Mini-Lesson

Equations, Formulae & Identities

7 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 6Cycle out at 10 km/h, back at 30 km/h. Is your average 20 km/h?
Mission 7Square any odd number and subtract 1. Always a multiple of 8?

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 (Higher)

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.

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

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

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.
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.
Reading the solution as the point where two lines intersect is Higher tier (2.6B).

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 6

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.

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 7

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 (Higher)

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 (Higher)

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).

Calculate

?

n

is an integer. Expand and simplify

(n + 5)^{2} - (n + 1)^{2}

and write it as

8 (n + a)

, which proves it is a multiple of 8. Enter

a

(exact).

Mission check

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

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.
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 in the form $a \pm b$ a ± √b.
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.
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.
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.

🏆

Mini-lesson complete!

⭐⭐⭐

You've worked through Equations, Formulae & Identities for Edexcel International GCSE Maths A (4MA1). 🎉

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 6Cycle out at 10 km/h, back at 30 km/h. Is your average 20 km/h?
Mission 7Square any odd number and subtract 1. Always a multiple of 8?

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

Equations, Formulae & Identities

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 (Higher)

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

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)

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)

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

Writing integers with algebra (Higher)

Show that, proof and counterexample (Higher)

Proving a result about consecutive integers (Higher)

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 (Higher)

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