⁿ⁻¹, e.g. $3 \times 2^{n - 1}$ 3 × 2ⁿ⁻¹.

Sort it

?Sort each sequence.

Tap a statement, then tap the group it belongs in. Then press Check.

Algebra · sequences

The nth term of a quadratic sequence (Extended, except the Core tip below)

If the second differences are constant, the nth term is $a n^{2} + bn + c$ an² + bn + c.
The coefficient of $n^{2}$ n² is half the second difference.
Subtract $a n^{2}$ an² from each term. What is left is linear: find its nth term and add it on.
Core: compare with the square or cube numbers. 2, 5, 10, 17, … is $n^{2} + 1$ n² + 1; 2, 9, 28, 65, … is $n^{3} + 1$ n³ + 1. Extended: any quadratic, and cubic sequences with constant third differences.

Worked example Algebra · sequences

Quadratic nth term (Extended)

Question

Find the nth term of 4, 11, 22, 37, ….

1
First differences: 7, 11, 15. Second differences: 4, 4, so it is quadratic
2
Halve the second difference: $\frac{4}{2} = 2$ 4/2 = 2, so it starts $2 n^{2}$ 2n²
3
Subtract $2 n^{2}$ 2n² (2, 8, 18, 32) from each term: 2, 3, 4, 5
4
The leftover 2, 3, 4, 5 has nth term $n + 1$ n + 1

Answer

nth term

= 2 n^{2} + n + 1

(check:

n = 1

gives

2 + 1 + 1 = 4

)

⚠️ Watch out

the coefficient of

n^{2}

is half the second difference, not the second difference itself.

Calculate

?Extended. Find the 10th term of the quadratic sequence 3, 9, 19, 33, 51, ….

Mission check

?On the chessboard, square 1 has 1 grain and each square doubles the one before. How many grains are on square 64?

Mission 9

Two runners' distance–time graphs cross. What does the crossing mean?

Make a guess: Two runners' distance–time graphs cross. What does the crossing mean?

Algebra · real-life graphs

Distance–time graphs

The gradient of a distance–time graph is the speed: distance ÷ time.
Steeper means faster. A horizontal line means stopped.
Where two graphs cross, the two objects are at the same place at the same time. If the lines do not have the same gradient there, one is overtaking the other.
Use the units on the axes: metres per minute, km/h, m/s.

Worked example Algebra · real-life graphs

Reading two runners' graphs

Question

Runner A runs 2000 m in 10 minutes at a steady pace. Runner B covers 1200 m in the first 4 minutes. Find each speed for the first 4 minutes, in m/s, and say who is faster at the start.

1
Gradient of B's first section: $\frac{1200}{4} = 300$ 1200/4 = 300 m per minute
2
Convert to m/s: $\frac{1200}{4 \times 60} = 5$ 1200/(4 × 60) = 5 m/s
3
Gradient of A's line: $\frac{2000}{10 \times 60} = 3.33$ 2000/(10 × 60) = 3.33 m/s (3 s.f.)
4
Compare: B's line is steeper at the start

Answer

B runs at 5 m/s and A at 3.33 m/s, so B is faster at the start.

⚠️ Watch out

change minutes to seconds before dividing, or the answer is in metres per minute.

Calculate

?Exam-style. A cyclist rides 15 km in 45 minutes, stops for 15 minutes, then rides 15 km back home in 30 minutes. Find the average speed for the whole journey.

km/h

Algebra · real-life graphs

Conversion graphs and rates

A conversion graph is a straight line through $(0, 0)$ (0, 0): the two quantities are in direct proportion.
The gradient is the rate: kilometres per mile, euros per pound.
A line that starts above $(0, 0)$ (0, 0) has a fixed charge. A plumber charging £40 call-out plus £30 per hour has intercept 40 and gradient 30 (£ per hour).
Read across from one axis to the line, then down (or across) to the other axis.

Calculate

?The conversion graph is a straight line through

(0, 0)

and (50 miles, 80 km). Use it to convert 28 miles to kilometres.

km

Algebra · real-life graphs

Speed–time graphs (Extended)

The gradient is the acceleration, in m/s². On this graph, $\frac{12}{4} = 3$ 12/4 = 3 m/s² for the first 4 s.
A horizontal line means constant speed; a downward slope means deceleration.
The area under the graph is the distance travelled: split it into triangles, rectangles and trapezia.

Calculate

?Extended. A car speeds up from rest to 12 m/s in 4 s, travels at 12 m/s for 6 s, then slows steadily to rest in 3 s. Find the total distance travelled.

m

Algebra · real-life graphs

Gradient of a curve (Extended)

On a curve the gradient changes. Draw a tangent: a straight line that just touches the curve at the point.
The gradient of the tangent is the instantaneous rate of change there, e.g. the speed at that moment on a distance–time graph.
Pick two points far apart on the tangent: gradient = change in $y$ y ÷ change in $x$ x.

Calculate

?Extended. A tangent is drawn to a distance–time curve at

t = 3

s. It passes through (1, 2) and (5, 14), with time in seconds and distance in metres. Estimate the speed at

t = 3

s.

m/s

Mission check

?Two runners start together. Their distance–time graphs cross at (8 minutes, 1600 m). Runner B's line is steeper at the start. What does the crossing mean?

Mission 10

Cycle out at 10 km/h, back at 30 km/h. Is your average 20 km/h?

Make a guess: Cycle out at 10 km/h, back at 30 km/h. Is your average 20 km/h?

Algebra · algebraic fractions

Simplifying algebraic fractions (Extended)

An algebraic fraction has algebra in the numerator, the denominator or both.
To simplify, factorise the numerator and denominator fully, then cancel common factors (whole brackets or whole factors).
$\frac{x ^{2} - 2 x}{x ^{2} - 5 x + 6} = \frac{x ( x - 2 )}{( x - 3 ) ( x - 2 )} = \frac{x}{x - 3}$ (x² − 2x)/(x² − 5x + 6) = (x(x − 2))/((x − 3)(x − 2)) = x/(x − 3)
You can only cancel factors, never single terms: in $\frac{x + 3}{x}$ (x + 3)/x nothing cancels.

Quick check

?Simplify fully

\frac{2 x ^{2} + 3 x}{4 x ^{2} - 9}

.

Algebra · algebraic fractions

The four operations with algebraic fractions (Extended)

Multiply: factorise, cancel, then multiply numerators and denominators.
Divide: flip the second fraction and multiply: $\frac{3 a}{4} \div \frac{9 a}{10} = \frac{3 a}{4} \times \frac{10}{9 a} = \frac{5}{6}$ 3a/4 ÷ 9a/10 = 3a/4 × 10/9a = 5/6.
Add or subtract: use a common denominator, just as $\frac{1}{2} + \frac{1}{3} = \frac{3 + 2}{6}$ 1/2 + 1/3 = (3 + 2)/6.
$\frac{1}{x} + \frac{1}{x + 1} = \frac{( x + 1 ) + x}{x ( x + 1 )} = \frac{2 x + 1}{x ( x + 1 )}$ 1/x + 1/(x + 1) = ((x + 1) + x)/(x(x + 1)) = (2x + 1)/(x(x + 1))
Never add the denominators: $\frac{1}{2} + \frac{1}{3}$ 1/2 + 1/3 is not $\frac{2}{5}$ 2/5.

Calculate

?Write

\frac{2}{x} + \frac{3}{x + 1}

as a single fraction

\frac{a x + b}{x ( x + 1 )}

. Enter the value of

a

(exact).

Worked example Algebra · algebraic fractions

Subtracting algebraic fractions (Extended)

Question

Write

\frac{3 x + 1}{x + 2} - \frac{x - 2}{x - 1}

as a single fraction.

1
Use the common denominator $(x + 2) (x - 1)$ (x + 2)(x − 1): $\frac{( 3 x + 1 ) ( x - 1 ) - ( x - 2 ) ( x + 2 )}{( x + 2 ) ( x - 1 )}$ ((3x + 1)(x − 1) − (x − 2)(x + 2))/((x + 2)(x − 1))
2
Expand the first product: $(3 x + 1) (x - 1) = 3 x^{2} - 2 x - 1$ (3x + 1)(x − 1) = 3x² − 2x − 1
3
Expand the second product: $(x - 2) (x + 2) = x^{2} - 4$ (x − 2)(x + 2) = x² − 4
4
Subtract the whole second product: $3 x^{2} - 2 x - 1 - (x^{2} - 4) = 2 x^{2} - 2 x + 3$ 3x² − 2x − 1 − (x² − 4) = 2x² − 2x + 3

Answer

\frac{2 x ^{2} - 2 x + 3}{( x + 2 ) ( x - 1 )}

⚠️ Watch out

the minus applies to all of

x^{2} - 4

, so

- (- 4) = + 4

. Writing

- x^{2} - 4

gives the wrong numerator

2 x^{2} - 2 x - 5

.

Worked example Algebra · algebraic fractions

Solving an equation with algebraic fractions (Extended)

Question

Solve

\frac{2}{x + 2} + \frac{3}{2 x - 1} = 1

.

1
Multiply every term by $(x + 2) (2 x - 1)$ (x + 2)(2x − 1): $2 (2 x - 1) + 3 (x + 2) = (x + 2) (2 x - 1)$ 2(2x − 1) + 3(x + 2) = (x + 2)(2x − 1)
2
Expand both sides: $7 x + 4 = 2 x^{2} + 3 x - 2$ 7x + 4 = 2x² + 3x − 2
3
Rearrange to $= 0$ = 0: $2 x^{2} - 4 x - 6 = 0$ 2x² − 4x − 6 = 0, so $x^{2} - 2 x - 3 = 0$ x² − 2x − 3 = 0
4
Factorise: $(x - 3) (x + 1) = 0$ (x − 3)(x + 1) = 0
5
Check neither value makes a denominator zero: both are fine

Answer

x = 3

or

x = - 1

⚠️ Watch out

multiply every term, including the 1 on the right, by the common denominator.

Calculate

?A walker goes 12 km uphill at

x

km/h, then 12 km back down at

(x + 2)

km/h. The whole walk takes 5 hours. Form an equation and solve it to find the uphill speed (exact).

km/h

Mission check

?Cycle a route at 10 km/h and ride back along it at 30 km/h. What is your average speed for the whole trip?

Mission 11

40 m of fence and a wall: which rectangle encloses the most area?

Make a guess: 40 m of fence and a wall: which rectangle encloses the most area?

Algebra · differentiation

The gradient function (Extended)

A curve's gradient changes along it. The derivative $\frac{d y}{d x}$ dy/dx gives the gradient of the tangent at any $x$ x.
Rule: if $y = a x^{n}$ y = axⁿ, then

\frac{d y}{d x} = na x^{n - 1}

Multiply by the power, then reduce the power by 1: $5 x^{3}$ 5x³ gives $15 x^{2}$ 15x², $7 x$ 7x gives $7$ 7, a constant gives $0$ 0.
Differentiate a sum term by term. Substitute an $x$ x value to get the gradient at that point.
Here $n$ n is a positive integer or zero. You also estimate gradients by drawing a tangent.

Worked example Algebra · differentiation

Gradient at a point (Extended)

Question

Find the gradient of the curve

y = 4 x^{3} - 5 x^{2} + 7

at the point where

x = 2

.

1
Differentiate each term: $4 x^{3} \to 12 x^{2}$ 4x³ → 12x², $- 5 x^{2} \to - 10 x$ −5x² → −10x, $7 \to 0$ 7 → 0
2
Write the derivative: $\frac{d y}{d x} = 12 x^{2} - 10 x$ dy/dx = 12x² − 10x
3
Substitute $x = 2$ x = 2: $12 \times 4 - 10 \times 2 = 48 - 20$ 12 × 4 − 10 × 2 = 48 − 20

Answer

the gradient is 28

⚠️ Watch out

substitute into

\frac{d y}{d x}

, not into

y

. Putting

x = 2

into

y

gives the height of the curve (19), not its gradient.

Quick check

?Find

\frac{d y}{d x}

when

y = 2 x^{5} - 3 x

.

Calculate

?Find the gradient of the curve

y = x^{3} - 6 x + 2

at the point where

x = - 1

.

Algebra · differentiation

Stationary points (Extended)

At a stationary point (turning point) the gradient is zero: solve $\frac{d y}{d x} = 0$ dy/dx = 0, then find $y$ y.
Maximum: the gradient goes positive, zero, negative. Minimum: negative, zero, positive.
Decide which is which from the shape of the graph: a cubic with positive $x^{3}$ x³ term rises, falls, then rises, so its left turning point is the maximum.
You may also use the gradient either side, or the second derivative: $\frac{d ^{2} y}{d x ^{2}} < 0$ d²y/dx² < 0 at a maximum and $> 0$ > 0 at a minimum.

Worked example Algebra · differentiation

Finding and classifying turning points (Extended)

Question

Find the turning points of

y = x^{3} - 12 x + 5

and say which is a maximum.

1
Differentiate: $\frac{d y}{d x} = 3 x^{2} - 12$ dy/dx = 3x² − 12
2
Set it to zero: $3 x^{2} = 12$ 3x² = 12, so $x^{2} = 4$ x² = 4 and $x = 2$ x = 2 or $x = - 2$ x = −2
3
Find each $y$ y: $x = - 2$ x = −2 gives $- 8 + 24 + 5 = 21$ −8 + 24 + 5 = 21; $x = 2$ x = 2 gives $8 - 24 + 5 = - 11$ 8 − 24 + 5 = −11
4
Use the shape: a positive cubic rises, falls, rises, so the left point is the maximum

Answer

maximum at

(- 2, 21)

, minimum at

(2, - 11)

⚠️ Watch out

x^{2} = 4

has two solutions,

\pm 2

. Missing

- 2

loses a turning point.

Calculate

?A ball is thrown upwards. Its height after

t

seconds is

h = 20 t - 5 t^{2}

metres. Use differentiation to find the greatest height the ball reaches.

m

Mission check

?40 m of fence makes three sides of a rectangular pen; a wall makes the fourth side. Which pen encloses the most area?

Recap

Big ideas

Collect only like terms (same letters, same powers); expand by multiplying every term in one bracket by every term in the other.
$(a + b)^{2} = a^{2} + 2 ab + b^{2}$ (a + b)² = a² + 2ab + b², never just $a^{2} + b^{2}$ a² + b²: the grid shows the two missing strips.
Factorising undoes expanding: common factor first, then look for $x^{2} + b x + c$ x² + bx + c or a difference of two squares, and check by expanding.
Substitute with brackets round negatives, then follow the order of operations.
Change the subject by undoing operations in reverse order; if the subject appears twice, collect those terms and factorise.
An expression has no equals sign, an equation is true for particular values, and a formula links quantities.
Expand brackets, collect the unknown on one side, and keep the equation balanced; check by substituting.
To compare two deals, write each as an expression, set them equal, and interpret what happens either side of the solution.
Inequalities solve like equations, except that multiplying or dividing by a negative reverses the sign.
Make one side 0, factorise, and set each bracket to 0: each bracket gives a root.
Extended: if it will not factorise, use $x = \frac{- b \pm b ^{2} - 4 a c}{2 a}$ x = (−b ± √(b² − 4ac))/2a.
Completing the square gives exact surd solutions and the turning point $(- p, q)$ (−p, q) of $y = (x + p)^{2} + q$ y = (x + p)² + q.
A quadratic graph can cross a line twice, so a quadratic equation can have two solutions.
Two unknowns need two equations; the solution fits both and is where their graphs intersect.
Eliminate by matching coefficients and adding or subtracting; substitute when one equation gives $y = \dots$ y = ….
Extended: for a line and a curve, substitute to get a quadratic, then find a $y$ y for each $x$ x.
The highest power of $x$ x and its sign set the shape: line, parabola, S-shaped cubic, or two-branch reciprocal.
Roots are where $y = 0$ y = 0; to solve $f (x) = k$ f(x) = k from a graph, read off where the line $y = k$ y = k meets the curve.
Higher/Extended: $y = k^{x}$ y = k^x passes through $(0, 1)$ (0, 1) and never touches the $x$ x-axis.
Higher/Extended: $sin x$ sin x and $cos x$ cos x repeat every $36 0^{\circ}$ 360°, $tan x$ tan x every $18 0^{\circ}$ 180°; use symmetry for second solutions.
$f g (x)$ fg(x) means do $g$ g first, then $f$ f; order matters.
To find $f^{- 1} (x)$ f⁻¹(x), undo the steps in reverse order, or rearrange $y = f (x)$ y = f(x) to make $x$ x the subject.
The domain excludes inputs that would divide by zero or take the square root of a negative.
A linear sequence adds the same amount each time; its nth term is (difference) $\times n$ × n + adjustment.
A geometric sequence multiplies by the same amount each time, so it grows much faster.
Higher/Extended: constant second differences mean a quadratic nth term; the $n^{2}$ n² coefficient is half the second difference.
Gradient of a distance–time graph = speed; steeper means faster; a crossing means the same place at the same time.
A straight-line graph's gradient is a rate (km per mile, £ per hour); its intercept is a fixed amount.
On a velocity–time (speed–time) graph, gradient = acceleration and area = distance.
Higher/Extended: the tangent's gradient gives the rate of change at an instant.
Factorise first, then cancel whole factors; never cancel single terms.
Add or subtract algebraic fractions over a common denominator, and multiply every term by it to solve an equation.
If $y = a x^{n}$ y = axⁿ, then $\frac{d y}{d x} = na x^{n - 1}$ dy/dx = naxⁿ⁻¹; substitute $x$ x into $\frac{d y}{d x}$ dy/dx for the gradient at a point.
Stationary points: solve $\frac{d y}{d x} = 0$ dy/dx = 0, find $y$ y, then decide maximum or minimum.
To maximise a quantity, write it in one variable, differentiate and set the derivative to zero.

🏆

Mini-lesson complete!

⭐⭐⭐

You've worked through Algebra & Graphs for Cambridge IGCSE Maths (0580). 🎉

Mission 1A square grows 3 cm each way. Is its new area just $x^{2} + 9$ x² + 9?
Mission 2Is there one temperature that reads the same in °C and °F?
Mission 3When does the pricier phone plan become the cheaper one?
Mission 4A thrown ball is 10 m up at two different times. When, and why?
Mission 53 coffees + 2 cakes = £13.50; 2 coffees + 3 cakes = £13. A cake?
Mission 6Four graphs, four equations: can you match them without plotting?
Mission 7A machine doubles your number, then adds 3. How do you undo it?
Mission 8Chessboard: 1 grain, then double each square. How many on square 64?
Mission 9Two runners' distance–time graphs cross. What does the crossing mean?
Mission 10Cycle out at 10 km/h, back at 30 km/h. Is your average 20 km/h?
Mission 1140 m of fence and a wall: which rectangle encloses the most area?

Your stars: 0 / 0

Next: test yourself in the Verify stage.

📣 Smashed it? Share your score

Challenge a mate to beat your stars, or show a parent how you got on.

→ Back to IGCSE Mathematics · Cambridge

Specification points and sources for this lesson are listed at the bottom of the page.

Algebra & Graphs

A square grows 3 cm each way. Is its new area just x2+9x² + 9?

Terms, expressions and like terms

Expanding two brackets

Expanding and simplifying two brackets

Factorising: expanding in reverse

Factorising ax2+bx+cax² + bx + c (Extended)

Is there one temperature that reads the same in °C and °F?

Expressions, equations, formulae

Substituting into formulae

Substituting into a kinematics formula

Changing the subject

Rearranging the temperature formula

When the subject appears twice (Extended)

When does the pricier phone plan become the cheaper one?

Solving linear equations

Brackets and the unknown on both sides

Forming an equation from words

When do two costs match?

Linear inequalities

Regions on a graph (Extended)

A thrown ball is 10 m up at two different times. When, and why?

Solving by factorising

Solving x2+2x−15=0x² + 2x − 15 = 0 (Extended)

The quadratic formula (Extended)

Using the formula (Extended)

Completing the square (Extended)

3 coffees + 2 cakes = £13.50; 2 coffees + 3 cakes = £13. A cake?

Two unknowns, two equations

Solving by elimination

Solving by substitution

One linear, one quadratic (Extended)

A line meeting a curve (Extended)

Four graphs, four equations: can you match them without plotting?

Lines and parabolas

Reading a quadratic graph

Cubics and reciprocals

Exponential graphs (Extended)

Trigonometric graphs (Extended)

A machine doubles your number, then adds 3. How do you undo it?

Function notation (Extended)

Composite functions (Extended)

Inverse functions (Extended)

Domain and range (Extended)

Chessboard: 1 grain, then double each square. How many on square 64?

Two kinds of rule

Finding the nth term of a linear sequence

Special sequences

The nth term of a quadratic sequence (Extended, except the Core tip below)

Quadratic nth term (Extended)

Two runners' distance–time graphs cross. What does the crossing mean?

Distance–time graphs

Reading two runners' graphs

Conversion graphs and rates

Speed–time graphs (Extended)

Gradient of a curve (Extended)

Cycle out at 10 km/h, back at 30 km/h. Is your average 20 km/h?

Simplifying algebraic fractions (Extended)

The four operations with algebraic fractions (Extended)

Subtracting algebraic fractions (Extended)

Solving an equation with algebraic fractions (Extended)

40 m of fence and a wall: which rectangle encloses the most area?

The gradient function (Extended)

Gradient at a point (Extended)

Stationary points (Extended)

Finding and classifying turning points (Extended)

Big ideas

Mini-lesson complete!

📣 Smashed it? Share your score

A square grows 3 cm each way. Is its new area just $x^{2} + 9$ x² + 9?

Factorising $a x^{2} + b x + c$ ax² + bx + c (Extended)

Solving $x^{2} + 2 x - 15 = 0$ x² + 2x − 15 = 0 (Extended)