Edexcel International GCSE Mathematics A 4MA1: Inequalities – Study Guide

This study guide covers section 2.8 Inequalities in Topic 2 (Equations, formulae and identities) of the Pearson Edexcel International GCSE Mathematics A (4MA1) specification, Issue 2 (November 2017), for the January and June series examined on that specification. Statements 2.8A–E are Foundation tier content, so you need them on both tiers. The Higher tier adds two more statements: quadratic inequalities and harder regions. These are labelled Higher tier only below. A calculator may be used on every 4MA1 paper.

The rest of Topic 2 is in the use of symbols and algebraic manipulation guide, its revision notes and its practice questions. The Higher tier work here needs its factorising skills. The Edexcel IGCSE Mathematics course hub links the whole course, and the printable checklist lists every specification statement.

What this section covers

Statement	What you must be able to do	Tier
2.8A (Foundation)	Understand and use the symbols >, <, ≥ and ≤, including double-ended inequalities such as 1 < x ≤ 5	Both tiers
2.8B (Foundation)	Use the convention for open and closed intervals on a number line	Both tiers
2.8C (Foundation)	Solve simple linear inequalities in one variable and show the solution set on a number line	Both tiers
2.8D (Foundation)	Represent simple linear inequalities on Cartesian graphs (shade a region)	Both tiers
2.8E (Foundation)	Identify regions on Cartesian graphs defined by simple linear inequalities	Both tiers
2.8A (Higher)	Solve quadratic inequalities in one unknown and show the solution set on a number line	Higher tier only
2.8B (Higher)	Identify harder examples of regions defined by linear inequalities	Higher tier only

2.8A The four symbols

Symbol	Meaning	Example	Is 3 allowed?
>	greater than	x > 3	No
<	less than	x < 3	No
≥	greater than or equal to	x ≥ 3	Yes
≤	less than or equal to	x ≤ 3	Yes

The narrow end of the symbol points at the smaller quantity, so 2 < 7 and 7 > 2 say the same thing.

A double-ended inequality gives a lower and an upper limit together. The statement −2 < x ≤ 6 means x is greater than −2 and x is less than or equal to 6.

Worked example 1: integers in a range

List the integers n such that −3 ≤ n < 2.

−3 ≤ n  → −3 is included
n < 2   → 2 is not included

The integers are −3, −2, −1, 0, 1.

2.8B Open and closed circles on a number line

On a number line:

• a closed (filled) circle ● means the end value is included (≥ or ≤)
• an open (empty) circle ○ means the end value is not included (> or <)
• an arrow shows the solution carries on for ever in that direction
• a line segment between two circles shows a double-ended inequality.

x > −1          ○──────────→      open at −1
x ≤ 4      ←──────────●           closed at 4
−2 < x ≤ 3      ○──────────●      open at −2, closed at 3

2.8C Solving linear inequalities

Solve a linear inequality exactly like a linear equation, with one extra rule:

If you multiply or divide both sides by a negative number, reverse the inequality sign.

Adding, subtracting, and multiplying or dividing by a positive number keep the sign. The answer is an inequality, not a single value.

Worked example 2: unknown on both sides

Solve 5x + 3 > 2x − 9.

5x + 3 > 2x − 9
3x + 3 > −9        subtract 2x from both sides
    3x > −12       subtract 3
     x > −4        divide by 3 (positive, so the sign stays)

Solution: x > −4. On a number line: an open circle at −4 with an arrow to the right.

Worked example 3: dividing by a negative

Solve 4 − 3x ≥ 19.

4 − 3x ≥ 19
   −3x ≥ 15        subtract 4
     x ≤ −5        divide by −3, so ≥ becomes ≤

Solution: x ≤ −5. Check: x = −6 gives 22, and 22 ≥ 19 is true.

Alternatively, move the x-term across: −15 ≥ 3x, so −5 ≥ x, which is the same as x ≤ −5.

Worked example 4: a double-ended inequality

Solve −7 ≤ 2x − 3 < 5 and list the integer solutions.

Do the same operation to all three parts:

−7 ≤ 2x − 3 < 5
−4 ≤ 2x     < 8        add 3 to every part
−2 ≤  x     < 4        divide every part by 2

Solution: −2 ≤ x < 4. Integer solutions: −2, −1, 0, 1, 2, 3. On a number line: a closed circle at −2, an open circle at 4, joined by a line.

Worked example 5: a fraction

Solve (x + 4)/3 ≤ x − 2.

(x + 4)/3 ≤ x − 2
   x + 4  ≤ 3(x − 2)     multiply both sides by 3 (positive)
   x + 4  ≤ 3x − 6
      10  ≤ 2x           subtract x, add 6
       5  ≤ x

Solution: x ≥ 5.

2.8D Representing inequalities as regions on a graph

Every straight line splits the coordinate plane into two halves. An inequality in x and y describes one of those halves.

• x ≥ a: the region to the right of the vertical line x = a
• x ≤ a: the region to the left of x = a
• y ≥ b: the region above the horizontal line y = b
• y ≤ mx + c: the region below the line y = mx + c
• for a line such as x + y ≤ 6, use a test point (the origin is easiest if it is not on the line): 0 + 0 = 0 and 0 ≤ 6, so the side containing (0, 0) is the required side.

The specification notes that conventions for showing whether a boundary is included are not required. Many textbooks use solid lines for ≤ or ≥ and dashed lines for < or >; what matters is the right lines and the right region. Always follow the question’s instruction about shading and labelling (for example “label the region R”).

Worked example 6: shading a region

Show the region defined by x ≥ 1, y ≥ x and x + y ≤ 6. Label it R.

1. Draw the three boundary lines:
   • x = 1 is vertical through (1, 0)
   • y = x passes through (0, 0), (2, 2), (4, 4)
   • x + y = 6 passes through (0, 6) and (6, 0)
2. Decide the side of each line:
   • x ≥ 1: to the right of x = 1
   • y ≥ x: above y = x (test (0, 3): 3 ≥ 0, true, and (0, 3) is above the line)
   • x + y ≤ 6: test (0, 0): 0 ≤ 6, true, so the side containing the origin
3. R is the triangle where all three conditions hold. Its vertices are where pairs of lines meet:
   • x = 1 and y = x: (1, 1)
   • x = 1 and x + y = 6: (1, 5)
   • y = x and x + y = 6: 2x = 6, so (3, 3)

2.8E Identifying the inequalities for a given region

Here you are given the region and must write the inequalities. Work one boundary line at a time:

1. Find the equation of each boundary line (vertical x = a, horizontal y = b, or y = mx + c).
2. Pick a point clearly inside the region.
3. Substitute it to decide which symbol is needed.

Worked example 7

A triangular region R has vertices (−2, −1), (3, −1) and (3, 4). Its sides lie on y = −1, x = 3 and y = x + 1. Write down the three inequalities that define R.

Choose the test point (1, 0), which is inside the triangle.

Line	Value at (1, 0)	Inequality
y = −1	y = 0, which is above −1	y ≥ −1
x = 3	x = 1, which is left of 3	x ≤ 3
y = x + 1	0 compared with 1 + 1 = 2: 0 is less	y ≤ x + 1

Higher tier only: 2.8A Quadratic inequalities

The method is: find the critical values, then decide whether you want the inside or the outside of them.

1. Rearrange so one side is 0 (if needed): ax² + bx + c < 0, > 0, ≤ 0 or ≥ 0.
2. Solve the equation ax² + bx + c = 0 to get the critical values. Factorise if you can; otherwise use the quadratic formula or completing the square (statement 2.7B).
3. Sketch the parabola. For a positive x² coefficient it is U-shaped, below the x-axis between the roots and above it outside them.
4. < 0 or ≤ 0: the part between the roots (one inequality, e.g. 2 < x < 7). > 0 or ≥ 0: the two parts outside the roots (two inequalities, joined by “or”).

Worked example 8: the x² ≤ k type

Solve (a) x² < 36, (b) 9x² ≥ 16.

(a) The critical values are x = ±6. Squares less than 36 come from numbers between −6 and 6, so −6 < x < 6.

(b) x² ≥ 16/9, critical values x = ±4/3. Squares at least 16/9 come from numbers at least 4/3 away from zero: x ≤ −4/3 or x ≥ 4/3.

Writing only x < 6 in (a) wrongly includes x = −10, whose square is 100.

Worked example 9: factorising, “greater than”

Solve x² − 2x − 15 > 0.

x² − 2x − 15 = (x − 5)(x + 3)
critical values: x = 5 and x = −3

The graph is U-shaped and crosses the x-axis at −3 and 5. It is above the axis (> 0) outside these values:

x < −3 or x > 5

On a number line: an open circle at −3 with an arrow to the left, and an open circle at 5 with an arrow to the right.

Worked example 10: a non-unit x² coefficient

Solve 2x² + 5x − 3 ≤ 0.

2x² + 5x − 3 = (2x − 1)(x + 3)
critical values: x = 1/2 and x = −3

“≤ 0” means on or below the axis, which is between the roots:

−3 ≤ x ≤ 1/2

On a number line: closed circles at −3 and 1/2, joined by a line.

Check: x = 0 gives −3 ≤ 0, true.

Worked example 11: rearrange first

Solve x(x + 4) ≥ 12.

Do not set each factor ≥ 12. Expand and rearrange so the right-hand side is zero:

x² + 4x ≥ 12
x² + 4x − 12 ≥ 0
(x + 6)(x − 2) ≥ 0
critical values: x = −6 and x = 2

“≥ 0” is outside the roots: x ≤ −6 or x ≥ 2.

Higher tier only: 2.8B Harder regions

Harder examples can include lines written in the form ax + by = c, more than one sloping line, and vertices with non-integer coordinates. The method is the same as 2.8D and 2.8E.

Worked example 12

R is the region defined by x ≥ 0, y > x − 2 and 2x + 3y ≤ 12. Find the coordinates of the vertices of R and decide whether (3, 3) lies in R.

Boundary lines:

• x = 0 is the y-axis.
• y = x − 2 passes through (0, −2) and (2, 0).
• 2x + 3y = 12: when x = 0, y = 4; when y = 0, x = 6. Plot (0, 4) and (6, 0).

Sides: x ≥ 0 is right of the y-axis. y > x − 2 is above that line (test (0, 0): 0 > −2, true). 2x + 3y ≤ 12 contains the origin (0 ≤ 12).

Vertices:

x = 0 and y = x − 2:        (0, −2)
x = 0 and 2x + 3y = 12:     (0, 4)
y = x − 2 and 2x + 3y = 12:
   2x + 3(x − 2) = 12
   5x − 6 = 12
   x = 18/5 = 3.6,  y = 1.6     (3.6, 1.6)

Test (3, 3): x ≥ 0 is true and 3 > 1 is true, but 2(3) + 3(3) = 15, which is not ≤ 12. So (3, 3) is not in R.

The third vertex is not on grid points, so find it by solving the equations simultaneously (statement 2.6), not by reading the graph.

Common errors

• Forgetting to reverse the sign after multiplying or dividing by a negative number.
• Giving x = 3 instead of x < 3 as the final answer.
• Using a filled circle for < or >, or an open circle for ≤ or ≥.
• In a double-ended inequality, applying an operation to only two of the three parts.
• Listing integers and including the end value that has a strict symbol, or leaving out 0.
• Shading the wrong side of a sloping line because you guessed rather than used a test point.
• Higher: writing x² > 25 as x > 5 only, or x² < 25 as x < ±5.
• Higher: giving “outside” solutions as one combined statement such as 5 < x < −3, which no number satisfies.

Next steps

Move on to the revision notes, then the practice questions. To find gaps elsewhere, take a free diagnostic.

Official syllabus

Pearson Edexcel International GCSE in Mathematics (Specification A) (4MA1), Specification, Issue 2, November 2017, Pearson Education Limited (first assessment June 2018). Topic 2, Equations, formulae and identities, section 2.8 Inequalities (Foundation tier statements 2.8A–E and Higher tier statements 2.8A–B).