Inequalities · Study Notes · Edexcel IGCSE 4MA1 Higher

Linear and quadratic inequalities. Number-line representation. Critical: flip when multiplying/dividing by a negative.

At a glance

Solve like equations, but with $<, >, \le, \ge$ .
FLIP when × or ÷ by a NEGATIVE.
Quadratic: factor, find roots, test regions.
Number line: open circle for strict, closed for inclusive.
Integer solutions: list ALL integers in range.
Set notation $\{x : ...\}$ also acceptable.

What you’ll learn

Mapped to the Pearson Edexcel IGCSE 4MA1 syllabus (2026 onwards).

2.8 — Solve linear inequalities.
2.8 — Solve quadratic inequalities.
2.8 — Represent solutions on number lines.
2.8 — Find integer solutions in a range.

Linear inequalities

▼

Solve like equations. FLIP when × or ÷ by negative.

Standard methods (same as for equations) EXCEPT:

When MULTIPLYING or DIVIDING both sides by a NEGATIVE, FLIP THE INEQUALITY.
Adding/subtracting: NO flip.
Multiplying/dividing by POSITIVE: NO flip.

Example. $3x - 5 < 13$ .

Add $5$ : $3x < 18$ .
Divide by $3$ : $x < 6$ . (Positive; no flip.)

Example with flip. $-2x + 7 \le 1$ .

Subtract $7$ : $-2x \le -6$ .
Divide by $-2$ AND FLIP: $x \ge 3$ .

Verify. Test $x = 3$ : $-2(3) + 7 = 1 \le 1$ ✓. Test $x = 4$ : $-1 \le 1$ ✓. Test $x = 2$ : $3 \le 1$ ✗. So $x \ge 3$ is correct.

Double inequalities.

$-7 < 2x - 3 \le 5$ . Apply each operation to ALL parts.

Add $3$ : $-4 < 2x \le 8$ .
Divide by $2$ : $-2 < x \le 4$ .

Why does multiplying by a negative flip the sign?

$2 < 3$ is true. Multiply both by $-1$ : $-2$ vs $-3$ . Now $-2 > -3$ . The inequality flipped.

Edexcel tip. When you flip, write 'flipping inequality due to negative divisor' as a side note. This earns the M1.

Linear: same as equations except for negatives.
FLIP when × or ÷ by negative.
Double inequality: apply to all three parts.
Test values to verify.

Quadratic inequalities

▼

Factor, find roots, test regions. 'Between' = product negative.

Method:

Rearrange to $ax^{2} + bx + c < 0$ (or $> 0$ , $\le$ , $\ge$ ).
Factor (or use quadratic formula).
Find the ROOTS (where the quadratic = 0).
Test a value in each region of the number line.
State the solution as an interval.

Example. $x^{2} - 5x - 14 < 0$ .

Already in standard form.
Factor: $(x - 7)(x + 2) < 0$ .
Roots: $x = -2, x = 7$ . These divide the number line into three regions.

Region	Test value	$(x-7)(x+2)$	Sign
$x < -2$	$-3$	$(-10)(-1) = 10$	+
$-2 < x < 7$	$0$	$(-7)(2) = -14$	−
$x > 7$	$8$	$(1)(10) = 10$	+

We want NEGATIVE, so the answer is the middle region: $-2 < x < 7$ .

Sketch method (for visual learners).

Plot the parabola $y = x^{2} - 5x - 14$ :

Roots at $x = -2, 7$ .
Up-opening parabola (positive $x^{2}$ coefficient).
BELOW the $x$ -axis BETWEEN the roots.
ABOVE the $x$ -axis OUTSIDE the roots.

For x² − 5x − 14 < 0, the curve is below the axis between the roots, giving −2 < x < 7.

Match to the inequality direction.

Common patterns:

Form	Solution shape
$(x - a)(x - b) < 0$ , $a < b$	$a < x < b$ (between)
$(x - a)(x - b) > 0$ , $a < b$	$x < a$ OR $x > b$ (outside)

Worked qualitative. Why does 'between' correspond to negative product?

Two factors with opposite signs multiply to negative.
One factor positive (right of one root) AND one negative (left of other) → between the roots.
Both positive or both negative → outside.

Edexcel tip. Always state the FACTORS and the ROOTS explicitly. Then state the solution interval. Mark schemes credit each step.

Factor and find roots.
Test regions on number line.
Sketch parabola if helps.
Negative product: between roots.
Positive product: outside roots.

Number-line representation

▼

Open circle for strict ( $<, >$ ), closed for inclusive ( $\le, \ge$ ).

Conventions:

$<$ or $>$ : OPEN CIRCLE (boundary excluded).
$\le$ or $\ge$ : CLOSED CIRCLE (boundary included).
Arrow shows the direction.

Example. $x < 6$ :

←——————————o
            6

Open circle at $6$ , arrow extending left.

Example. $-2 < x \le 4$ :

       o——————•
       -2     4

Open at $-2$ (excluded), closed at $4$ (included), shaded between.

Open circle for a strict bound, closed circle for an inclusive bound.

Set notation.

Sometimes Edexcel uses set-builder: $\{x : -2 < x \le 4\}$ .

Or interval notation: $(-2, 4]$ . Round bracket = excluded; square = included.

Worked qualitative. What's the difference between $\{x : x \le 5\}$ and $\{x : x < 5\}$ for INTEGERS?

$x \le 5$ : $..., 3, 4, 5$ .
$x < 5$ : $..., 3, 4$ .
$5$ included for $\le$ , excluded for $<$ .

Edexcel tip. Match the bracket type to the inequality. Strict → open. Inclusive → closed.

Open circle: strict.
Closed circle: inclusive.
Number line shows direction.
Set notation: $\{x : ...\}$ .

Listing integer solutions

▼

Solve, then list integers within range.

Method:

Solve the inequality (all the rules apply).
List all INTEGERS within the resulting range.

Example. $-3 < 2x - 1 \le 5$ .

Add $1$ : $-2 < 2x \le 6$ . Divide by $2$ : $-1 < x \le 3$ .
Integers: $0, 1, 2, 3$ .

(Note: $-1$ excluded by strict inequality; $3$ included.)

Quadratic example. Find integer values of $x$ for which $x^{2} - 4x \le 0$ .

Factor: $x(x - 4) \le 0$ .
Roots: $0, 4$ . Between them = product non-positive.
So $0 \le x \le 4$ .
Integers: $0, 1, 2, 3, 4$ .

Worked qualitative. What if the range includes a fractional boundary?

$-1.5 < x \le 3.7$ : integers are $-1, 0, 1, 2, 3$ .
Round upward for the lower bound (next integer at or above $-1.5$ is $-1$ ).
Round downward for the upper bound (last integer at or below $3.7$ is $3$ ).

Edexcel tip. Write all integers explicitly. Don't shortcut to $\{0, 1, 2, 3\}$ — list each separately for clarity.

Solve first.
List integers in range.
Strict bounds: exclude.
Inclusive bounds: include.

Quick recap

Linear: solve like equations, FLIP for negative.
Quadratic: factor, roots, test regions.
'Between' for $< 0$ . 'Outside' for $> 0$ .
Number line: open vs closed circles.
Integers: list all in range.

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

FLIP when × or ÷ by negative.
$<, >$ : strict (excluded boundary).
$\le, \ge$ : inclusive.
Quadratic $< 0$ : between roots.
Number line: open ○ for strict, closed ● for inclusive.

How it’s examined

Inequalities appear Paper 1H + 2H (3-5 marks). Linear (2-3 marks), quadratic for A* candidates (4-5 marks). Examiner reports flag (1) not flipping for negative, (2) wrong region for quadratic, (3) confusion about strict vs inclusive boundaries.

Sources: Pearson Edexcel International GCSE Mathematics A 4MA1 specification (2026 onwards); 4MA1 Examiner Reports — Jan and May/Jun 2022-2024 series; Past Papers — 4MA1/1H Jan 2024, 4MA1/1H May/Jun 2024. Last reviewed 2026-05-07.

Step-by-step solutions to past-paper-style questions on inequalities, written exactly the way a tutor would explain them at the board.

1Solve a linear inequality
Foundation• inequality, linear
▼
Question
Solve $3x - 5 < 13$ .
Step-by-step solution
1. Step 1
  Add $5$ : $3x < 18$ .
2. Step 2
  Divide by $3$ : $x < 6$ .
Answer
$x < 6$
2Flip when multiplying/dividing by negative
Foundation• Adapted from 4MA1/1H Jan 2024 Q7• inequality, flip
▼
Question
Solve $-2x + 7 \le 1$ .
Step-by-step solution
1. Step 1
  Subtract $7$ : $-2x \le -6$ .
2. Step 2
  Divide by $-2$ . FLIP THE INEQUALITY when dividing by negative:
  $x \ge 3$
Answer
$x \ge 3$
Examiner tip
ALWAYS flip the inequality when multiplying or dividing both sides by a NEGATIVE number. Mark scheme awards M1 for correct flipping.
3Double inequality
Foundation• double inequality
▼
Question
Solve $-7 < 2x - 3 \le 5$ .
Step-by-step solution
1. Step 1
  Add $3$ to all three parts: $-4 < 2x \le 8$ .
2. Step 2
  Divide by $2$ : $-2 < x \le 4$ .
Answer
$-2 < x \le 4$
Examiner tip
Apply each operation to ALL THREE parts simultaneously.
4Quadratic inequality
Higher• Adapted from 4MA1/1H May/Jun 2024 Q19• quadratic inequality
▼
Question
Solve $x^{2} - 5x - 14 < 0$ .
Step-by-step solution
1. Step 1
  Factor: $(x - 7)(x + 2) < 0$ .
2. Step 2
  Roots: $x = 7$ and $x = -2$ . These divide the number line into three intervals.
3. Step 3
  Test a value in each interval:
4. Step 4
  $x = -3$ : $(-10)(-1) = +10 > 0$ . $x = 0$ : $(-7)(2) = -14 < 0$ ✓. $x = 8$ : $(1)(10) = +10 > 0$ .
5. Step 5
  So $(x - 7)(x + 2) < 0$ when $-2 < x < 7$ .
Answer
$-2 < x < 7$
Examiner tip
For quadratic inequalities, factor THEN sketch a parabola or test values. The 'between' answer occurs when product is negative.
5Integer solutions
Higher• integer solutions
▼
Question
List the integer solutions of $-3 < 2x - 1 \le 5$ .
Step-by-step solution
1. Step 1
  Add $1$ : $-2 < 2x \le 6$ . Divide by $2$ : $-1 < x \le 3$ .
2. Step 2
  Integer values: $x = 0, 1, 2, 3$ .
Answer
$x = 0, 1, 2, 3$
Examiner tip
Integer solutions: list ALL integers in the range. $-1 < x$ : doesn't include $-1$ . $x \le 3$ : includes $3$ .
6Checking a value at the boundary [1 mark]
Getting started• Paper 1H/2H style — 1 mark• inequalities, boundary values
▼
Question
Does $x = -2$ satisfy the inequality $3 - 2x \leqslant 7$ ?

Give a reason for your answer. [1]
Step-by-step solution
1. Step 1
  Substitute $x = -2$ into the left-hand side. Take care with the double negative.
  $3 - 2(-2) = 3 + 4 = 7$
2. Step 2
  B1 The inequality allows the left-hand side to equal $7$ , because $\leqslant$ means "less than or equal to". So $x = -2$ does satisfy it.
Answer
Yes, because $3 - 2(-2) = 7$ and $7 \leqslant 7$ is true.
Examiner tip
Examiner tip: The mark is for the reason, so show the substitution and the comparison $7 \leqslant 7$ . A bare "Yes" scores 0.

Common mistake: Working out $3 - 2(-2)$ as $3 - 4 = -1$ , or saying "No, because $7$ is not less than $7$ " and ignoring the "or equal to" part of the sign.
7Collecting terms to avoid dividing by a negative [2 marks]
Getting started• Paper 1H/2H style — 2 marks• inequalities, linear, reversing the sign
▼
Question
Solve $9 - 4x \geqslant 2x + 21$ . [2]
Step-by-step solution
1. Step 1
  M1 Collect the $x$ terms on the right, where the coefficient becomes positive ( $+2x + 4x = 6x$ ), and the numbers on the left.
  $9 - 21 \geqslant 2x + 4x \quad\Rightarrow\quad -12 \geqslant 6x$
2. Step 2
  A1 Divide by $6$ (positive, so no change) and write $x$ first.
  $-2 \geqslant x \quad\Rightarrow\quad x \leqslant -2$
Answer
$x \leqslant -2$
Examiner tip
Examiner tip: Moving the $x$ terms to the side where they end up positive means you never divide by a negative, so there is no sign to reverse. Collecting on the left instead gives $-6x \geqslant 12$ , and then you must reverse.

Common mistake: Reaching $-6x \geqslant 12$ and writing $x \geqslant -2$ . That scores M1 A0. Check with $x = 0$ : $9 \geqslant 21$ is false, so $x = 0$ must not be in the answer.
8Double inequality with a fraction [3 marks]
Building confidence• Paper 1H/2H style — 3 marks• inequalities, double inequality, fractions
▼
Question
Solve $4 < \dfrac{2x + 5}{3} \leqslant 7$ . [3]
Step-by-step solution
1. Step 1
  M1 Multiply all three parts by $3$ .
  $12 < 2x + 5 \leqslant 21$
2. Step 2
  M1 Subtract $5$ from all three parts.
  $7 < 2x \leqslant 16$
3. Step 3
  A1 Divide all three parts by $2$ .
  $3.5 < x \leqslant 8$
Answer
$3.5 < x \leqslant 8$
Examiner tip
Examiner tip: Whatever you do to the middle, do to both ends. Keep each end's sign exactly as it was: strict on the left, inclusive on the right.

Common mistake: Multiplying only the middle by $3$ , giving $4 < 2x + 5 \leqslant 7$ , which leads to $-0.5 < x \leqslant 1$ . That scores 0.
9Quadratic inequality with a leading coefficient of 2 [4 marks]
Building confidence• Paper 1H/2H style — 4 marks• inequalities, quadratic inequality
▼
Question
Solve $2x^{2} - 7x + 3 < 0$ .

Show clear algebraic working. [4]
Step-by-step solution
1. Step 1
  M1 Factorise. Check by expanding: $2x^{2} - 6x - x + 3$ .
  $(2x - 1)(x - 3) < 0$
2. Step 2
  A1 Critical values, from $2x - 1 = 0$ and $x - 3 = 0$ .
  $x = \frac{1}{2},\quad x = 3$
3. Step 3
  M1 The parabola is U-shaped, so it is below the axis between the critical values. Check $x = 1$ : $2 - 7 + 3 = -2 < 0$ .
4. Step 4
  A1 Write the answer as one double inequality.
  $\frac{1}{2} < x < 3$
Answer
$\dfrac{1}{2} < x < 3$
Examiner tip
Examiner tip: Test one value between the roots. If it makes the expression negative, the answer is "inside"; if positive, it is "outside".

Common mistake: Giving the critical value as $x = 2$ from the bracket $(2x - 1)$ , or as $x = -\dfrac{1}{2}$ . Solve $2x - 1 = 0$ properly to get $x = \dfrac{1}{2}$ .
10Expanding before solving a quadratic inequality [5 marks]
Stretch• Paper 1H/2H style — 5 marks• inequalities, quadratic inequality, expanding brackets
▼
Question
Find the set of values of $x$ for which $(x + 1)^{2} \geqslant 5x + 11$ .

Show clear algebraic working. [5]
Step-by-step solution
1. Step 1
  M1 Expand the square as two brackets (three terms).
  $(x + 1)^{2} = x^{2} + 2x + 1$
2. Step 2
  M1 Bring everything to one side so the right-hand side is zero.
  $x^{2} + 2x + 1 - 5x - 11 \geqslant 0 \quad\Rightarrow\quad x^{2} - 3x - 10 \geqslant 0$
3. Step 3
  M1 Factorise to find the critical values.
  $(x - 5)(x + 2) \geqslant 0 \quad\Rightarrow\quad x = 5,\ x = -2$
4. Step 4
  A1 Both critical values correct. The parabola is on or above the axis outside them; check $x = 0$ : $1 \geqslant 11$ is false, so the middle is not wanted.
5. Step 5
  A1 Give both parts, with the end points included.
  $x \leqslant -2 \quad\text{or}\quad x \geqslant 5$
Answer
$x \leqslant -2$ or $x \geqslant 5$
Examiner tip
Examiner tip: Check an end point in the original: $x = 5$ gives $36 \geqslant 36$ , true, so $5$ is included.

Common mistake: Expanding $(x + 1)^{2}$ as $x^{2} + 1$ . That gives $x^{2} - 5x - 10 \geqslant 0$ , which does not factorise, so the correct critical values and the final answer cannot be reached.

Question 1
Paper 1H/2H style — 1 mark1 mark
$n$ is greater than $-3$ and $n$ is at most $4$ .

Write this information as a single inequality. [1]
Model answer
Greater than $-3$ is strict; at most $4$ includes $4$ .

Answer: $-3 < n \leqslant 4$
Why this scores
Mark scheme:
- B1 for $-3 < n \leqslant 4$ oe
Why this scores full marks: Each phrase is turned into the correct sign. "At most" means "less than or equal to", which many candidates write as $<$ .
Question 2
Paper 1H/2H style — 2 marks2 marks
Solve $\dfrac{3x - 4}{5} \geqslant 2$ . [2]
Model answer
$3x - 4 \geqslant 10$

$3x \geqslant 14$

Answer: $x \geqslant \dfrac{14}{3}$
Why this scores
Mark scheme:
- M1 for $3x - 4 \geqslant 10$ or $3x \geqslant 14$ oe
- A1 for $x \geqslant \dfrac{14}{3}$ oe (accept $x \geqslant 4\dfrac{2}{3}$ or $x \geqslant 4.67$ or better)
Why this scores full marks: Multiplying the whole of both sides by $5$ clears the fraction in one step. The exact answer $\dfrac{14}{3}$ is left as a fraction rather than rounded too early.
Question 3
Paper 1H/2H style — 2 marks2 marks
$n$ is a positive integer such that $4n - 1 < 17$ .

Work out the sum of all the possible values of $n$ . [2]
Model answer
$4n < 18 \Rightarrow n < 4.5$

$n$ is a positive integer, so $n = 1, 2, 3, 4$

$1 + 2 + 3 + 4 = 10$

Answer: $10$
Why this scores
Mark scheme:
- M1 for $n < 4.5$ oe, or listing $1, 2, 3, 4$
- A1 for $10$ cao
Why this scores full marks: Solving first shows exactly where the integers stop: $4.5$ is not an integer, so the largest value is $4$ ( $4(4) - 1 = 15 < 17$ , but $4(5) - 1 = 19$ is not). The word "positive" supplies the lower end, so $0$ and the negative integers are excluded, and the question asks for the sum, not the list.
Question 4
Paper 1H/2H style — 3 marks3 marks
Solve $5(2 - x) > 3(x + 6)$ . [3]
Model answer
$10 - 5x > 3x + 18$

$10 - 18 > 3x + 5x$

$-8 > 8x$

Answer: $x < -1$
Why this scores
Mark scheme:
- M1 for expanding both brackets correctly, $10 - 5x$ and $3x + 18$
- M1 for collecting terms correctly, e.g. $-8 > 8x$ or $-8x > 8$
- A1 for $x < -1$ cao
Why this scores full marks: The $-5x$ from the first bracket keeps its sign, and the $x$ terms are collected on the right so the final division is by $+8$ . A candidate who reaches $-8x > 8$ must reverse the sign to get the same answer.
Question 5
Paper 1H/2H style — 3 marks3 marks
Solve $3 - 2x \leqslant x + 12 < 4x$ . [3]
Model answer
Split into two inequalities.

$3 - 2x \leqslant x + 12 \Rightarrow -9 \leqslant 3x \Rightarrow x \geqslant -3$

$x + 12 < 4x \Rightarrow 12 < 3x \Rightarrow x > 4$

Both must be true, and every number greater than $4$ is also at least $-3$ .

Answer: $x > 4$
Why this scores
Mark scheme:
- M1 for one of the two inequalities solved correctly, $x \geqslant -3$ or $x > 4$
- M1 for both solved correctly
- A1 for $x > 4$ cao
Why this scores full marks: The $x$ appears in all three parts, so the usual double-inequality method does not work. Splitting into two separate inequalities is the key step, and the final answer is the overlap, not $-3 \leqslant x < 4$ .
Question 6
Paper 1H/2H style — 3 marks3 marks
Solve $x^{2} \leqslant 6x$ .

Show clear algebraic working. [3]
Model answer
$x^{2} - 6x \leqslant 0$

$x(x - 6) \leqslant 0$

Critical values $x = 0$ and $x = 6$ ; a U-shaped parabola is on or below the axis between them.

Answer: $0 \leqslant x \leqslant 6$
Why this scores
Mark scheme:
- M1 for $x^{2} - 6x \leqslant 0$ and $x(x - 6)$ oe
- A1 for critical values $0$ and $6$
- A1 for $0 \leqslant x \leqslant 6$ cao
Why this scores full marks: The candidate does not divide both sides by $x$ . Dividing gives $x \leqslant 6$ , which loses the lower limit $0$ and wrongly includes every negative number (for example $x = -1$ gives $1 \leqslant -6$ , false).
Question 7
Paper 1H/2H style — 4 marks4 marks
Solve $x^{2} + 4x - 7 < 0$ .

Give the values in your answer correct to $3$ significant figures. [4]
Model answer
$x^{2} + 4x - 7$ does not factorise, so use the formula for the critical values.

$x = \frac{-4 \pm \sqrt{4^{2} - 4(1)(-7)}}{2(1)} = \frac{-4 \pm \sqrt{44}}{2}$

$x = 1.3166\ldots$ or $x = -5.3166\ldots$

The quadratic is negative between these values.

Answer: $-5.32 < x < 1.32$
Why this scores
Mark scheme:
- M1 for correct substitution into the quadratic formula (allow one sign error), or $(x + 2)^{2} - 11$
- A1 for critical values $-5.32$ and $1.32$ (awrt)
- M1 for choosing the region between the critical values
- A1 for $-5.32 < x < 1.32$ (awrt)
Why this scores full marks: The candidate recognises that the quadratic does not factorise and goes straight to the formula, writing the discriminant $16 + 28 = 44$ with the sign of $c$ handled correctly. The critical values are kept unrounded until the final line.
Question 8
Paper 1H/2H style — 4 marks4 marks
$n$ is an integer such that $3n + 10 > -5$ and $n^{2} < 12$ .

Find all the possible values of $n$ . [4]
Model answer
$3n > -15 \Rightarrow n > -5$

$n^{2} < 12 \Rightarrow -\sqrt{12} < n < \sqrt{12}$ , that is $-3.46\ldots < n < 3.46\ldots$

Both together: $-3.46\ldots < n < 3.46\ldots$

Answer: $-3, -2, -1, 0, 1, 2, 3$
Why this scores
Mark scheme:
- B1 for $n > -5$
- M1 for critical values $\pm\sqrt{12}$ or $\pm 3.46$ (awrt)
- A1 for $-3.46 < n < 3.46$ oe
- A1 for $-3, -2, -1, 0, 1, 2, 3$ and no others
Why this scores full marks: Both square roots of $12$ are used. A candidate who writes only $n < 3.46$ combines it with $n > -5$ and wrongly includes $-4$ ; but $(-4)^{2} = 16$ , which is not less than $12$ .
Question 9
Paper 1H/2H style — 5 marks5 marks
The sum of an integer $n$ and its square is less than $30$ .

Find all the possible values of $n$ .

Show clear algebraic working. [5]
Model answer
$n^{2} + n < 30$

$n^{2} + n - 30 < 0$

$(n + 6)(n - 5) < 0$

Critical values $n = -6$ and $n = 5$ , so $-6 < n < 5$

Answer: $-5, -4, -3, -2, -1, 0, 1, 2, 3, 4$
Why this scores
Mark scheme:
- M1 for $n^{2} + n < 30$ oe
- M1 for $(n + 6)(n - 5)$ or critical values $-6$ and $5$
- A1 for $-6 < n < 5$
- M1 for listing integers within their range
- A1 for $-5, -4, -3, -2, -1, 0, 1, 2, 3, 4$ and no others
Why this scores full marks: The words are turned into an inequality first, then solved like any quadratic inequality. The negative integers are not forgotten: $(-5)^{2} + (-5) = 20 < 30$ . The end values are excluded because $5^{2} + 5 = 30$ and $(-6)^{2} + (-6) = 30$ , and $30$ is not less than $30$ . Listing only $1$ to $4$ from trial is the usual weak answer.
Question 10
Paper 1H/2H style — 5 marks5 marks
The region $R$ is defined by the inequalities

$y \geqslant 2 \qquad y \leqslant 2x - 1 \qquad x + y \leqslant 8$

(a) Find the coordinates of the point where the lines $y = 2x - 1$ and $x + y = 8$ meet. [2]

(b) Work out the number of points with integer coordinates that lie in $R$ . [3]
Model answer
(a) $x + (2x - 1) = 8$

$3x = 9$ , so $x = 3$ and $y = 2(3) - 1 = 5$

Answer: $(3, 5)$

(b) $R$ runs from $y = 2$ up to the corner at $y = 5$ . For each $y$ : $x \geqslant \dfrac{y + 1}{2}$ and $x \leqslant 8 - y$ .

$y = 2$ : $1.5 \leqslant x \leqslant 6$ , so $x = 2, 3, 4, 5, 6$ (5 points)

$y = 3$ : $2 \leqslant x \leqslant 5$ , so $x = 2, 3, 4, 5$ (4 points)

$y = 4$ : $2.5 \leqslant x \leqslant 4$ , so $x = 3, 4$ (2 points)

$y = 5$ : $3 \leqslant x \leqslant 3$ , so $x = 3$ (1 point)

Answer: $5 + 4 + 2 + 1 = 12$ points
Why this scores
Mark scheme:
- M1 for substituting one equation into the other, e.g. $x + 2x - 1 = 8$
- A1 for $(3, 5)$
- M1 for a systematic method, e.g. the range of $x$ for at least two values of $y$
- A1 for at least three rows correct
- A1 for $12$ cao
Why this scores full marks: Part (a) gives the top corner of the region, which tells the candidate where to stop counting. Rearranging each boundary to give the range of $x$ for each row makes the count reliable, and the boundary points are included because every sign is $\leqslant$ or $\geqslant$ .

4MA1 Inequalities

At a glance

What you’ll learn

Linear inequalities

Quadratic inequalities

Number-line representation

Listing integer solutions

Quick recap

Memorise this

How it’s examined

Worked ExamplesStep-by-step solutions to past-paper-style questions

1Solve a linear inequality

2Flip when multiplying/dividing by negative

3Double inequality

4Quadratic inequality

5Integer solutions

6Checking a value at the boundary [1 mark]

7Collecting terms to avoid dividing by a negative [2 marks]

8Double inequality with a fraction [3 marks]

9Quadratic inequality with a leading coefficient of 2 [4 marks]

10Expanding before solving a quadratic inequality [5 marks]

Model AnswersFull mark-scheme-style answers

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10