Extended. An inequality in \(x\) and \(y\) splits the grid in two along a straight line. Several of them together fence off a region, and every point inside it is a solution.
Draw the boundary as if it were an equation: SOLID for \(\le\) or \(\ge\), BROKEN for \(<\) or \(>\). Test a point that is not on the line to decide which side is wanted, then shade the side you do NOT want.
Fig. 1: R satisfies y ≥ 1, x + y ≤ 6 and y < 2x. The two ≤ / ≥ boundaries are solid; y = 2x is broken because the inequality is strict. In the exam you shade the three unwanted sides and leave R clear.
Section 01
Lines become boundaries.
In the lab, type an inequality such as \(y < 2x + 1\). Watch the boundary line appear, switch between \(<\) and \(\le\) to see it change from broken to solid, then drag a test point across the line and see the inequality go from true to false.
Boundary: replace the inequality sign by \(=\) and draw that straight line.
Solid line for \(\le\) or \(\ge\): points on the line are included.
Broken (dashed) line for \(<\) or \(>\): points on the line are NOT included.
Shade the unwanted region, unless the question tells you otherwise. The answer is the region left clear.
Worked example Vertical and horizontal lines
Show the region \(x < 3\) and, on another grid, the region \(y \ge -1\).
\(x < 3\): the boundary \(x = 3\) is a VERTICAL line through 3 on the \(x\)-axis. Strict sign, so draw it broken.
Wanted points have \(x\) less than 3, which is the left side. Shade the right side.
\(y \ge -1\): the boundary \(y = -1\) is a HORIZONTAL line. Inclusive sign, so draw it solid.
Wanted points are on or above it. Shade below.
Answer: Broken line \(x = 3\), shade \(x > 3\); solid line \(y = -1\), shade \(y < -1\).
Examiner trap
\(x = 3\) is a vertical line, not a horizontal one. Students often draw \(x = 3\) across the page because '3 on the \(x\)-axis' feels horizontal. A line \(x = k\) is always vertical; \(y = k\) is always horizontal.
Section 02
Which side?
For a sloping boundary, pick a test point that is not on the line, often \((0, 0)\). Substitute it into the inequality. If the statement is true, the test point's side is wanted; if false, the other side is.
Worked example Test the origin
Show the region \(y > 2x - 1\).
Boundary \(y = 2x - 1\): gradient 2, \(y\)-intercept \(-1\). The sign is strict, so draw it broken.
Test \((0, 0)\): is \(0 > 2(0) - 1 = -1\)? Yes, true.
So the side containing the origin (above the line) is wanted. Shade below the line.
Answer: Broken line \(y = 2x - 1\); shade the side not containing \((0, 0)\).
Worked example Using intercepts
Show the region \(2x + 3y \le 12\).
Boundary \(2x + 3y = 12\). Put \(x = 0\): \(y = 4\). Put \(y = 0\): \(x = 6\). Join \((0, 4)\) and \((6, 0)\) with a SOLID line.
Test \((0, 0)\): \(0 \le 12\), true.
The origin's side is wanted, so shade the side away from the origin.
Answer: Solid line through \((0, 4)\) and \((6, 0)\); shade the side not containing \((0, 0)\).
Your turn
Does the region \(x + 2y > 8\) contain the origin?
Test \((0, 0)\): \(0 + 0 > 8\) is false. So the origin is NOT in the region; the wanted side is the one away from the origin.
Examiner trap
If the line passes through \((0, 0)\), such as \(y < 2x\), the origin cannot be a test point: it is ON the line. Use \((1, 0)\) instead: \(0 < 2\) is true, so the side containing \((1, 0)\) is wanted.
Stage 1 · Learn
Check your understanding
Four quick checks. Each correct answer earns XP and records this skill on your profile.
Quick check+10 XP
Which boundary do you draw for \(y < 3\)?
Quick check+10 XP
Which point satisfies \(y > 2x + 1\)?
Quick check+10 XP
For the region \(x + y \le 4\), which point is on the wanted side of the line \(x + y = 4\)?
Quick check+10 XP
Unless a question says otherwise, what do you shade?
Section 03
Many inequalities, one region.
Draw every boundary, shade the unwanted side of each, and the region R that stays clear satisfies all the inequalities at once. You may be asked to go the other way too: find the inequalities that define a region you are shown.
Worked example Integer points in a region
R is defined by \(y \ge 1\), \(x + y \le 6\) and \(y < 2x\) (Fig. 1). List the points in R with integer coordinates.
\(y = 1\): \(x\) must satisfy \(1 < 2x\) and \(x \le 5\), so \(x = 1, 2, 3, 4, 5\).
\(y = 2\): \(2 < 2x\) gives \(x > 1\), and \(x \le 4\), so \(x = 2, 3, 4\).
\(y = 3\): \(x > 1.5\) and \(x \le 3\), so \(x = 2, 3\).
\(y = 4\): \(x > 2\) and \(x \le 2\), impossible. \((2, 4)\) is on the BROKEN line, so it is excluded.
A region is bounded by the solid line \(x = 1\), the solid line \(y = 5\) and the broken line \(y = x\). The point \((2, 4)\) is inside it. Write down the three inequalities.
\(x = 1\), solid: \((2, 4)\) has \(x = 2\), which is bigger than 1, so \(x \ge 1\).
\(y = 5\), solid: \(4\) is less than 5, so \(y \le 5\).
\(y = x\), broken: \(4 > 2\), so \(y > x\).
Answer: \(x \ge 1,\ y \le 5,\ y > x\)
Your turn
List the integer points that satisfy \(x \ge 1\), \(y \ge x\) and \(x + y < 6\).
\(x = 1\): \(y = 1, 2, 3, 4\). \(x = 2\): \(y = 2, 3\) (\(y = 4\) gives \(x + y = 6\), on the broken line). \(x = 3\): \(y \ge 3\) gives \(x + y \ge 6\), none. So \((1,1), (1,2), (1,3), (1,4), (2,2), (2,3)\).
Examiner trap
Points on a broken line are NOT in the region; points on a solid line ARE. Check every boundary point against the sign before listing it.
Stage 2 · Exam
Exam-style questions
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Finish the four checks above to unlock the exam questions
Exam style+20 XP
How many points with integer coordinates lie in the region defined by \(y \ge 1\), \(x + y \le 6\) and \(y < 2x\)?
Exam style+20 XP
A region is bounded by the solid line \(x = 2\), the broken line \(y = 4\) and the solid line \(y = x\). Its corners are \((2, 2)\), \((2, 4)\) and \((4, 4)\). Which inequalities define it?
Exam style+20 XP
A solid line passes through \((0, 6)\) and \((3, 0)\). The wanted region is the side containing the origin. Which inequality describes it?
Exam style+20 XP
Which point lies in the region \(y > x - 1\), \(x + 2y < 8\), \(x \ge 0\)?
Skill unlocked
Regions, fenced off.
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