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.
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.
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.
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.
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.
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 $y \leq mx + c$.
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.
Linear inequality in two variables: boundary line is y = mx + c (solid for ≤ or ≥; dashed for < or >). Test (0,0): if it satisfies the inequality, shade the side containing it; otherwise shade the other side.
Pause, copy the two-variable inequality graphing rules: solid boundary line for ≤ or ≥ (boundary included); dashed for < or > (boundary excluded); shade the region containing (0, 0) if it satisfies the inequality, otherwise shade the opposite side into your book.
We just saw drawing the boundary line (solid for ≤ or ≥, dashed for < or >), testing the point (0, 0) to identify which side to shade, and shading the solution region. That raises a question: given a completed two-variable inequality graph, how do you reverse the process and write the inequality it represents? This card answers it → read the boundary line's gradient and y-intercept, determine solid/dashed → choose ≤/≥ or , then test the shaded side with (0, 0) to confirm the direction of the inequality.
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.
From a graph of an inequality: read the boundary line’s equation (identify slope and y-intercept), determine solid or dashed → write ≤/≥ or , then check the shaded side with a test point to confirm the correct direction of the symbol.
Pause, copy the three-step process for reading an inequality from a two-variable graph: (1) identify the boundary line's equation (y = mx + c); (2) solid → ≤/≥, dashed → ; (3) test (0, 0) in the inequality, if the shaded side contains (0, 0), the symbol points toward (0, 0) into your book.
We just saw reading the inequality back from a two-variable graph by identifying the boundary line, solid/dashed status, and shaded side. That raises a question: after identifying the solution region, can you test any specific point to confirm whether it is a solution? This card answers it → substitute the point's coordinates into the original inequality; if both sides satisfy the inequality, the point is a solution, for points on the boundary, the solid/dashed distinction determines inclusion.
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, $\leq$).
Any point in the shaded solution region satisfies the inequality. Test specific points by substituting (a, b) into the original inequality. Boundary points: on solid line → satisfies ≤ or ≥; on dashed line → does not satisfy < or >.
Pause, copy the point-testing rule: substitute (a, b) into the inequality; if both sides balance, the point is a solution; and the boundary inclusion check (solid line → point on boundary is a solution; dashed line → it is not) into your book.
Worked examples · 3 in a row, reveal as you go
Graph the solution region for $y > x + 1$ on the Cartesian plane.
Graph the solution region for $y \leq -2x + 4$.
A graph shows a solid line through $(0, -1)$ and $(3, 2)$ with shading above the line. Write the inequality.
- 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.
$y > x$ 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.
Pick your answer, then rate your confidence.
Q1. Graph the solution region for $y \geq x - 2$. Describe the boundary type, the test point, and the shaded side. (4 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)
Q3. Explain why the boundary line is dashed for a strict inequality but solid for a non-strict inequality. (2 marks)
📖 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].
Identify and shade the correct solution region for two-variable inequalities. Beat the boss to bank a tier.
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