Section 1
What Are Inequalities?
An inequality is like an equation, but instead of saying two things are equal, it says one is greater than or less than the other. You have already met inequalities on a number line — graphical inequalities extend this idea into two dimensions, creating regions on a coordinate grid.
| Symbol | Meaning | Example | Line type on graph |
|---|---|---|---|
| \(<\) | strictly less than | \(y < 3\) | dashed — boundary not included |
| \(>\) | strictly greater than | \(x > -2\) | dashed — boundary not included |
| \(\leq\) | less than or equal to | \(y \leq x+1\) | solid — boundary is included |
| \(\geq\) | greater than or equal to | \(y \geq 2x-3\) | solid — boundary is included |
🔑 Key idea
A graphical inequality describes a region (an area of the coordinate plane), not just a line.
The boundary line separates the plane into two halves — you shade the half that satisfies the inequality.
⚠ Common mistake: dashed vs solid
Strict inequalities (\(<\) and \(>\)) use a dashed line — the boundary itself is not part of the region.
Non-strict inequalities (\(\leq\) and \(\geq\)) use a solid line — the boundary is included.
Section 2
Representing Inequalities on a Graph
The boundary line
Every graphical inequality has a boundary line. To draw it, replace the inequality symbol with \(=\) and draw that line exactly as you would for a straight-line graph.
- For \(y \geq 2x + 1\), draw the line \(y = 2x + 1\) (solid, because \(\geq\)).
- For \(y < x - 3\), draw the line \(y = x - 3\) (dashed, because \(<\)).
Which side do you shade?
Once you have drawn the boundary line, you need to decide which side to shade. The most reliable method is the test point method:
✅ The test point method
Pick any point that is not on the boundary line — the origin \((0, 0)\) is usually easiest.
Substitute it into the inequality. If it makes the inequality true, shade the side containing
that point. If it makes it false, shade the other side.
Shading conventions at GCSE
GCSE questions often ask you to shade the region that does not satisfy the inequality (leaving the wanted region clear), or to shade the region that does satisfy it — always read the question carefully. In this guide we shade the region that satisfies the inequality.
Simple one-line examples
Example: \(y > 2\)
The boundary is the horizontal line \(y = 2\) (dashed). The region \(y > 2\) is above this line.
Test \((0,0)\): is \(0 > 2\)? No — so shade above the line, not below.
Example: \(x \leq 3\)
The boundary is the vertical line \(x = 3\) (solid). The region \(x \leq 3\) is to the left (and on) this line.
Test \((0,0)\): is \(0 \leq 3\)? Yes — shade the side containing the origin, which is the left side.
Section 3
The Step-by-Step Method
Follow these four steps for every graphical inequality question:
Draw the boundary line. Replace the inequality sign with \(=\) and plot the straight line. Make it dashed for strict inequalities (\(<\) or \(>\)) and solid for non-strict (\(\leq\) or \(\geq\)).
Choose a test point. Pick a point clearly on one side of the line. The origin \((0,0)\) is ideal unless the line passes through the origin — in that case use \((0,1)\) or \((1,0)\).
Substitute the test point. Put the coordinates of your test point into the inequality (using \(x\) and \(y\) values). Check whether the result is true or false.
Shade the correct region. If the test point made the inequality true, shade the region containing the test point. If it was false, shade the region on the other side.
💡 Tip: a quick shortcut for \(y\)-inequalities
If the inequality is in the form \(y > \text{something}\) or \(y \geq \text{something}\), the region is above the line.
If it is \(y < \text{something}\) or \(y \leq \text{something}\), the region is below the line.
This works as long as the inequality is already solved for \(y\).
Section 4
Worked Examples
Worked Example 1
A simple diagonal inequality
Show the region satisfied by \(y \leq x + 2\). Shade the region.
Step 1
Draw the boundary line \(y = x + 2\).
This has gradient \(1\) and \(y\)-intercept \(2\). Plot points: when \(x=0\), \(y=2\); when \(x=2\), \(y=4\); when \(x=-2\), \(y=0\).
The inequality is \(\leq\) (non-strict), so draw a solid line.
This has gradient \(1\) and \(y\)-intercept \(2\). Plot points: when \(x=0\), \(y=2\); when \(x=2\), \(y=4\); when \(x=-2\), \(y=0\).
The inequality is \(\leq\) (non-strict), so draw a solid line.
Step 2
Choose the test point \((0,0)\).
The origin is not on the line \(y=x+2\), so it is a valid test point.
The origin is not on the line \(y=x+2\), so it is a valid test point.
Step 3
Substitute \((0,0)\) into \(y \leq x+2\).
\(0 \leq 0 + 2\) gives \(0 \leq 2\). This is TRUE.
\(0 \leq 0 + 2\) gives \(0 \leq 2\). This is TRUE.
Step 4
Shade the region containing \((0,0)\).
Since the test point satisfies the inequality, shade the region below and on the line \(y=x+2\).
Since the test point satisfies the inequality, shade the region below and on the line \(y=x+2\).
Answer
Draw a solid line through \(y = x + 2\). Shade the region below (and including) the line — this is the set of all points where \(y \leq x+2\).
Worked Example 2
Strict inequality with a negative gradient
Show the region satisfied by \(y > -2x + 4\).
Step 1
Draw the boundary line \(y = -2x + 4\).
Gradient \(= -2\), \(y\)-intercept \(= 4\). Points: \((0,4)\), \((1,2)\), \((2,0)\).
The inequality is \(>\) (strict), so draw a dashed line.
Gradient \(= -2\), \(y\)-intercept \(= 4\). Points: \((0,4)\), \((1,2)\), \((2,0)\).
The inequality is \(>\) (strict), so draw a dashed line.
Step 2–3
Test \((0,0)\): is \(0 > -2(0)+4\)?
\(0 > 4\) — this is FALSE.
\(0 > 4\) — this is FALSE.
Step 4
Since the origin fails the test, shade the region not containing the origin — that is, the region above the dashed line.
Answer
Dashed line for \(y = -2x+4\). Shade the region above the line.
Worked Example 3
Vertical boundary line
Represent the inequality \(x > -1\) on a coordinate grid.
Step 1
Draw the vertical line \(x = -1\).
This is a vertical line crossing the \(x\)-axis at \(-1\). Because the inequality is strict (\(>\)), the line is dashed.
This is a vertical line crossing the \(x\)-axis at \(-1\). Because the inequality is strict (\(>\)), the line is dashed.
Step 2–3
Test \((0,0)\): is \(0 > -1\)? TRUE.
Step 4
Shade the region containing the origin — that is, everything to the right of the line \(x = -1\).
Answer
Dashed vertical line at \(x = -1\). Shade the region to the right.
Worked Example 4
Line that passes through the origin
Show the region \(y \geq 3x\).
Step 1
Draw the line \(y = 3x\).
This passes through the origin with gradient \(3\). Points: \((0,0)\), \((1,3)\), \((-1,-3)\). Use a solid line (\(\geq\)).
This passes through the origin with gradient \(3\). Points: \((0,0)\), \((1,3)\), \((-1,-3)\). Use a solid line (\(\geq\)).
Step 2–3
The origin lies on the boundary line, so we can't use \((0,0)\).
Use \((0,1)\) instead: is \(1 \geq 3(0)\)? Is \(1 \geq 0\)? TRUE.
Use \((0,1)\) instead: is \(1 \geq 3(0)\)? Is \(1 \geq 0\)? TRUE.
Step 4
Shade the region containing \((0,1)\) — this is above and to the left of the line \(y=3x\).
Answer
Solid line \(y=3x\). Shade above the line. Note: we needed to avoid the origin as a test point because it lies on the boundary.
Section 5
Multiple Inequalities — Finding a Region
Many GCSE questions give you two or three inequalities and ask you to identify (or shade) the region that satisfies all of them simultaneously. This region is called the feasible region or simply "the required region".
✅ How to find the intersection region
- Draw each boundary line separately (dashed or solid as appropriate).
- For each inequality, work out which side of its line satisfies it (using the test-point method).
- The required region is where all the satisfied sides overlap.
- You can label this region R, or shade only this region (leaving the rest unshaded), or shade everything outside it — follow the wording of the question.
Worked Example 5
Two simultaneous inequalities
Show the region satisfied by both \(y < x + 3\) and \(y \geq 1\).
Inequality 1
\(y < x+3\): draw a dashed line \(y=x+3\) (gradient 1, intercept 3).
Test \((0,0)\): \(0 < 3\) ✓ — shade below the dashed line.
Test \((0,0)\): \(0 < 3\) ✓ — shade below the dashed line.
Inequality 2
\(y \geq 1\): draw a solid horizontal line \(y=1\).
Test \((0,0)\): \(0 \geq 1\) ✗ — shade above the solid line.
Test \((0,0)\): \(0 \geq 1\) ✗ — shade above the solid line.
Overlap
The required region is below the dashed line \(y=x+3\) AND above (or on) the solid line \(y=1\).
It is the band-shaped region between these two lines.
Answer
The region satisfying both inequalities is the area where \(1 \leq y < x+3\) — above or on the line \(y=1\) and strictly below the line \(y=x+3\).
Worked Example 6
Three simultaneous inequalities
Identify and label the region \(R\) satisfying all three: \(x \geq 0\), \(y \geq 0\), and \(x + y \leq 5\).
Ineq. 1
\(x \geq 0\): solid vertical line \(x=0\) (the \(y\)-axis). Region is to the right.
Ineq. 2
\(y \geq 0\): solid horizontal line \(y=0\) (the \(x\)-axis). Region is above.
Ineq. 3
\(x+y \leq 5\): rearrange to \(y \leq -x+5\). Draw a solid line through \((0,5)\) and \((5,0)\).
Test \((0,0)\): \(0+0 \leq 5\) ✓ — shade below this line.
Test \((0,0)\): \(0+0 \leq 5\) ✓ — shade below this line.
Region R
\(R\) is the triangle with vertices at \((0,0)\), \((5,0)\), and \((0,5)\) — the area satisfying all three inequalities simultaneously.
Answer
Label the triangular region bounded by the \(y\)-axis, the \(x\)-axis, and the line \(x+y=5\) as \(R\). All three boundary lines are solid.
Section 6
Reading Inequalities from a Graph
Sometimes a question shows you a shaded region and asks you to write down the inequalities that describe it. Here is how to approach this:
Identify each boundary line. Find the equation of each line forming the boundary (using \(y=mx+c\) for diagonal lines, or \(x=k\) / \(y=k\) for vertical/horizontal lines).
Check solid or dashed. A solid line means \(\leq\) or \(\geq\). A dashed line means \(<\) or \(>\).
Use a test point in the shaded region. Pick a point clearly inside the shaded region and substitute into the equation of each line to determine whether the inequality is \(<\)/\(\leq\) or \(>\)/\(\geq\).
Write each inequality. Combine the equation with the correct inequality sign. Check your answer by verifying the test point satisfies all the inequalities you have written.
💡 Tip: finding the gradient and intercept
For a diagonal boundary line, identify two points on the line from the grid, then use:
\[ m = \frac{y_2 - y_1}{x_2 - x_1}, \qquad c = y\text{-intercept} \]
Then write the line as \(y = mx + c\) and attach the correct inequality sign.
Section 7
Quick Reference Summary
| Inequality | Boundary line type | Region to shade | Test point check |
|---|---|---|---|
| \(y > k\) | dashed horizontal | Above the line | Substitute \((0,0)\): is \(0 > k\)? |
| \(y \leq k\) | solid horizontal | Below (and on) the line | Substitute \((0,0)\): is \(0 \leq k\)? |
| \(x > k\) | dashed vertical | Right of the line | Substitute \((0,0)\): is \(0 > k\)? |
| \(x \leq k\) | solid vertical | Left of (and on) the line | Substitute \((0,0)\): is \(0 \leq k\)? |
| \(y \geq mx+c\) | solid diagonal | Above (and on) the line | Sub \((0,0)\): is \(0 \geq c\)? |
| \(y < mx+c\) | dashed diagonal | Below the line | Sub \((0,0)\): is \(0 < c\)? |
| Multiple inequalities | Draw each separately | Overlap of all satisfied sides | Test a single point in the overlap region against every inequality |
⚠ Watch out: when the line passes through the origin
If the boundary line passes through \((0,0)\), the origin cannot be your test point — it lies on the boundary, not on either side. Choose a different point such as \((0, 1)\), \((1, 0)\), or \((2, 0)\).
Practice Questions
Try each question first. Click "Show worked solution" when you're ready to check your answer.
Question 1
On a coordinate grid, show the region satisfied by \(y \geq 4\). Label the region \(R\).
[2 marks]
Step 1
Draw the horizontal line \(y = 4\). Because the inequality is \(\geq\) (non-strict), the line is solid.
Step 2–3
Test \((0,0)\): is \(0 \geq 4\)? FALSE.
Step 4
Since the origin fails, shade the region on the other side — above the line \(y=4\). Label this region \(R\).
Answer
A solid horizontal line at \(y=4\). Region \(R\) is above the line (including the line itself).
Question 2
Represent the inequality \(x < 2\) on a coordinate grid.
[2 marks]
Step 1
Draw the vertical line \(x = 2\). The inequality is strict (\(<\)), so the line is dashed.
Step 2–3
Test \((0,0)\): is \(0 < 2\)? TRUE.
Step 4
Shade the region containing the origin — everything to the left of the line \(x=2\).
Answer
A dashed vertical line at \(x=2\). Shade to the left (the boundary line is not included).
Question 3
On a grid, shade the region satisfied by \(y < 2x - 1\).
[3 marks]
Step 1
Draw the line \(y = 2x - 1\). Gradient \(2\), \(y\)-intercept \(-1\). Plot: \((0,-1)\), \((1,1)\), \((2,3)\). Draw as a dashed line (strict inequality).
Step 2–3
Test \((0,0)\): is \(0 < 2(0)-1\)? Is \(0 < -1\)? FALSE.
Step 4
Since the origin fails, shade the other side — below the dashed line.
Answer
Dashed line \(y=2x-1\). Shade below the line.
Question 4
Show the region \(y \geq -x + 3\) on a coordinate grid.
[3 marks]
Step 1
Draw the line \(y = -x+3\). Gradient \(-1\), \(y\)-intercept \(3\). Points: \((0,3)\), \((3,0)\), \((1,2)\). Draw as a solid line (\(\geq\)).
Step 2–3
Test \((0,0)\): is \(0 \geq -0+3\)? Is \(0 \geq 3\)? FALSE.
Step 4
Shade the region not containing the origin — above the solid line \(y=-x+3\).
Answer
Solid line \(y=-x+3\). Shade above (and including) the line.
Question 5
Find the region satisfying both \(y \leq x + 1\) and \(y \geq -2\). Label the region \(R\).
[4 marks]
Line 1
\(y = x+1\) — draw as a solid line. Test \((0,0)\): \(0 \leq 1\) ✓. Region is below this line.
Line 2
\(y = -2\) — draw as a solid horizontal line. Test \((0,0)\): \(0 \geq -2\) ✓. Region is above this line.
Overlap
Region \(R\) is the strip between the two solid lines: below \(y=x+1\) and above (or on) \(y=-2\). Check a point in \(R\), say \((0,-1)\): \(-1 \leq 0+1\) ✓ and \(-1 \geq -2\) ✓.
Answer
Both lines solid. Region \(R\) is below \(y=x+1\) and above \(y=-2\) (a diagonal strip that widens to the right).
Question 6
Identify the region \(R\) satisfying all three inequalities: \(x \geq 0\), \(y \geq 0\), \(y \leq -\frac{1}{2}x + 3\).
[4 marks]
Line 1
\(x=0\) (the \(y\)-axis) — solid. Region: to the right (\(x \geq 0\)).
Line 2
\(y=0\) (the \(x\)-axis) — solid. Region: above (\(y \geq 0\)).
Line 3
\(y = -\frac{1}{2}x+3\) — gradient \(-\tfrac{1}{2}\), intercept \(3\). Points: \((0,3)\) and \((6,0)\). Draw solid. Test \((0,0)\): \(0 \leq 3\) ✓ — shade below.
Region R
A triangle with vertices at \((0,0)\), \((6,0)\), and \((0,3)\).
Answer
Region \(R\) is the triangle bounded by the \(x\)-axis, the \(y\)-axis, and the line \(y=-\frac{1}{2}x+3\), with vertices at \((0,0)\), \((6,0)\), \((0,3)\). All three boundaries are solid.
Question 7
A shaded region on a grid is bounded by three solid lines: the \(y\)-axis (\(x=0\)), the line \(y=2\), and the line \(y=x+4\). The shaded region contains the point \((1, 3)\). Write down the three inequalities that define this region.
[3 marks]
Line 1: \(x=0\)
Test point \((1,3)\): \(x=1\). Is \(1 \geq 0\)? ✓ So the inequality is \(x \geq 0\).
Line 2: \(y=2\)
Test point \((1,3)\): \(y=3\). Is \(3 \geq 2\)? ✓ So the inequality is \(y \geq 2\).
Line 3: \(y=x+4\)
Test point \((1,3)\): is \(3 \leq 1+4=5\)? ✓ So the inequality is \(y \leq x+4\).
Answer
\[x \geq 0, \quad y \geq 2, \quad y \leq x + 4\]
(All boundary lines are solid, so \(\leq\) and \(\geq\) throughout.)
Question 8
Find all integer coordinate pairs \((x, y)\) that satisfy all three of: \(x \geq 1\), \(y \geq 1\), \(x + y \leq 4\).
[3 marks]
Setup
We need integer points with \(x \geq 1\), \(y \geq 1\), and \(x+y \leq 4\). Try each valid \(x\):
\(x=1\)
\(y \geq 1\) and \(1+y \leq 4 \Rightarrow y \leq 3\). Integer \(y\)-values: \(1, 2, 3\). Points: \((1,1)\), \((1,2)\), \((1,3)\).
\(x=2\)
\(y \geq 1\) and \(2+y \leq 4 \Rightarrow y \leq 2\). Points: \((2,1)\), \((2,2)\).
\(x=3\)
\(y \geq 1\) and \(3+y \leq 4 \Rightarrow y \leq 1\). Point: \((3,1)\).
\(x=4\)
\(4+y \leq 4 \Rightarrow y \leq 0\), but \(y \geq 1\) — no solution.
Answer — 6 integer points
\((1,1),\ (1,2),\ (1,3),\ (2,1),\ (2,2),\ (3,1)\)
Question 9
The region \(R\) is defined by \(y > 2x - 3\) and \(y < 5\). Is the point \((3, 4)\) inside region \(R\)? Show your working.
[2 marks]
Check ineq. 1
Substitute \((3,4)\) into \(y > 2x-3\):
\(4 > 2(3)-3 = 6-3 = 3\). Is \(4 > 3\)? TRUE ✓
\(4 > 2(3)-3 = 6-3 = 3\). Is \(4 > 3\)? TRUE ✓
Check ineq. 2
Substitute \((3,4)\) into \(y < 5\):
Is \(4 < 5\)? TRUE ✓
Is \(4 < 5\)? TRUE ✓
Answer
Both inequalities are satisfied, so \((3,4)\) is inside region \(R\).
Question 10
On a coordinate grid, draw and label the region \(R\) satisfying all of:
\[y \leq 2x + 1, \quad y \geq -x + 2, \quad x \leq 3\]
Find the coordinates of the three vertices of region \(R\).
[5 marks]
Three lines
Draw three solid boundary lines (all inequalities are non-strict):
— \(y = 2x+1\) (gradient 2, intercept 1)
— \(y = -x+2\) (gradient \(-1\), intercept 2)
— \(x = 3\) (vertical line)
— \(y = 2x+1\) (gradient 2, intercept 1)
— \(y = -x+2\) (gradient \(-1\), intercept 2)
— \(x = 3\) (vertical line)
Sides
Test \((0,0)\) in each:
\(y \leq 2x+1\): \(0 \leq 1\) ✓ → below/on line 1
\(y \geq -x+2\): \(0 \geq 2\) ✗ → above/on line 2
\(x \leq 3\): \(0 \leq 3\) ✓ → left of/on line 3
\(y \leq 2x+1\): \(0 \leq 1\) ✓ → below/on line 1
\(y \geq -x+2\): \(0 \geq 2\) ✗ → above/on line 2
\(x \leq 3\): \(0 \leq 3\) ✓ → left of/on line 3
Vertex A
Intersection of \(y=2x+1\) and \(y=-x+2\):
\(2x+1 = -x+2 \Rightarrow 3x=1 \Rightarrow x=\tfrac{1}{3}\), \(y=\tfrac{5}{3}\). Vertex: \(\left(\tfrac{1}{3},\, \tfrac{5}{3}\right)\).
\(2x+1 = -x+2 \Rightarrow 3x=1 \Rightarrow x=\tfrac{1}{3}\), \(y=\tfrac{5}{3}\). Vertex: \(\left(\tfrac{1}{3},\, \tfrac{5}{3}\right)\).
Vertex B
Intersection of \(y=2x+1\) and \(x=3\):
\(y=2(3)+1=7\). Vertex: \((3,\ 7)\).
\(y=2(3)+1=7\). Vertex: \((3,\ 7)\).
Vertex C
Intersection of \(y=-x+2\) and \(x=3\):
\(y=-3+2=-1\). Vertex: \((3,\ -1)\).
\(y=-3+2=-1\). Vertex: \((3,\ -1)\).
Answer
Region \(R\) is the triangle with vertices:
\[\left(\tfrac{1}{3},\, \tfrac{5}{3}\right), \quad (3,\ 7), \quad (3,\ -1)\]
All three boundary lines are solid. Region \(R\) lies below \(y=2x+1\), above \(y=-x+2\), and to the left of \(x=3\).