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

This guide is for OxfordAQA International GCSE Mathematics (9260), aligned to the specification Version 5.1 (for exams May/June 2018 onwards). It covers references A19 and A22 in section 3.2.3 (Solving equations and inequalities). Every statement in these two references is Core content, so it is also assessable on the Extension papers. There is no Extension-only content to separate out here.

Related pages: revision notes, practice questions, algebra notation and manipulation, the course hub, the printable checklist and the free 10-minute diagnostics.

What this unit covers

Ref	What you must be able to do	Tier
A19	Solve linear equations in one unknown algebraically, including brackets and the unknown on both sides	Core
A19	Find approximate solutions using a graph	Core
A22	Translate simple situations or procedures into algebraic expressions or formulae	Core
A22	Derive an equation (or two simultaneous equations), solve it and interpret the solution	Core
A22	Apply this to geometrical problems and problems set in context	Core

Solving quadratic equations, the full methods for simultaneous equations, and inequalities each have their own pages. Here, simultaneous equations appear only as something you form from a problem.

1. Solving a linear equation (A19)

In a linear equation the unknown appears only to the first power. Treat the equation like a pair of scales: any operation applied to one side must be applied to the other. Your aim is to isolate the unknown.

Method:

1. Expand any brackets.
2. Clear any fractions by multiplying every term by a common multiple of the denominators.
3. Collect the unknown terms on one side and the numbers on the other.
4. Divide by the coefficient of the unknown.
5. Check by substituting back into the original equation.

Worked example 1: unknown on both sides

Solve 5x − 7 = 3x + 11.

5x − 7 = 3x + 11
5x − 3x = 11 + 7      (subtract 3x, add 7)
2x = 18
x = 9

Check: left side 5(9) − 7 = 38, right side 3(9) + 11 = 38.

Worked example 2: brackets on both sides

Solve 4(2x − 3) = 3(x + 6) − 5.

8x − 12 = 3x + 18 − 5
8x − 12 = 3x + 13
5x = 25
x = 5

Check: left side 4(7) = 28, right side 3(11) − 5 = 28.

Worked example 3: a negative coefficient

Solve 7 − 3x = 2x + 22.

7 − 22 = 2x + 3x      (add 3x, subtract 22)
−15 = 5x
x = −3

Check: left side 7 + 9 = 16, right side −6 + 22 = 16. Moving the x terms to the side where they are positive avoids sign slips, but either route works.

Worked example 4: fractions

Solve (2x + 1)/3 − (x − 4)/2 = 2.

Both denominators divide into 6, so multiply every term by 6, including the 2 on the right:

2(2x + 1) − 3(x − 4) = 12
4x + 2 − 3x + 12 = 12
x + 14 = 12
x = −2

The bracket (x − 4) is subtracted, so its minus sign changes both terms inside. Check: (−3)/3 − (−6)/2 = −1 + 3 = 2.

2. Approximate solutions from a graph (A19)

An equation can be solved by drawing a graph for each side and reading the x-coordinate where the two graphs meet. Because you are reading a scale, the answer is approximate. Say “x ≈ 2.3” and give it to the accuracy the scale allows.

Method:

1. Write each side of the equation as y = (something).
2. Make a table of values, or use the gradient and intercept, and draw both lines.
3. Find where they cross.
4. Read off the x-coordinate (not the y-coordinate) of the crossing point.

Worked example 5

Use a graph to find an approximate solution of 4 − x = 2x − 3.

Draw y = 4 − x and y = 2x − 3 for x from 0 to 4.

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

At x = 2 the first line is above the second (2 against 1). At x = 3 it is below (1 against 3). The lines therefore cross between x = 2 and x = 3, about one third of the way along. On a carefully drawn graph you would read x ≈ 2.3.

Solving algebraically gives 3x = 7, so x = 7/3 ≈ 2.33. The two agree, which is a useful check on a graph question that also asks for the algebra.

A shortcut: to solve 3x − 5 = 4 from the graph of y = 3x − 5, draw the horizontal line y = 4 and read the x-coordinate where it meets the graph. The graph meets y = 4 at x = 3. To solve 3x − 5 = 0, read where the graph meets the x-axis: x = 5/3 ≈ 1.7.

3. Translating situations into expressions and formulae (A22)

Choose a letter for the unknown quantity, say what it stands for, and build the expression from the words.

Words	Algebra
5 more than n	n + 5
5 less than n	n − 5
three times the sum of n and 4	3(n + 4)
half of n, then add 6	n/2 + 6
the next two even numbers after 2k	2k + 2 and 2k + 4
a number 4 less than n, then multiplied by 3	3(n − 4)

A formula links a quantity you want to a letter you choose.

Worked example 6. A garden designer charges £40 for a first visit plus £26 for each hour worked.

• Let C be the total charge in pounds and h the number of hours. Then C = 40 + 26h.
• For 3.5 hours: C = 40 + 26 × 3.5 = 40 + 91 = 131, so the charge is £131.

Always state what each letter means. An examiner reading your working should never have to guess what x represents.

4. Forming and solving an equation (A22)

Follow the same steps every time:

1. Define the unknown (“Let x be …”, with units if there are any).
2. Write expressions for the other quantities.
3. Find the fact that gives an equation: two things are equal, or parts sum to a total.
4. Solve.
5. Interpret: answer the question asked, put units on it, and check it makes sense in the situation.

Geometrical problems

Worked example 7 (angles). The angles of a triangle are (2x + 10)°, (3x − 5)° and (x + 25)°. Find the three angles.

(2x + 10) + (3x − 5) + (x + 25) = 180     (angles in a triangle)
6x + 30 = 180
6x = 150
x = 25

The angles are 60°, 70° and 50°. Check: 60 + 70 + 50 = 180.

Worked example 8 (perimeter). A rectangle has length (3x − 2) cm and width (x + 4) cm. Its perimeter is 60 cm. Find the length and width.

2[(3x − 2) + (x + 4)] = 60
2(4x + 2) = 60
8x + 4 = 60
8x = 56
x = 7

Length = 3(7) − 2 = 19 cm and width = 7 + 4 = 11 cm. Check: 2(19 + 11) = 60.

Problems set in context

Worked example 9 (comparing costs). Shop A hires out a bike for a £15 deposit plus £6 per day. Shop B charges £9 per day with no deposit. For how many days is the total cost the same?

Let d be the number of days.

15 + 6d = 9d
15 = 3d
d = 5

Interpretation: after 5 days both shops have charged £45 (15 + 30 = 45 and 9 × 5 = 45). For a shorter hire Shop B is cheaper; for a longer hire Shop A is cheaper.

Worked example 10 (consecutive numbers). Three consecutive odd numbers add up to 111. Find them.

Let the smallest be x. The numbers are x, x + 2 and x + 4.

x + (x + 2) + (x + 4) = 111
3x + 6 = 111
x = 35

The numbers are 35, 37 and 39. Check: 35 + 37 + 39 = 111.

Interpreting the solution

A solution of the equation is not automatically a solution of the problem. Ask:

• Is the value possible? A length, a number of people or a time cannot be negative.
• Must it be a whole number? You cannot buy 2.5 tickets.
• Does it answer the question? Often x is only a step, and you must go on to find a length, an angle or a cost.
• Does it satisfy the original situation? Substitute into the words, not only into your equation.

5. Forming two simultaneous equations (A22)

Use two unknowns when a problem gives two separate facts. Write one equation per fact, then solve the pair. The full methods for solving simultaneous equations are covered on their own page, so only a short solution is shown here.

Worked example 11. Two adult tickets and three child tickets cost £46.50. One adult ticket and five child tickets cost £49.50. Find the price of an adult ticket and of a child ticket.

Let a be the price of an adult ticket and c the price of a child ticket, both in pounds.

2a + 3c = 46.50     (1)
a + 5c = 49.50      (2)

Multiply (2) by 2: 2a + 10c = 99 (3). Subtract (1) from (3):

7c = 52.50
c = 7.50

Substitute into (2): a + 37.50 = 49.50, so a = 12.

An adult ticket costs £12 and a child ticket costs £7.50. Check in (1): 24 + 22.50 = 46.50.

Common errors

• Subtracting a bracket and changing only the first term: −(x − 4) is −x + 4, not −x − 4.
• Multiplying only some terms by the common denominator. Every term on both sides must be multiplied.
• Moving a term across the equals sign without changing its sign.
• Reading the y-coordinate of the crossing point when the question asks for the solution of the equation. The solution is the x-value.
• Giving a graph answer as an exact value such as 2.3333 when the graph can only support x ≈ 2.3.
• Starting “x = …” with no definition of x. Say what the letter stands for.
• Stopping at x when the question asks for a length, an age or a cost.
• Forming an equation in which the two sides do not describe equal quantities, such as adding a width to a perimeter.

Next steps

Revise the key results in the revision notes, then test yourself with the practice questions. Tick off the content on the printable checklist and find gaps with the free diagnostics.

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.