6.04b, 7.02a - Regions defined by inequalities | OCR GCSE Mathematics

An inequality can describe a whole side of a line rather than one line. In 7.02a you will draw solid and dashed boundaries, then in 6.04b you will combine conditions to find a common region and its integer points. These are Higher-tier graph skills.

Drawing inequality boundaries

How can a point lie on a drawn edge but still fail the inequality? The line's style records whether its points are included.

The inequality $y\le2x+1$ allows many points. Its boundary occurs where equality holds, so start with $y=2x+1$ . A point such as $(1,3)$ satisfies equality and is included. For $y<2x+1$ , that same point is excluded even though the boundary equation stays the same.

Boundary line

The boundary line separates the two sides of a linear inequality. Draw it solid for $\le$ or $\ge$ , because equality is allowed. Draw it dashed for $<$ or $>$ , because equality is excluded.

To draw an oblique boundary, replace the inequality sign by equals, find two distinct points, and join them with a ruled line. A vertical boundary $x=a$ has fixed horizontal coordinate; a horizontal boundary $y=b$ has fixed vertical coordinate.

Worked example 1: an included boundary

Question. Draw the boundary for $y\le2x+1$ . Give two points on it and explain the line style.

Plan. Plot the equality line, then use the inequality sign to decide whether it is included.

\begin{aligned} y&=2x+1 &&\text{boundary equation}\\ x=0&\implies y=1 &&\text{first point }(0,1)\\ x=1&\implies y=3 &&\text{second point }(1,3) \end{aligned}

Answer. Draw a solid straight line through $(0,1)$ and $(1,3)$ , because $\le$ includes equality.

Check. At $(1,3)$ , $3\le2(1)+1$ becomes $3\le3$ , which is true. A solid line correctly includes that point.

Change only the inequality from non-strict to strict.

Worked example 2: an excluded boundary

Question. Draw the boundary for $y<2x+1$ . Give two points on it and explain the line style.

Plan. Keep the equality line but exclude its points.

\begin{aligned} y&=2x+1 &&\text{same boundary}\\ x=0&\implies(0,1) &&\text{first plot}\\ x=1&\implies(1,3) &&\text{second plot} \end{aligned}

Answer. Draw a dashed line through the same points, because $<$ excludes equality.

Check. At $(1,3)$ , $3<3$ is false. The dashed style correctly excludes the boundary point.

The position alone does not tell you whether the boundary belongs to the answer.

[DIAGRAM: asset_name: Solid and dashed boundaries; asset_slug: inequality-boundary-styles; caption: The equality line stays fixed while the inequality sign changes whether it is included.; description: Two identical coordinate grids show the line y equals two x plus one through zero comma one and one comma three. The left line is solid for y less than or equal to two x plus one, and the right is dashed for y strictly less than two x plus one. No half-plane shading is used yet.]
The equality line stays fixed while the inequality sign changes whether it is included.

The same inclusion rule applies to a boundary without a finite gradient.

Worked example 3: a vertical boundary

Question. Draw the boundary for $x\ge2$ .

Plan. Hold the horizontal coordinate fixed.

\begin{aligned} x&=2 &&\text{boundary equation}\\ (2,0),\ (2,3)&\text{ lie on it} &&\text{same horizontal coordinate} \end{aligned}

Answer. Draw the solid vertical line $x=2$ .

Check. Both listed points satisfy $x\ge2$ with equality, so the boundary is included.

Change only the inclusion of the boundary.

Worked example 4: a strict vertical boundary

Question. Draw the boundary for $x>2$ .

Plan. Use the same vertical line and exclude equality.

\begin{aligned} x&=2 &&\text{unchanged boundary}\\ 2>2&\text{ is false} &&\text{boundary points excluded} \end{aligned}

Answer. Draw the dashed vertical line $x=2$ .

Check. The point $(3,0)$ satisfies $x>2$ , whereas $(2,0)$ does not. The next part chooses which side of the dashed line to keep.

A horizontal boundary works in the corresponding way: $y\le3$ has the solid horizontal boundary $y=3$ .

Watch out

Dashed does not mean “below”, and solid does not mean “above”. Line style records inclusion. The inequality itself decides the side.

State the boundary information without needing a drawing tool.

Choosing the correct half-plane

A boundary splits the plane into two half-planes, and a point away from the boundary tells you which one satisfies the inequality.

Choose a simple test point, often $(0,0)$ , and substitute both coordinates into the original inequality. If it gives a true statement, keep the side containing that point. If it gives a false statement, keep the opposite side. Do not use a point on the boundary: it cannot distinguish the two sides.

Problem-solving

For an oblique boundary written $y=mx+c$ , $y>mx+c$ selects above the line and $y<mx+c$ selects below it. A test point also works when the inequality is not yet in this form.

Shade the solution side clearly and label it. Some questions shade unwanted regions instead: follow their instruction and make the region you mean unambiguous.

Worked example 5: a test point that works

Question. Identify the half-plane satisfying $y<2x+1$ .

Plan. Draw the dashed boundary and test the origin, which is not on it.

\begin{aligned} (x,y)&=(0,0) &&\text{test point}\\ 0&<2(0)+1 &&\text{substitute}\\ 0&<1 &&\text{true} \end{aligned}

Answer. Keep the side containing the origin: below the dashed line $y=2x+1$ .

Check. Another point below the line, $(1,0)$ , gives $0<3$ , true. A point above it, $(0,2)$ , gives $2<1$ , false.

Reverse only the inequality. The test changes truth value, so the solution side reverses.

Worked example 6: a test point that fails

Question. Identify the half-plane satisfying $y>2x+1$ .

Plan. Use the same boundary and the same test point.

\begin{aligned} (x,y)&=(0,0) &&\text{same test point}\\ 0&>2(0)+1 &&\text{substitute}\\ 0&>1 &&\text{false} \end{aligned}

Answer. Keep the side not containing the origin: above the dashed line.

Check. $(0,2)$ now works, since $2>1$ , while $(1,0)$ does not, since $0>3$ is false.

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6.04b, 7.02a - Regions defined by inequalities

Drawing inequality boundaries

Choosing the correct half-plane

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