Mini-Lesson

Algebra

13 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 4Rise 1 for every 12 along: which of three ramps is legal?
Mission 5A thrown ball is 10 m up at two different times. When, and why?
Mission 63 coffees + 2 cakes = £13.50; 2 coffees + 3 cakes = £13. A cake?
Mission 7Four graphs, four equations: can you match them without plotting?
Mission 8A machine doubles your number, then adds 3. How do you undo it?
Mission 9Chessboard: 1 grain, then double each square. How many on square 64?
Mission 10Two runners' distance–time graphs cross. What does the crossing mean?
Mission 11Square any odd number and subtract 1. Always a multiple of 8?
Mission 12Cycle out at 10 km/h, back at 30 km/h. Is your average 20 km/h?
Mission 13A sprinkler reaches 13 m. Does it wet a plant at (12, 6)?

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.
Higher: 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.

Quick check

?Factorise

x^{2} + 2 x - 24

.

Worked example Algebra · expressions

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

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.
An identity is true for every value of the letter. It is often written with $\equiv$ ≡: $2 (x + 3) \equiv 2 x + 6$ 2(x + 3) ≡ 2x + 6.

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.
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

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.

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.
Higher: in set notation, $x < 4$ x < 4 is ${x : x < 4}$ {x : x < 4} (or ${x ∣ x < 4}$ {x | x < 4}).

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 (Higher)

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.

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

Rise 1 for every 12 along: which of three ramps is legal?

Make a guess: Rise 1 for every 12 along: which of three ramps is legal?

Algebra · straight-line graphs

Gradient measures steepness

Gradient = change in $y$ y ÷ change in $x$ x ("rise over run").

m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}

Uphill from left to right: positive. Downhill: negative. Horizontal: 0.
The bigger the gradient (ignoring its sign), the steeper the line.
"Rise 1 for every 12 along" is a gradient of $\frac{1}{12}$ 1/12.

Algebra · straight-line graphs

The equation $y = m x + c$ y = mx + c

In $y = m x + c$ y = mx + c, $m$ m is the gradient and $c$ c is the $y$ y-intercept: the line crosses the $y$ y-axis at $(0, c)$ (0, c).
Rearrange to $y = \dots$ y = … first: $2 y = 6 x - 4$ 2y = 6x − 4 gives $y = 3 x - 2$ y = 3x − 2, so $m = 3$ m = 3 and $c = - 2$ c = −2.
To plot: make a table of values (three or more $x$ x values), plot the points, rule the line.
$y = k$ y = k is a horizontal line; $x = k$ x = k is a vertical line.

Quick check

?What is the gradient of the line

2 y = 12 - 3 x

?

Worked example Algebra · straight-line graphs

Equation of the line through two points

Question

Find the equation of the line through

(2, 7)

and

(5, 13)

.

1
Find the gradient: $m = \frac{13 - 7}{5 - 2} = \frac{6}{3} = 2$ m = (13 − 7)/(5 − 2) = 6/3 = 2
2
Substitute one point into $y = 2 x + c$ y = 2x + c: $7 = 2 \times 2 + c$ 7 = 2 × 2 + c
3
Solve for $c$ c: $c = 3$ c = 3
4
Check with the other point: $2 \times 5 + 3 = 13$ 2 × 5 + 3 = 13 ✓

Answer

y = 2 x + 3

⚠️ Watch out

subtract the coordinates in the same order on the top and the bottom, or the sign of the gradient flips.

Calculate

?A taxi fare is a fixed charge plus a set amount per mile, so fare against miles is a straight line. A 3-mile trip costs £8.40 and an 8-mile trip costs £15.40. Find the fixed charge. Give your answer as a decimal.

£

Algebra · straight-line graphs

Parallel and perpendicular lines

Parallel lines have equal gradients. The line parallel to $y = 4 x - 1$ y = 4x − 1 through $(1, - 3)$ (1, −3): $- 3 = 4 \times 1 + c$ −3 = 4 × 1 + c, so $y = 4 x - 7$ y = 4x − 7.
Higher: perpendicular lines have gradients that multiply to $- 1$ −1. Use the negative reciprocal: gradient 2 becomes $- \frac{1}{2}$ −1/2.
Higher: $2 y = 3 x + 1$ 2y = 3x + 1 has gradient $\frac{3}{2}$ 3/2, so a perpendicular line has gradient $- \frac{2}{3}$ −2/3.

Calculate

?(Higher) Line

L

has equation

y = 2 x + 5

. Line

M

is perpendicular to

L

and passes through

(4, 1)

. Find the

y

-intercept of line

M

.

Algebra · straight-line graphs

Midpoint and length of a line segment

The midpoint averages the coordinates:

(\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})

The length comes from Pythagoras: $(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$ √((x₂ − x₁)² + (y₂ − y₁)²).
For $A (1, 2)$ A(1, 2) and $B (7, 10)$ B(7, 10): the midpoint is $(4, 6)$ (4, 6) and the length is $6^{2} + 8^{2} = 10$ √(6² + 8²) = 10.

Calculate

?Point

P

is

(- 3, 8)

and point

Q

is

(9, - 2)

. Find the length of

PQ

. Give your answer to 3 significant figures.

Mission check

?The rule: no steeper than a rise of 1 for every 12 along. Ramp P rises 0.6 m over 9 m, ramp Q is the line

y = 0.1 x

, ramp R rises 0.25 m over 3 m. Which ramps are legal?

Mission 5

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.
Foundation: $x^{2} + b x + c = 0$ x² + bx + c = 0. Higher: also $a x^{2} + b x + c = 0$ ax² + bx + c = 0 and equations you must rearrange first.

Worked example Algebra · quadratic equations

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

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 (Higher)

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 (Higher)

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

?(Higher) Solve

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

. Give the positive solution to 3 significant figures.

Algebra · quadratic equations

Completing the square (Higher)

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 6

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 (Higher)

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 (Higher)

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

?Higher. 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 7

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.

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.

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 (Higher)

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 (Higher)

$y = sin x$ 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 angles of any size, including negative angles.

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 8

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 (Higher)

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.
OCR questions may use function machines or $y = \dots$ y = … instead of $f (x)$ f(x) notation.

Worked example Algebra · functions

Composite functions (Higher)

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 (Higher)

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

Transforming graphs (Higher)

$y = f (x) + a$ y = f(x) + a: translation $(0 a)$ ( 0 ; a ), up by $a$ a.
$y = f (x + a)$ y = f(x + a): translation $(- a 0)$ ( −a ; 0 ), left by $a$ a. So $f (x - 3)$ f(x − 3) moves the graph right 3: you need $x = 3$ x = 3 to get the old $f (0)$ f(0).
$y = - f (x)$ y = −f(x): reflection in the $x$ x-axis. $y = f (- x)$ y = f(−x): reflection in the $y$ y-axis.
OCR writes these as equations, e.g. $y = (x + 2)^{2} - 1$ y = (x + 2)² − 1 is $y = x^{2}$ y = x² moved left 2, down 1.

Quick check

?The graph of

y = f (x)

has a turning point at

(2, - 3)

. Where is the turning point of

y = f (x - 4) + 1

?

Mission check

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

Mission 9

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, ….

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 (Higher)

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.

Worked example Algebra · sequences

Quadratic nth term (Higher)

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

?Higher. 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 10

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

Velocity–time graphs

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.
Higher: the area under the graph is the distance travelled. For a curve, estimate it with trapezia of equal width; if the curve is shaped like a ∩ (its gradient keeps decreasing), the estimate is too small.

Calculate

?Higher. A cyclist's velocity is 0, 7, 11, 13 and 14 m/s at

t = 0

, 2, 4, 6 and 8 s; the graph is a smooth curve. Use four trapezia of width 2 s to estimate the distance travelled in the 8 seconds.

m

Algebra · real-life graphs

Gradient of a curve (Higher)

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.
The gradient of a chord between two points is the average rate of change between them.

Calculate

?Higher. 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 11

Square any odd number and subtract 1. Always a multiple of 8?

Make a guess: Square any odd number and subtract 1. Always a multiple of 8?

Algebra · proof

Writing integers with algebra

Let $n$ n be any integer. Then $2 n$ 2n is always even and $2 n + 1$ 2n + 1 is always odd.
Consecutive integers: $n$ n, $n + 1$ n + 1, $n + 2$ n + 2. Consecutive even numbers: $2 n$ 2n, $2 n + 2$ 2n + 2. Consecutive odd numbers: $2 n + 1$ 2n + 1, $2 n + 3$ 2n + 3.
A multiple of 3 can be written $3 n$ 3n; any expression of the form $3 \times (integer)$ 3 × (integer) is a multiple of 3.
The goal is to write the result in a form like $4 (\dots)$ 4(…) or $2 (\dots) + 1$ 2(…) + 1 where the bracket is an integer.

Algebra · proof

Show that, proof and counterexample

Show that two expressions are equivalent: start from one side and work to the other. $(x + 1)^{2} + 2 = x^{2} + 2 x + 1 + 2 = x^{2} + 2 x + 3$ (x + 1)² + 2 = x² + 2x + 1 + 2 = x² + 2x + 3.
Checking examples is not a proof: the next number might fail.
Higher: a proof uses general expressions ( $n$ n, $2 n + 1$ 2n + 1, ...), simplifies them, and ends with a sentence saying why the result follows.
One counterexample is enough to prove a statement false.

Worked example Algebra · proof

Proving a result about consecutive integers (Higher)

Question

Prove that the sum of any three consecutive integers is a multiple of 3.

1
Write three consecutive integers in general: $n$ n, $n + 1$ n + 1, $n + 2$ n + 2 where $n$ n is an integer
2
Add them: $n + (n + 1) + (n + 2) = 3 n + 3$ n + (n + 1) + (n + 2) = 3n + 3
3
Factorise to show the factor 3: $3 n + 3 = 3 (n + 1)$ 3n + 3 = 3(n + 1)
4
Conclude in words: $n + 1$ n + 1 is an integer, so the sum is 3 times an integer

Answer

The sum is

3 (n + 1)

, which is a multiple of 3 for every integer

n

.

⚠️ Watch out

the final sentence earns a mark. "

= 3 (n + 1)

" on its own doesn't say why that is a multiple of 3.

Quick check

?

n

is an integer. Which working proves that the sum of two consecutive odd numbers is always a multiple of 4?

Sort it

?Is each statement true for every integer

n

, or can you find a counterexample?

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

Calculate

?

n

is an integer. Expand and simplify

(2 n + 3)^{2} - (2 n - 3)^{2}

to the form

kn

. Enter

k

(exact).

Algebra · proof

Finding roots by iteration (Higher)

Some equations, such as $x^{3} + 4 x - 1 = 0$ x³ + 4x − 1 = 0, can't be solved exactly by the usual methods. Find the root approximately.
Sign change: if $f (x) = x^{3} + 4 x - 1$ f(x) = x³ + 4x − 1 changes sign between $x = a$ x = a and $x = b$ x = b, there is a root between them: $f (0) = - 1$ f(0) = −1 and $f (1) = 4$ f(1) = 4. Narrowing the interval (decimal search or interval bisection) traps the root.
Iteration: rearrange to $x = g (x)$ x = g(x) and use $x_{n + 1} = g (x_{n})$ xₙ₊₁ = g(xₙ). Start with $x_{0}$ x₀, feed each answer back in, and stop when the values agree to the accuracy needed.

Worked example Algebra · proof

Using an iterative formula (Higher)

Question

Show that

x^{3} + 4 x - 1 = 0

has a root between 0 and 1. Use

x_{n + 1} = \frac{1 - x _{n 3}}{4}

with

x_{0} = 0

to find the root to 3 decimal places.

1
Sign change: $f (0) = - 1 < 0$ f(0) = −1 < 0 and $f (1) = 4 > 0$ f(1) = 4 > 0, so a root lies between 0 and 1
2
Iterate once: $x_{1} = \frac{1 - 0 ^{3}}{4} = 0.25$ x₁ = (1 − 0³)/4 = 0.25
3
Iterate again: $x_{2} = \frac{1 - 0.2 5 ^{3}}{4} = 0.24609...$ x₂ = (1 − 0.25³)/4 = 0.24609...
4
Keep going: $x_{3} = 0.24627...$ x₃ = 0.24627... and $x_{4} = 0.24626...$ x₄ = 0.24626..., which agree to 3 decimal places
5
Check with a sign change: $f (0.2455) < 0$ f(0.2455) < 0 and $f (0.2465) > 0$ f(0.2465) > 0

Answer

x = 0.246

to 3 decimal places

⚠️ Watch out

use the full calculator value each time (the ANS key), not a rounded one, and state the accuracy you were asked for.

Calculate

?Use

x_{n + 1} = 3 3 x_{n} + 5

with

x_{0} = 2

to find

x_{3}

. Give your answer to 3 decimal places.

Mission check

?Square any odd number and subtract 1. Is the answer always a multiple of 8?

Mission 12

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 (Higher)

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 (Higher)

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 (Higher)

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 (Higher)

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

Algebra · algebraic fractions

Quadratic inequalities (Higher)

Solve $x^{2} - x - 6 < 0$ x² − x − 6 < 0. First solve the equation $x^{2} - x - 6 = 0$ x² − x − 6 = 0: $(x - 3) (x + 2) = 0$ (x − 3)(x + 2) = 0, so the critical values are $- 2$ −2 and $3$ 3.
Sketch $y = x^{2} - x - 6$ y = x² − x − 6: a U-shaped curve crossing the $x$ x-axis at $- 2$ −2 and $3$ 3.
" $< 0$ < 0" means below the axis: one interval, $- 2 < x < 3$ −2 < x < 3.
" $> 0$ > 0" means above the axis: two separate parts, $x < - 2$ x < −2 or $x > 3$ x > 3.
In set notation, $x < - 2$ x < −2 or $x > 3$ x > 3 is ${x : x < - 2} \cup {x : x > 3}$ {x : x < −2} ∪ {x : x > 3}.

Quick check

?Solve

x^{2} > 25

.

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 13

A sprinkler reaches 13 m. Does it wet a plant at (12, 6)?

Make a guess: A sprinkler reaches 13 m. Does it wet a plant at (12, 6)?

Algebra · circles on a grid

The equation of a circle (Higher)

A point $(x, y)$ (x, y) on a circle with centre the origin and radius $r$ r makes a right-angled triangle with legs $x$ x and $y$ y.
By Pythagoras, every point on the circle satisfies

x^{2} + y^{2} = r^{2}

$x^{2} + y^{2} = 25$ x² + y² = 25 has radius 5: $(3, 4)$ (3, 4) is on it because $9 + 16 = 25$ 9 + 16 = 25.
If $x^{2} + y^{2} < r^{2}$ x² + y² < r² the point is inside; if $x^{2} + y^{2} > r^{2}$ x² + y² > r² it is outside.

Quick check

?What is the radius of the circle

x^{2} + y^{2} = 49

?

Worked example Algebra · circles on a grid

Finding a missing coordinate (Higher)

Question

The point

(p, 4)

lies on the circle

x^{2} + y^{2} = 41

, and

p < 0

. Find

p

.

1
Substitute the point: $p^{2} + 4^{2} = 41$ p² + 4² = 41
2
Simplify: $p^{2} = 25$ p² = 25
3
Square-root, keeping both signs: $p = 5$ p = 5 or $p = - 5$ p = −5
4
Use the condition $p < 0$ p < 0

Answer

p = - 5

⚠️ Watch out

the square root has two signs. Read the question to choose the right one.

Sort it

?Is each point inside, on or outside the circle

x^{2} + y^{2} = 100

?

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

Algebra · circles on a grid

The tangent at a point (Higher)

A tangent touches the circle at one point and is perpendicular to the radius at that point.
The radius from $O (0, 0)$ O(0, 0) to $P (a, b)$ P(a, b) has gradient $\frac{b}{a}$ b/a.
So the tangent at $P$ P has gradient $- \frac{a}{b}$ −a/b, the negative reciprocal.
Substitute $P$ P into $y = m x + c$ y = mx + c to find $c$ c.

Worked example Algebra · circles on a grid

Equation of a tangent (Higher)

Question

Find the equation of the tangent to

x^{2} + y^{2} = 25

at the point

(3, 4)

.

1
Gradient of the radius to $(3, 4)$ (3, 4): $\frac{4}{3}$ 4/3
2
Tangent gradient, the negative reciprocal: $- \frac{3}{4}$ −3/4
3
Substitute $(3, 4)$ (3, 4) into $y = - \frac{3}{4} x + c$ y = −3/4x + c: $4 = - \frac{9}{4} + c$ 4 = −9/4 + c, so $c = \frac{25}{4}$ c = 25/4
4
Multiply by 4 to clear fractions: $4 y = - 3 x + 25$ 4y = −3x + 25

Answer

y = - \frac{3}{4} x + \frac{25}{4}

, or

3 x + 4 y = 25

⚠️ Watch out

the radius gradient

\frac{4}{3}

is not the tangent gradient. Flip it and change the sign.

Quick check

?The point

(- 2, 4)

lies on

x^{2} + y^{2} = 20

. What is the gradient of the tangent at

(- 2, 4)

?

Calculate

?The point

(1, 3)

lies on the circle

x^{2} + y^{2} = 10

. The tangent to the circle at

(1, 3)

crosses the

x

-axis at

Q

. Find the

x

-coordinate of

Q

.

Calculate

?The line

y = x + 1

meets the circle

x^{2} + y^{2} = 25

at two points. Find the positive

x

-coordinate of the points of intersection.

Mission check

?A sprinkler at the origin sprays a circle of radius 13 m. Is the plant at

(12, 6)

inside, on or outside the wet circle?

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 equation is true for some values, an identity ( $\equiv$ ≡) for every value, 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.
Gradient = change in $y$ y ÷ change in $x$ x; in $y = m x + c$ y = mx + c, $m$ m is the gradient and $(0, c)$ (0, c) the $y$ y-intercept.
To find a line's equation: gradient first, then substitute a point to find $c$ c.
Parallel lines share a gradient; perpendicular gradients multiply to $- 1$ −1 (Higher).
Midpoint: average the coordinates. Length: Pythagoras on the changes in $x$ x and $y$ y.
Make one side 0, factorise, and set each bracket to 0: each bracket gives a root.
Higher: 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 = ….
Higher: 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.
$f (x) + a$ f(x) + a moves the graph up $a$ a; $f (x + a)$ f(x + a) moves it left $a$ a; $- f (x)$ −f(x) and $f (- x)$ f(−x) reflect in the $x$ x- and $y$ y-axes.
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.
Write general numbers with algebra: even $2 n$ 2n, odd $2 n + 1$ 2n + 1, consecutive $n$ n, $n + 1$ n + 1.
A proof simplifies the general expression into the form you need and finishes with a sentence; examples never prove, but one counterexample disproves.
A sign change traps a root; iterate $x_{n + 1} = g (x_{n})$ xₙ₊₁ = g(xₙ) until the values agree to the accuracy required.
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.
For a quadratic inequality, find the critical values, sketch the curve, and read off where it is above or below the axis.
A circle with centre the origin and radius $r$ r has equation $x^{2} + y^{2} = r^{2}$ x² + y² = r²; compare $x^{2} + y^{2}$ x² + y² with $r^{2}$ r² to test a point.
The tangent at $P$ P is perpendicular to the radius $OP$ OP: use the negative reciprocal of the radius gradient.
To find where a line meets the circle, substitute the line into $x^{2} + y^{2} = r^{2}$ x² + y² = r² and solve the quadratic.

🏆

Mini-lesson complete!

⭐⭐⭐

You've worked through Algebra for OCR GCSE Maths (J560). 🎉

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 4Rise 1 for every 12 along: which of three ramps is legal?
Mission 5A thrown ball is 10 m up at two different times. When, and why?
Mission 63 coffees + 2 cakes = £13.50; 2 coffees + 3 cakes = £13. A cake?
Mission 7Four graphs, four equations: can you match them without plotting?
Mission 8A machine doubles your number, then adds 3. How do you undo it?
Mission 9Chessboard: 1 grain, then double each square. How many on square 64?
Mission 10Two runners' distance–time graphs cross. What does the crossing mean?
Mission 11Square any odd number and subtract 1. Always a multiple of 8?
Mission 12Cycle out at 10 km/h, back at 30 km/h. Is your average 20 km/h?
Mission 13A sprinkler reaches 13 m. Does it wet a plant at (12, 6)?

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

Algebra

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 (Higher)

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

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 (Higher)

Rise 1 for every 12 along: which of three ramps is legal?

Gradient measures steepness

The equation y=mx+cy = mx + c

Equation of the line through two points

Parallel and perpendicular lines

Midpoint and length of a line segment

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

The quadratic formula (Higher)

Using the formula (Higher)

Completing the square (Higher)

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 (Higher)

A line meeting a curve (Higher)

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

Lines and parabolas

Reading a quadratic graph

Cubics and reciprocals

Exponential graphs (Higher)

Trigonometric graphs (Higher)

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

Function notation (Higher)

Composite functions (Higher)

Inverse functions (Higher)

Transforming graphs (Higher)

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 (Higher)

Quadratic nth term (Higher)

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

Distance–time graphs

Reading two runners' graphs

Conversion graphs and rates

Velocity–time graphs

Gradient of a curve (Higher)

Square any odd number and subtract 1. Always a multiple of 8?

Writing integers with algebra

Show that, proof and counterexample

Proving a result about consecutive integers (Higher)

Finding roots by iteration (Higher)

Using an iterative formula (Higher)

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

Simplifying algebraic fractions (Higher)

The four operations with algebraic fractions (Higher)

Subtracting algebraic fractions (Higher)

Solving an equation with algebraic fractions (Higher)

Quadratic inequalities (Higher)

A sprinkler reaches 13 m. Does it wet a plant at (12, 6)?

The equation of a circle (Higher)

Finding a missing coordinate (Higher)

The tangent at a point (Higher)

Equation of a tangent (Higher)

Big ideas

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 (Higher)

The equation $y = m x + c$ y = mx + c

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