OxfordAQA International GCSE Mathematics: Linear equations and forming equations from problems (9260) – Revision Notes

These notes summarise references A19 and A22 of section 3.2.3 in the OxfordAQA International GCSE Mathematics (9260) specification, Version 5.1 (for exams May/June 2018 onwards). All of this content is Core, so it is also assessable on the Extension papers. Longer explanations and more worked examples are in the study guide. Test yourself with the practice questions, see the course hub, tick off the checklist and take a free diagnostic to find weak spots.

At a glance

Ref	Skill	Key idea
A19	Solve linear equations	Keep the equation balanced; leave the unknown alone
A19	Brackets, unknown on both sides	Expand first, then collect terms
A19	Approximate solution from a graph	Draw each side as a line; read the x-value where they cross
A22	Expressions and formulae	Define a letter, then build from the words
A22	Form and solve an equation	Equal quantities, or parts that add to a total
A22	Two simultaneous equations	One unknown per fact; one equation per fact
A22	Interpret	Units, sense check, answer the actual question

Solving a linear equation

Steps:

1. Expand brackets (watch a minus sign in front of a bracket).
2. Multiply every term by the common denominator to clear fractions.
3. Collect unknown terms on one side, numbers on the other.
4. Divide by the coefficient.
5. Substitute back to check.

Reminder A (unknown on both sides): 6x + 5 = 2x − 19.

6x − 2x = −19 − 5
4x = −24
x = −6

Reminder B (brackets, with a subtraction): 2(3x − 1) − (x − 7) = 20.

6x − 2 − x + 7 = 20
5x + 5 = 20
x = 3

Reminder C (fractions on both sides): (x + 3)/4 = (2x − 1)/3. Multiply both sides by 12.

3(x + 3) = 4(2x − 1)
3x + 9 = 8x − 4
13 = 5x
x = 13/5 = 2.6

Reminder D (fractions added): x/3 + x/4 = 14. Multiply by 12.

4x + 3x = 168
7x = 168
x = 24

Check every solution by substituting it into the original equation, not into a line of your own working. This catches both slips in expanding and slips in moving terms. Leave an answer as a fraction or a decimal as the question asks. A non-integer answer is not a sign of an error.

Approximate solutions from a graph

• Solve f(x) = g(x) by drawing y = f(x) and y = g(x). The solution is the x-coordinate of each crossing point.
• Solve f(x) = a number k by drawing y = f(x) and the horizontal line y = k.
• Solve f(x) = 0 by reading where y = f(x) meets the x-axis.
• Give answers to the accuracy the scale allows (one decimal place is typical). A graph answer is approximate; an algebraic answer is exact.

Reminder E: 2x − 1 = 5 − x. Table for x = 0, 1, 2, 3:

x	0	1	2	3
y = 2x − 1	−1	1	3	5
y = 5 − x	5	4	3	2

The values are equal at x = 2 (both 3), so the lines cross at (2, 3) and the solution is x = 2. Algebra confirms it: 3x = 6.

Forming expressions and formulae

Words	Algebra
a number, doubled, then 7 less	2n − 7
4 more than half of a number	n/2 + 4
the sum of three consecutive integers starting at n	n + (n + 1) + (n + 2) = 3n + 3
cost of t tickets at £8.50 plus a £2 fee	C = 8.5t + 2
a rectangle 5 cm longer than it is wide, width w	length w + 5, perimeter 2(2w + 5)

Method for forming an equation

1. Let x be … (say what, with units).
2. Write the other quantities in terms of x.
3. Find the fact that gives equality: angle sum, perimeter, equal cost, total.
4. Solve.
5. Interpret and check against the original words.

Useful facts for geometrical problems:

• Angles in a triangle add to 180°. Angles in a quadrilateral add to 360°. Angles on a straight line add to 180°.
• Opposite sides of a rectangle are equal. Perimeter = sum of all sides.
• Equal sides or equal angles in an isosceles triangle give an equation by setting them equal.

Reminder F (triangle). Angles x°, 2x° and (x + 20)° in a triangle:

x + 2x + (x + 20) = 180
4x + 20 = 180
x = 40

Angles 40°, 80°, 60°.

Reminder G (rectangle). Width x cm, length (x + 6) cm, perimeter 44 cm:

2(2x + 6) = 44
4x + 12 = 44
x = 8

Width 8 cm, length 14 cm.

Reminder H (consecutive integers). Three consecutive integers sum to 87:

x + (x + 1) + (x + 2) = 87
3x + 3 = 87
x = 28

The integers are 28, 29, 30.

Two simultaneous equations from a problem

Define two letters. Write one equation for each fact. Solve the pair, then answer in context with units.

Reminder I. Two pens and one notebook cost £7.50. One pen and two notebooks cost £6.60. Let p be the price of a pen and n the price of a notebook, in pounds.

2p + n = 7.50     (1)
p + 2n = 6.60     (2)

Doubling (2) gives 2p + 4n = 13.20, and subtracting (1) leaves 3n = 5.70, so n = 1.90. Substitute into (1): 2p = 5.60, so p = 2.80. A pen costs £2.80 and a notebook costs £1.90. Check in (2): 2.80 + 3.80 = 6.60.

Comparing two costs

When two rates are compared, form each cost, set them equal, and then interpret.

Reminder J. Firm A charges £3 plus £2 per km. Firm B charges £6 plus £1.50 per km. Let d be the distance in km.

3 + 2d = 6 + 1.5d
0.5d = 3
d = 6

Both firms charge £15 for 6 km. For a shorter journey Firm A is cheaper because its fixed charge is lower; for a longer journey Firm B is cheaper because its rate per km is lower. State this in words when the question asks you to interpret.

Rejecting an impossible solution

Reminder K. A rectangle has width x cm and length (x + 5) cm, and a perimeter of 4x + 10. A student sets the perimeter equal to 3x + 4 and finds x = −6. A width cannot be negative, so no such rectangle exists. When you reach an impossible value, say so in a sentence and check the equation for a slip before concluding.

Must-know distinctions

• Expression versus equation: an expression has no equals sign and cannot be “solved”; an equation can.
• Equation versus formula: an equation is solved to find the value of the unknown; a formula is a rule linking letters, such as C = 40 + 26h.
• Solution versus answer: x = 4 solves the equation, but the question may ask for a cost, a length or a number of items.
• Graph versus algebra: graphs give approximate values read from a scale; algebra gives exact values.
• x-value versus y-value: the solution of an equation from a crossing point is the x-coordinate.

Quick self-test

1. Solve 3x + 8 = 5x − 6.
2. Solve 5(x − 2) = 3(x + 4).
3. Solve (x + 3)/4 = (2x − 1)/3.
4. Solve 7 − 2x = 3x + 22.
5. Write an expression for the perimeter of a rectangle with width w and length 3w − 1.
6. The angles of a triangle are x°, 2x° and 3x°. Find x.
7. A taxi charges £4 plus £1.50 per km. Write a formula for the cost C of a journey of d km.
8. Using the formula in question 7, find the distance for a cost of £16.
9. Using graphs of y = 2x + 1 and y = 7, state the solution of 2x + 1 = 7.
10. The sum of three consecutive integers is 66. Find them.

Answers

1. 3x + 8 = 5x − 6 gives 14 = 2x, so x = 7.
2. 5x − 10 = 3x + 12 gives 2x = 22, so x = 11.
3. x = 13/5 = 2.6, as in reminder C.
4. −15 = 5x, so x = −3.
5. 2(w + 3w − 1) = 8w − 2.
6. 6x = 180, so x = 30 (angles 30°, 60°, 90°).
7. C = 4 + 1.5d.
8. 4 + 1.5d = 16 gives 1.5d = 12, so d = 8 km.
9. The graphs meet where y = 7, at x = 3.
10. 3n + 3 = 66 gives n = 21, so 21, 22, 23.

Where marks are usually lost

• Leaving a whole-number term unmultiplied when clearing fractions.
• A sign error when expanding −(x − 7) or −2(3 − x).
• Forgetting to change sign when moving a term to the other side.
• Giving the y-coordinate of the crossing point instead of the x-coordinate.
• Writing an equation without defining the letter.
• Solving for x and stopping, when the question asks for an angle, length or cost.
• Not checking whether the answer is sensible (negative length, 2.5 people).
• Rounding too early. Keep exact values until the final answer.

Official syllabus

OxfordAQA International GCSE Mathematics (9260) specification, Version 5.1, for exams May/June 2018 onwards, published by OxfordAQA. Section 3.2.3, references A19 and A22.