Maths · 2 Algebra

2.6 Inequalities

Before you start

Solving linear equations by doing the same thing to both sides (2.5).
Ordering negative numbers: $- 5$ is smaller than $- 2$ because it is further left on the number line (1.5).
The equation of a straight line in the form $y = m x + c$ , and finding a gradient from two points (4.2 and 4.4).
Drawing two graphs on a graphic display calculator and finding where they intersect (3.2).

Syllabus objectives covered (5)

Represent and interpret inequalities, including on a number line.
Construct, solve and interpret linear inequalities.
Solve inequalities using a graphic display calculator.
Represent and interpret linear inequalities in two variables graphically.
List inequalities that define a given region.

Part 1 of 8

What an inequality says

An equation such as $x = 3$ fixes $x$ at one value. An inequality such as $x > 3$ allows a whole range of values. $3.1$ , $4$ and $100$ all satisfy it, but $3$ does not, because $3$ is not greater than $3$ .
There are four symbols. $<$ means 'is less than' and $>$ means 'is greater than'. $⩽$ means 'is less than or equal to' and $⩾$ means 'is greater than or equal to'. The extra line under the symbol is the 'or equal to' part.
$x < 3 x > 3 x ⩽ 3 x ⩾ 3$
$<$ and $>$ are strict inequalities: the end value is not included. $⩽$ and $⩾$ are inclusive: the end value is included. Every number-line and graph convention in this subtopic depends on this one difference.
Read the statement from the letter. The open side of the symbol faces the larger quantity. So $3 < x$ says '3 is less than $x$ ', which is the same statement as $x > 3$ . Turn it round in your head before you draw or list anything.
A double inequality such as $- 3 ⩽ x < 1$ says two things at once: $x ⩾ - 3$ and $x < 1$ . So $x$ is trapped between $- 3$ and $1$ . The value $- 3$ is allowed, but the value $1$ is not.
To list the integers (whole numbers) that satisfy $- 3 ⩽ x < 1$ , decide each end separately. $- 3$ is in, because the symbol there is $⩽$ . $1$ is out, because the symbol there is $<$ . So the list is $- 3, - 2, - 1, 0$ .

Part 2 of 8

Showing an inequality on a number line

On a number line you draw a circle at the end value, then a line over every value that is allowed. The circle shows whether the end value itself is in or out.
The syllabus sets the convention. Use an open circle for a strict inequality ( $<$ or $>$ ), because the end value is not included. Use a closed (filled) circle for an inclusive inequality ( $⩽$ or $⩾$ ), because the end value is included.
$x > 2$ : an open circle at $2$ , and a line with an arrow to the right, because every larger number is allowed. $x ⩽ - 1$ : a filled circle at $- 1$ , and a line with an arrow to the left.
$- 3 ⩽ x < 1$ : a filled circle at $- 3$ , an open circle at $1$ , and a line between them only. There is no arrow, because $x$ cannot go past either end.
graph diagram — not drawn yet
Three horizontal number lines, one under the other, each from -5 to 5 with every integer marked and labelled. Top: an open circle at 2 with a thick line and arrow running right to the end, labelled 'x > 2'. Middle: a filled circle at -1 with a thick line and arrow running left to the end, labelled 'x ⩽ -1'. Bottom: a filled circle at -3 and an open circle at 1 joined by a thick line, with no arrows, labelled '-3 ⩽ x < 1'.
Open circle: the end value is not included. Filled circle: the end value is included. The thick line covers every value that is allowed.
To interpret a number line, work backwards. The circle type gives the symbol: open means $<$ or $>$ , filled means $⩽$ or $⩾$ . The direction of the line tells you which way the symbol points. An open circle at $4$ with the line going left is $x < 4$ .
Two circles joined by a line give a double inequality. Write it with the smaller number on the left: a filled circle at $- 2$ joined to an open circle at $5$ is $- 2 ⩽ x < 5$ .

Part 3 of 8

Solving a linear inequality

You solve a linear inequality with the same moves as a linear equation. If you add or subtract the same amount on both sides, the inequality stays true. If you multiply or divide both sides by the same positive number, it also stays true.
For $3 x < 2 x + 4$ , subtract $2 x$ from both sides. That leaves $x < 4$ , and that is the answer. Any number less than $4$ works; $4$ itself does not.
$3 x < 2 x + 4 \Rightarrow x < 4$
There is one rule that is different. When you multiply or divide both sides by a negative number, you must reverse the inequality sign.
Here is why. $2 < 5$ is true. Multiply both sides by $- 1$ and you get $- 2$ and $- 5$ . On the number line $- 2$ is to the right of $- 5$ , so now $- 2 > - 5$ . Multiplying by a negative reverses the order of the numbers, so the sign must reverse too.
Example: $7 - 2 x > 1$ . Subtract $7$ : $- 2 x > - 6$ . Divide by $- 2$ and reverse the sign: $x < 3$ .
$- 2 x > - 6 \Rightarrow x < 3$
You can avoid that division completely. Collect the $x$ terms on the side where they are positive. Add $2 x$ to both sides: $7 > 1 + 2 x$ , so $6 > 2 x$ , so $3 > x$ . That is the same statement as $x < 3$ , and no sign was reversed.
Your answer is an inequality, not a number. $x < 3$ is the solution; $x = 3$ is not, because $3$ does not satisfy $7 - 2 x > 1$ .
Check with two test values in the original inequality: one inside your answer and one outside it. For $x < 3$ : $x = 0$ gives $7 > 1$ , which is true. $x = 4$ gives $- 1 > 1$ , which is false. If the inside value fails, you probably reversed the sign when you should not have, or you did not reverse it when you should have.

Part 4 of 8

Double inequalities

A double inequality such as $- 3 ⩽ 3 x - 2 < 7$ has three parts. You solve it in one go: whatever you do, do it to all three parts.
Add $2$ to all three parts, then divide all three parts by $3$ .
$- 3 ⩽ 3 x - 2 < 7 \Rightarrow - 1 ⩽ 3 x < 9 \Rightarrow - \frac{1}{3} ⩽ x < 3$
Each symbol stays attached to its own end. The left end was $⩽$ and it is still $⩽$ ; the right end was $<$ and it is still $<$ .
If you divide by a negative, both signs reverse. For $1 < 5 - 2 x < 9$ , subtract $5$ : $- 4 < - 2 x < 4$ . Divide by $- 2$ and reverse both signs: $2 > x > - 2$ . Then rewrite it with the smaller number on the left: $- 2 < x < 2$ .
$- 4 < - 2 x < 4 \Rightarrow 2 > x > - 2 \Rightarrow - 2 < x < 2$
This method works only when $x$ is in the middle part and nowhere else. If $x$ is in more than one part, as in $x < 2 x + 1 < 7$ , split it into two separate inequalities. $x < 2 x + 1$ gives $x > - 1$ , and $2 x + 1 < 7$ gives $x < 3$ . Then join them: $- 1 < x < 3$ .

Part 5 of 8

Constructing an inequality and interpreting the answer

In a word problem, the inequality comes from a limit in the question. First write down what your letter means, for example 'let $b$ be the number of boxes'.
Then translate the limit. 'At least' means $⩾$ . 'At most' and 'no more than' mean $⩽$ . 'More than' means $>$ . 'Less than' and 'fewer than' mean $<$ . Do not mix up 'at least' and 'at most': they point in opposite directions.
Example: a lift can carry at most $600$ kg. A technician of mass $75$ kg goes up with $b$ boxes of $24$ kg each. The total mass must not go over the limit:
$75 + 24 b ⩽ 600$
Solve it: $24 b ⩽ 525$ , so $b ⩽ 21.875$ .
Now interpret it. $b$ is a number of boxes, so it must be a whole number. The greatest value is $21$ . You round down here, even though $21.875$ is close to $22$ : $22$ boxes give $75 + 528 = 603$ kg, which is over the limit.
For a least value, you round up. If $380$ students travel in coaches with $50$ seats each, then $50 c ⩾ 380$ , so $c ⩾ 7.6$ . $7$ coaches leave students behind, so the answer is $8$ coaches. The context decides the direction of rounding, not the decimal.

Part 6 of 8

Solving inequalities with a graphic display calculator

Some inequalities, such as $x^{2} > 2 x + 3$ or $2^{x} < x + 3$ , are not linear. The graphic display calculator method works for linear and non-linear inequalities alike.
Graph each side as its own function: $y = x^{2}$ and $y = 2 x + 3$ . Then use the intersect function to find every point where the graphs cross. The $x$ -coordinates of those points are the ends of your answer.
Here the graphs meet at $x = - 1$ and $x = 3$ . Do not stop after the first intersection: the calculator gives you only the crossing that you ask about.
Now read the inequality as a question about the graphs. $x^{2} > 2 x + 3$ asks: for which $x$ is the curve $y = x^{2}$ above the line $y = 2 x + 3$ ? The curve is above the line to the left of $x = - 1$ and to the right of $x = 3$ .
graph diagram — not drawn yet
Axes from x = -3 to 5 and y = -2 to 16. The parabola y = x^2 and the straight line y = 2x + 3, crossing at (-1, 1) and (3, 9), with both points marked and vertical dashed lines dropped from them to the x-axis. The parts of the parabola for x < -1 and x > 3 (where it is above the line) are drawn thicker, and the matching parts of the x-axis are highlighted with the labels 'x < -1' and 'x > 3'.
The solution of $x^{2} > 2 x + 3$ is the set of $x$ -values where the curve is above the line: two separate pieces, outside the two intersections.
So the solution is $x < - 1$ or $x > 3$ . These are two separate pieces, so write two inequalities joined by 'or'. Do not write $3 < x < - 1$ : no number is both greater than $3$ and less than $- 1$ .
The opposite inequality, $x^{2} < 2 x + 3$ , asks where the curve is below the line. That happens between the intersections, so the answer is one piece: $- 1 < x < 3$ .
Another method: rearrange so one side is zero, then graph that function on its own. $x^{2} - 2 x - 3 > 0$ asks where $y = x^{2} - 2 x - 3$ is above the $x$ -axis. The graph crosses the $x$ -axis at $- 1$ and $3$ , so you get the same answer.
The end values follow the symbol. For $<$ or $>$ they are not included; for $⩽$ or $⩾$ they are. If an intersection is not exact, give it to 3 significant figures, for example $x < 2.44$ .
Write down the graphs that you drew and the intersection values that you found. The calculator screen does not appear on your answer paper, so this is your working.

Part 7 of 8

Inequalities in two variables: regions on a graph

An inequality in $x$ and $y$ , such as $y > 2 x + 1$ , is true for a whole region of points, not only for points on a line. The boundary of that region is the line $y = 2 x + 1$ .
Step 1: draw the boundary line. The syllabus convention is a broken (dashed) line for a strict inequality ( $<$ or $>$ ), because points on the line are not included. Use a solid line for an inclusive inequality ( $⩽$ or $⩾$ ), because points on the line are included. This is the same idea as the open and filled circles on a number line.
Step 2: find which side you want. Choose a test point that is not on the line; $(0, 0)$ is the easiest when the line does not pass through it. For $y > 2 x + 1$ , put in $(0, 0)$ : is $0 > 1$ ? No. So the side that contains the origin is not wanted.
Step 3: shade. The syllabus convention is to shade the unwanted region, unless the question tells you to do something different. The region you want stays clear. Read the question each time: if it says 'shade the region that satisfies', do what it says.
Shading the unwanted side has a purpose. When there are three inequalities, you shade the unwanted side of each line in turn. The one region that stays clear satisfies all three. It is easy to see a clear region; it is hard to see where three shadings overlap.
Vertical and horizontal lines work in the same way. $x ⩾ 1$ has the solid vertical line $x = 1$ as its boundary, and the wanted side is to the right. $y < 3$ has the broken horizontal line $y = 3$ , and the wanted side is below.
A shortcut: when $y$ is on its own on the left with a positive sign, $y > m x + c$ is the region above the line and $y < m x + c$ is the region below. For anything else, such as $x + 2 y ⩽ 12$ , use a test point. A test point is never wrong.
To interpret a region, test points. A point is in the region only if it satisfies every inequality. A point on a solid boundary is in the region. A point on a broken boundary is not in the region.

Part 8 of 8

Listing the inequalities that define a region

This is the reverse of drawing a region. For each boundary line you need three things: the equation of the line, the direction of the inequality and whether it is strict or inclusive.
Example. A region R is bounded by three lines: the solid vertical line $x = 1$ , the solid line through $(0, 2)$ and $(4, 4)$ , and the broken line through $(0, 8)$ and $(8, 0)$ . R is the triangle between them, with vertices $(1, 2.5)$ , $(1, 7)$ and $(4, 4)$ .
graph diagram — not drawn yet
A grid with x from 0 to 9 and y from 0 to 9. Three lines: a solid vertical line x = 1; a solid line through (0, 2) and (4, 4), labelled y = x/2 + 2; a broken line through (0, 8) and (8, 0), labelled x + y = 8. Everything outside the triangle with vertices (1, 2.5), (1, 7) and (4, 4) is shaded grey, the triangle is left clear and labelled R, and the point (2, 5) is marked with a small cross inside R.
R is the clear triangle. Its two solid sides are included in R; its broken side is not.
Find each equation. A vertical line is $x = k$ , so the first line is $x = 1$ . The second line has gradient $\frac{4 - 2}{4 - 0} = \frac{1}{2}$ and $y$ -intercept $2$ , so it is $y = \frac{1}{2} x + 2$ . The third line has gradient $- 1$ and $y$ -intercept $8$ , so it is $y = - x + 8$ , which is $x + y = 8$ .
Choose a test point that is clearly inside R and not on any line: $(2, 5)$ . Put it into each line to find the direction. $x = 2$ , which is greater than $1$ , so R is $x ⩾ 1$ on that side. $\frac{1}{2} (2) + 2 = 3$ , and $5 > 3$ , so R is above the second line. $2 + 5 = 7$ , and $7 < 8$ , so R is on the side where $x + y$ is less than $8$ .
Now take the strictness from the line style. Solid lines give $⩽$ or $⩾$ ; the broken line gives $<$ or $>$ .
$x ⩾ 1, y ⩾ \frac{1}{2} x + 2, x + y < 8$
Check with a point that is clearly outside R, such as $(0, 0)$ . It must fail at least one inequality: $x ⩾ 1$ gives $0 ⩾ 1$ , which is false. Good.
A region with three sides needs three inequalities, and a region with four sides needs four. Count the sides before you start, so that you do not leave one out.

Worked examples

Example 1. Solve $4 - 3 x ⩽ 19$ and show your solution on a number line.

Show the working

Subtract $4$ from both sides, so that only the $x$ term is on the left.
$- 3 x ⩽ 15$
Divide both sides by $- 3$ . You are dividing by a negative number, so reverse the sign.
$x ⩾ - 5$
Check with the other route, where you keep the $x$ term positive. Add $3 x$ to both sides: $4 ⩽ 19 + 3 x$ . Subtract $19$ : $- 15 ⩽ 3 x$ . Divide by $3$ : $- 5 ⩽ x$ . This is the same answer.
Test in the original inequality. $x = 0$ is inside the answer: $4 ⩽ 19$ is true. $x = - 6$ is outside it: $4 + 18 = 22$ , and $22 ⩽ 19$ is false. Both tests agree with $x ⩾ - 5$ .
On the number line, use a filled circle at $- 5$ , because $⩾$ includes $- 5$ . Draw the line from $- 5$ to the right with an arrow, because every larger number is allowed.

Answer: $x ⩾ - 5$ : a filled circle at $- 5$ , with a line and arrow to the right.

Example 2. Find all the integers $n$ that satisfy $- 5 < 2 n + 3 ⩽ 9$ .

Show the working

Subtract $3$ from all three parts.
$- 8 < 2 n ⩽ 6$
Divide all three parts by $2$ . $2$ is positive, so the signs do not change.
$- 4 < n ⩽ 3$
Now decide each end. On the left the sign is $<$ , so $- 4$ is not included. On the right the sign is $⩽$ , so $3$ is included.
List the integers from just above $- 4$ up to $3$ : $- 3, - 2, - 1, 0, 1, 2, 3$ .
Check the two ends in the original. $n = - 4$ gives $2 (- 4) + 3 = - 5$ , and $- 5 < - 5$ is false, so $- 4$ is correctly left out. $n = 3$ gives $9$ , and $9 ⩽ 9$ is true, so $3$ is correctly included.

Answer: $n = - 3, - 2, - 1, 0, 1, 2, 3$ (seven integers).

Example 3. Use a graphic display calculator to solve $2^{x} < x + 3$ . Give the end values correct to 3 significant figures.

Show the working

Graph each side as its own function: $y = 2^{x}$ and $y = x + 3$ . A window of $- 5 ⩽ x ⩽ 5$ and $- 3 ⩽ y ⩽ 12$ shows the whole picture.
The curve and the line cross twice. Use the intersect function on each crossing. The calculator gives $x = - 2.8625 \dots$ and $x = 2.4449 \dots$ .
$2^{x} < x + 3$ asks where the curve is below the line. Look at the screen: the curve is below the line between the two intersections.
Confirm with test values. $x = 0$ gives $2^{0} = 1$ and $0 + 3 = 3$ , and $1 < 3$ is true. $x = 4$ gives $16 < 7$ , which is false. $x = - 4$ gives $0.0625 < - 1$ , which is false. So the answer is the piece between the intersections only.
The symbol is strict, so the end values are not included. Round each end value to 3 significant figures only at this last step.

Answer: $- 2.86 < x < 2.44$ (end values to 3 s.f.).

Example 4. On a grid with $0 ⩽ x ⩽ 7$ and $0 ⩽ y ⩽ 7$ , show the region R defined by $y ⩾ 1$ , $x + y < 6$ and $y ⩽ 2 x$ . Then list all the points with integer coordinates that lie in R.

Show the working

Draw the three boundary lines. $y = 1$ is horizontal and solid ( $⩾$ ). $x + y = 6$ passes through $(0, 6)$ and $(6, 0)$ and is broken ( $<$ ). $y = 2 x$ passes through $(0, 0)$ and $(3, 6)$ and is solid ( $⩽$ ).
$y ⩾ 1$ : the wanted side is above $y = 1$ , so shade below it. $x + y < 6$ : test $(0, 0)$ . $0 < 6$ is true, so the origin side is wanted; shade the other side.
$y ⩽ 2 x$ : the line passes through $(0, 0)$ , so you cannot use the origin as the test point. Use $(1, 0)$ : $0 ⩽ 2$ is true. So the wanted side is the side that contains $(1, 0)$ , below and to the right of the line; shade the other side.
graph diagram — not drawn yet
A grid with x from 0 to 7 and y from 0 to 7. A solid horizontal line y = 1, a broken line x + y = 6 through (0, 6) and (6, 0), and a solid line y = 2x through (0, 0) and (3.5, 7). Everything outside the triangle with vertices (0.5, 1), (5, 1) and (2, 4) is shaded grey; the triangle is clear and labelled R. The eight points (1,1), (2,1), (3,1), (4,1), (1,2), (2,2), (3,2) and (2,3) are marked with dots, and (2, 4) is marked with an open circle on the broken line.
R is the clear triangle. (2, 4) is a corner of R but lies on the broken line, so it is not in R.
R is the clear triangle with corners at $(0.5, 1)$ , $(5, 1)$ and $(2, 4)$ . The corner $(2, 4)$ is on the broken line, so it is not in R. The corner $(5, 1)$ is also on the broken line, so it is not in R either.
Check the integer points row by row. Row $y = 1$ : you need $x ⩾ 0.5$ (from $y ⩽ 2 x$ ) and $x < 5$ (from $x + y < 6$ ), so $x = 1, 2, 3, 4$ . Row $y = 2$ : $x ⩾ 1$ and $x < 4$ , so $x = 1, 2, 3$ . Row $y = 3$ : $x ⩾ 1.5$ and $x < 3$ , so $x = 2$ . Row $y = 4$ : $x ⩾ 2$ and $x < 2$ , which no value satisfies.

Answer: Eight points: $(1, 1)$ , $(2, 1)$ , $(3, 1)$ , $(4, 1)$ , $(1, 2)$ , $(2, 2)$ , $(3, 2)$ and $(2, 3)$ .

Where people go wrong

Mistake: Dividing $- 2 x > - 6$ by $- 2$ and writing $x > 3$ .

Instead: Reverse the sign when you multiply or divide by a negative: $x < 3$ . Or add $2 x$ to both sides at the start so that the $x$ term stays positive.

Multiplying by a negative reverses the order of numbers: $2 < 5$ but $- 2 > - 5$ . A test value shows the error: $x = 4$ satisfies $x > 3$ , but $- 2 (4) = - 8$ , and $- 8 > - 6$ is false.

Mistake: Drawing a filled circle for $x < 4$ , or an open circle for $x ⩾ - 5$ .

Instead: Open circle for $<$ and $>$ ; filled circle for $⩽$ and $⩾$ .

The circle is the only part of the drawing that shows whether the end value is included. If it is the wrong type, the number line shows a different inequality.

Mistake: In $- 3 ⩽ 3 x - 2 < 7$ , adding $2$ only to the middle part, or dividing only the middle part by $3$ .

Instead: Do every operation to all three parts: $- 1 ⩽ 3 x < 9$ , then $- \frac{1}{3} ⩽ x < 3$ .

A double inequality is two inequalities, $- 3 ⩽ 3 x - 2$ and $3 x - 2 < 7$ , that share a middle part. If you change the middle part, you must change both outer parts too, or neither inequality stays true.

Mistake: Drawing a solid line for $x + y < 6$ , or shading the region that you want.

Instead: Broken line for $<$ and $>$ ; solid line for $⩽$ and $⩾$ . Shade the unwanted side, unless the question says something different.

These are the syllabus conventions. A solid line says the points on the line are in the region. If you shade the wanted side, the 'clear region' is the wrong one.

Mistake: Using the graphic display calculator to solve $x^{2} > 2 x + 3$ , finding one intersection at $x = 3$ , and writing $x > 3$ .

Instead: Find every intersection ( $x = - 1$ and $x = 3$ ), then read where the curve is above the line: $x < - 1$ or $x > 3$ .

A curve and a line can cross more than once. Each crossing is the end of a piece of the answer, and the calculator gives only the crossing that you ask about.

Check yourself

A number line has a filled circle at $- 2$ and an open circle at $5$ , joined by a line. Write down the inequality it shows.
Answer
$- 2 ⩽ x < 5$ . The filled circle includes $- 2$ ; the open circle excludes $5$ .
List the integers $x$ such that $- 2 < x ⩽ 3$ .
Answer
$- 1, 0, 1, 2, 3$ . $- 2$ is excluded (strict), and $3$ is included.
Solve $2 - 5 x ⩾ 17$ .
Answer
$- 5 x ⩾ 15$ . Divide by $- 5$ and reverse the sign: $x ⩽ - 3$ .
Solve $1 < \frac{x + 4}{3} ⩽ 5$ .
Answer
Multiply all three parts by $3$ : $3 < x + 4 ⩽ 15$ . Subtract $4$ : $- 1 < x ⩽ 11$ .
Concert tickets cost $7 each, plus a booking fee of $4 for the whole order. Priya can spend no more than $50. Write an inequality for the number of tickets, $t$ , and find the greatest number she can buy.
Answer
$7 t + 4 ⩽ 50$ , so $7 t ⩽ 46$ and $t ⩽ 6.57 \dots$ . $t$ must be a whole number, so the greatest number is $6$ tickets ( $7$ tickets would cost $53).
Describe how you would use a graphic display calculator to solve $x^{2} + x < 6$ , and give the solution.
Answer
Graph $y = x^{2} + x$ and $y = 6$ and find both intersections: $x = - 3$ and $x = 2$ . The curve is below the line between them, so $- 3 < x < 2$ .
Is the point $(2, 1)$ in the region $y < 2 x - 3$ ? Is the point $(3, 2)$ in it? What type of line is the boundary?
Answer
$(2, 1)$ : $2 (2) - 3 = 1$ , and $1 < 1$ is false. The point is on the boundary, which is a broken line, so it is not in the region. $(3, 2)$ : $2 (3) - 3 = 3$ , and $2 < 3$ is true, so it is in the region.
A region is above the solid line $y = x$ , below the broken line $y = 4$ , and to the right of the $y$ -axis, which is drawn solid. Write down the three inequalities that define it.
Answer
$y ⩾ x$ , $y < 4$ and $x ⩾ 0$ .

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1.

(a) Solve $5 x + 3 > 2 x - 9$ . 2 marks No calculator

Mark scheme

M1 $3 x > - 12$ or equivalent rearrangement seen
A1 $x > - 4$ oe, e.g. $- 4 < x$ ; an answer of $x = - 4$ does not score this mark

(b) Solve $11 - 3 x ⩾ 2$ . 2 marks No calculator

Mark scheme

M1 $- 3 x ⩾ - 9$ or $9 ⩾ 3 x$ seen
A1 $x ⩽ 3$ oe, e.g. $3 ⩾ x$

(c) Draw a number line from $- 6$ to $6$ . On it, show the values of $x$ that satisfy both of your answers to (a) and (b). 2 marks No calculator

Mark scheme

B1 open circle at $- 4$ and filled circle at $3$ ; FT their (a) and (b)
B1 line joining the two circles only, with no arrows; FT

Question 2. A gym charges a joining fee of $25 and then $8 for each class. Lena has $100 to spend. She takes $n$ classes.

(a) Write an inequality in $n$ to show that Lena spends no more than $100. 1 mark No calculator

Mark scheme

B1 $25 + 8 n ⩽ 100$ oe

(b) Work out the greatest number of classes Lena can take. 2 marks No calculator

Mark scheme

M1 $8 n ⩽ 75$ or $n ⩽ 9.375$ oe, or $75 \div 8$ seen, or correct trials such as $25 + 8 \times 9 = 97$ and $25 + 8 \times 10 = 105$ ; FT their (a)
A1 $9$ cao

Question 3. The graphs of $y = x^{3} - 4 x$ and $y = x + 1$ are drawn on a graphic display calculator.

(a) Find the $x$ -coordinates of the three points where the graphs intersect. Give each answer correct to 3 significant figures. 3 marks

Mark scheme

B1 $- 2.13$ awrt
B1 $- 0.202$ awrt
B1 $2.33$ awrt

(b) Solve $x^{3} - 4 x < x + 1$ . 2 marks

Mark scheme

B1 $x < - 2.13$ awrt; FT their smallest value from (a)
B1 $- 0.202 < x < 2.33$ awrt; FT their other two values from (a). Accept the two pieces given together, e.g. joined by 'or'

Question 4. A region R is drawn on a grid. R is the triangle with vertices $(0, 1)$ , $(6, 1)$ and $(2, 5)$ , and R is not shaded. The side from $(0, 1)$ to $(2, 5)$ is a broken line. The other two sides are solid lines.

(a) Find the equation of the line through $(0, 1)$ and $(2, 5)$ . 2 marks No calculator

Mark scheme

M1 gradient $= \frac{5 - 1}{2 - 0} = 2$
A1 $y = 2 x + 1$ oe

(b) Find the three inequalities that define R. 3 marks No calculator

Mark scheme

B1 $y ⩾ 1$
B1 $y < 2 x + 1$ oe; FT their line from (a)
B1 $x + y ⩽ 7$ oe

(c) State whether the point $(3, 4)$ lies in R. Give a reason for your answer. 1 mark No calculator

Mark scheme

B1 Yes, with a reason: e.g. $3 + 4 = 7$ , so the point is on the solid line $x + y = 7$ , and $4 ⩾ 1$ and $4 < 2 (3) + 1$ also hold

How this is examined

Showing the region for three inequalities in x and y, with the right line style for each boundary and the unwanted side shaded: A Draw question on a printed grid: draw the boundary lines for inequalities such as $x ⩾ 1$ , $y > 2$ and $x + 2 y ⩽ 12$ , and label the region R that satisfies all three What the marker needs to see: each boundary line in the correct position, drawn across the grid, with the correct style (broken for $<$ or $>$ , solid for $⩽$ or $⩾$ ), and the correct region left clear and labelled R. The line style is part of the syllabus convention, so draw it correctly: a solid line for $x + y < 6$ shows a different inequality, even when its position is correct.
Writing down the inequalities that define a region that is already drawn: A Write down or Find question below a grid that shows a shaded region bounded by three or four lines, which can follow a part that asks for the equation of one sloping line One correct inequality for each boundary. Each one needs the correct line equation, the correct direction (check with a point inside R) and the correct strictness from the line style. An equation such as $y = 2 x + 1$ is not an inequality, so it does not answer the question.
Solving an inequality on the calculator from the intersections of two graphs: A calculator-paper Solve, which can follow a part that asks you to find where two graphs intersect, with the end values to 3 significant figures The $x$ -coordinate of every intersection to 3 s.f., then the solution in the correct form: one piece such as $a < x < b$ , or two pieces joined by 'or'. Write down which graphs you drew, because the calculator screen is not part of your answer.
Writing an inequality from a word problem and giving a whole-number answer that fits the context: A Write part for the inequality, then a Work out part for the greatest or least number of items, people or trips The inequality with the symbol that matches the words ('at most' is $⩽$ , 'at least' is $⩾$ ). Then the solved inequality, such as $n ⩽ 9.375$ . Then the whole number that fits the context: round down for a greatest number, round up for a least number.
Solving a linear inequality, including a step where you divide by a negative number: A 2-mark Solve such as $11 - 3 x ⩾ 2$ , or a double inequality with $x$ in the middle part M1 for a correct rearrangement such as $- 3 x ⩾ - 9$ or $9 ⩾ 3 x$ . A1 for the final inequality with the sign the correct way round, such as $x ⩽ 3$ . A final answer of $x = 3$ is not the solution.
Showing a solution on a number line, or listing the integers that satisfy it: A 1- or 2-mark Draw on a number line, or a Write down for the integers that satisfy an inequality An open circle at each strict end and a filled circle at each inclusive end, with the line over only the allowed values. For a list of integers: every integer in the range, with each end value included only if its symbol is $⩽$ or $⩾$ .

Words to know

Inequality: A statement that compares two quantities with one of the symbols $<$ , $>$ , $⩽$ or $⩾$ , such as $x > 3$ or $2 x + 1 ⩽ 9$ .
Strict inequality: An inequality that uses $<$ or $>$ . The end value is not included.
Inclusive inequality: An inequality that uses $⩽$ or $⩾$ . The end value is included.
Double inequality: Two inequalities written together with the unknown in the middle, such as $- 3 ⩽ x < 1$ .
Integer: A whole number: positive, negative or zero.
Boundary line: The line at the edge of a region, found by replacing the inequality sign with $=$ . It is broken for a strict inequality and solid for an inclusive one.
Region: The set of all points on a graph that satisfy one or more inequalities. By the syllabus convention, the unwanted side of each boundary is shaded, so the region is the part left clear.

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