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Module 1 · L19 of 20 ~45 min ⚡ +90 XP available

Inequalities and the Number Plane

Graph linear inequalities in two variables on the Cartesian plane. Draw the boundary line (dashed for strict, solid for non-strict), test the point $(0,0)$, and shade the correct half-plane.

Today's hook — What does $y > x$ mean for points in the Cartesian plane? It means: all points where the $y$-value exceeds the $x$-value — that is an entire half of the plane, split diagonally by the line $y = x$.

0/5QUESTS

Recall — your gut answer first

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What does $y > x$ mean for points in the Cartesian plane? Is $(3, 5)$ in the solution region? Is $(5, 3)$? Without graphing, describe which side of the line $y = x$ the solutions lie on.

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Graphing inequalities on the Cartesian plane

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A two-variable inequality such as $y > x + 1$ divides the plane into two regions. The boundary is the line $y = x + 1$. The solution region is the set of all $(x, y)$ pairs that satisfy the inequality.

Step 1: Draw the boundary line. Dashed for strict ($<$, $>$). Solid for non-strict ($\leq$, $\geq$). Step 2: Test $(0,0)$. Step 3: Shade the correct side.

For $y > 2x + 1$: test $(0,0)$: $0 > 1$? No → shade the side NOT containing $(0,0)$.

Line type = boundary inclusion

Dashed boundary = points ON the line are NOT in the solution set ($<$, $>$). Solid boundary = boundary points ARE included ($\leq$, $\geq$).

Test (0,0) when possible

$(0,0)$ is the easiest test point. Substitute into the original inequality. If true, shade the side containing $(0,0)$; if false, shade the other side.

Avoid (0,0) on the boundary

If the boundary passes through $(0,0)$, choose another test point such as $(1,0)$ or $(0,1)$.

What you'll master

Know

Key facts

A two-variable inequality's solution is a region (half-plane) on the Cartesian plane.
The boundary line is dashed for strict, solid for non-strict inequalities.
The test-point method determines which half-plane to shade.

Understand

Concepts

Why the boundary line divides the plane into exactly two regions.
How substituting a test point into the original inequality identifies the correct region.
How dashed versus solid lines connect to the strict/non-strict distinction from number lines.

Can do

Skills

Draw the boundary line for a linear inequality in two variables.
Use the test-point method to identify the solution region.
Describe or shade the solution region for $y > mx + c$ and $y \leq mx + c$.

Key terms

Boundary lineThe line separating the solution region from the non-solution region. Dashed for strict, solid for non-strict inequalities.

Solution regionThe set of all points $(x, y)$ in the Cartesian plane that satisfy the inequality — a shaded half-plane.

Test pointA point substituted into the original inequality to determine which half-plane to shade. Usually $(0, 0)$.

Half-planeOne of the two regions created by a line dividing the Cartesian plane. A linear inequality's solution is always a half-plane (plus possibly the boundary).

Dashed lineA boundary line drawn with dashes to indicate points on the line are NOT included in the solution set (strict inequality).

Solid lineA boundary line drawn as a solid line to indicate points on the line ARE included in the solution set (non-strict inequality).

Drawing the boundary line and testing a point

core concept

For $y > x + 1$: the boundary line is $y = x + 1$. Because the symbol is $>$ (strict), draw it as a dashed line. Then test $(0,0)$: $0 > 0 + 1 = 1$? No → shade the side that does NOT contain $(0,0)$, which is above the line.

For $y \leq -2x + 4$: solid boundary line, test $(0,0)$: $0 \leq 4$? Yes → shade the side containing $(0,0)$, which is below the line.

Quick memory rule: For $y >$ or $y \geq$: if test passes, shade containing side. If fails, shade other side. Same logic for $y <$ and $y \leq$.

Quick check: for $y < 2x + 1$, is $(0,0)$ in the solution region?

What to write in your book

Step 1: Draw the boundary line (from its equation). Dashed = strict; solid = non-strict.
Step 2: Test $(0,0)$ by substituting into the original inequality.
Step 3: If test is TRUE → shade the side containing $(0,0)$. If FALSE → shade the other side.

Reading the solution region from a graph

core concept

Given a graph of an inequality, identify: the slope and intercept of the boundary line (to write its equation), whether the line is dashed or solid (to choose the symbol type), and which side is shaded (to determine direction).

If the shaded region is above the line, the inequality uses $>$ or $\geq$. If below, it uses $<$ or $\leq$. The line type determines whether strict or non-strict.

Which is NOT a characteristic of a strict inequality boundary on the Cartesian plane?

What to write in your book

To write an inequality from a graph: identify the boundary line equation, line type (dashed/solid), and shaded region (above/below).
Above the line → $y >$ or $y \geq$; below → $y <$ or $y \leq$.
Always verify with a test point from the shaded region.

Interpreting which points are solutions

core concept

A shaded solution region contains infinitely many points. Any specific point $(a, b)$ can be tested by substituting into the original inequality. If the inequality is satisfied, the point is in the solution region.

For $y \leq 3x - 2$: is $(2, 4)$ a solution? $4 \leq 3(2) - 2 = 4$? Yes — the point is on the boundary which is included (solid, $\leq$).

Fill the blank: the boundary line for $y \geq 3x - 2$ is drawn as a _______ line.

What to write in your book

To check a specific point: substitute into the original inequality. True = in solution region.
Points on a solid boundary are included. Points on a dashed boundary are excluded.
Test the origin first — it is usually the simplest calculation.

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · GRAPH y > x + 1

Graph the solution region for $y > x + 1$ on the Cartesian plane.

Boundary line: $y = x + 1$. The symbol is $>$ (strict) → draw a dashed line through $(0, 1)$ and $(2, 3)$.

Dashed = points on the line are not solutions.

PROBLEM 2 · GRAPH y ≤ −2x + 4

Graph the solution region for $y \leq -2x + 4$.

Boundary line: $y = -2x + 4$. Symbol is $\leq$ (non-strict) → draw a solid line through $(0, 4)$ and $(2, 0)$.

Solid = points on the boundary are included in the solution.

PROBLEM 3 · IDENTIFY THE REGION FROM A GRAPH

A graph shows a solid line through $(0, -1)$ and $(3, 2)$ with shading above the line. Write the inequality.

Gradient $= \dfrac{2 - (-1)}{3 - 0} = \dfrac{3}{3} = 1$. $y$-intercept $= -1$. Boundary: $y = x - 1$.

Find gradient and intercept from two points on the line.

What to write in your book

To graph: boundary line (solid/dashed) → test $(0,0)$ → shade correct half-plane.
To read: boundary equation → line type (solid/dashed) → shading direction (above/below).
Verify with at least one test point inside the shaded region and one outside.

Quick-fire practice

Describe the graph of $y < 3x - 2$ (boundary type, which side is shaded).
Test whether $(-1, 2)$ satisfies $y \geq 2x + 1$.
A graph shows a dashed line with equation $y = x + 2$ and shading below. Write the inequality.
Why must you use a different test point if the boundary passes through $(0,0)$? Give an example.

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Revisit y > x

$y > x$ divides the plane along the dashed line $y = x$. Points above the line (where the $y$-value exceeds $x$) are in the solution region. $(3, 5)$: $5 > 3$ ✓ — in the region. $(5, 3)$: $3 > 5$ — not in the region.

Earlier you described which side the solutions lie on. Now explain the full graphing process for $y > x$ including the test point method.

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Multiple choice

+5 XP per correct · +25 XP all-correct

Pick your answer, then rate your confidence.

Short answer

ApplyBand 44 marks

Q1. Graph the solution region for $y \geq x - 2$. Describe the boundary type, the test point, and the shaded side. (4 marks)

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ApplyBand 33 marks

Q2. A graph has a dashed line $y = -x + 3$ with the region above shaded. Write the inequality and verify using two test points. (3 marks)

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UnderstandBand 32 marks

Q3. Explain why the boundary line is dashed for a strict inequality but solid for a non-strict inequality. (2 marks)

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📖 Comprehensive answers (click to reveal)

Practice: 1. Dashed line, shade below (test (0,0): $0 < -2$? No → shade side not containing origin). Wait — $3(0)-2=-2$; $0<-2$? No, shade above. 2. $(-1,2)$: $2 \geq 2(-1)+1 = -1$ ✓ yes. 3. $y < x+2$. 4. If boundary passes through $(0,0)$, substituting gives $0$ on both sides and cannot determine the region; e.g. for $y > x$, use $(1,0)$.

Q1 (4 marks): Boundary $y = x-2$, solid (non-strict $\geq$) [1]. Test $(0,0)$: $0 \geq 0-2 = -2$? Yes [1]. Shade side containing $(0,0)$, which is above the line [1]. Check: $(0,2)$: $2 \geq -2$ ✓ [1].

Q2 (3 marks): Dashed line + shading above → $y > -x+3$ [1]. Test $(0,4)$: $4 > 0+3 = 3$ ✓ (inside) [1]. Test $(0,0)$: $0 > 3$? No ✓ (outside) [1].

Q3 (2 marks): Strict ($<$, $>$): points on the boundary do NOT satisfy the inequality, so the line is dashed to show exclusion [1]. Non-strict ($\leq$, $\geq$): points on the boundary DO satisfy it, so the line is solid to show inclusion [1].

Boss battle · Half-Plane Hunt

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Identify and shade the correct solution region for two-variable inequalities. Beat the boss to bank a tier.

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Module overview · Mathematics Standard