Solutions of a linear inequality in two variables, boundary lines and half-planes, dashed against solid boundaries, the two-step graphing procedure with a test point, one-variable inequalities graphed in the plane, and modelling a real constraint as a shaded region.
Subject: Algebra 2 · 65 slides · symbolic lesson
Open the interactive version of this deck
Title
Algebra 2 · Chapter 2 — Linear Equations and Functions
Graph Linear Inequalities in Two Variables
Objectives
Five outcomes. The third is the whole method, and it is two steps long.
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.8 Graph Linear Inequalities in Two Variables §2.8, pp. 132-137 — the lesson these objectives are drawn from
Warm-up
Lesson 1.6 solved inequalities in one variable and shaded a ray. This lesson adds a second variable and shades half the plane.
Discussion prompt
In one dimension, x less than 2 shaded everything to the left of an open dot at 2. If x and y are both allowed to vary, what does x less than 2 describe now, and what happened to the open dot?
Hint: Ask what values of y are allowed when x is 1.
Answer:
Every y is allowed, because the inequality says nothing about y at all. So the solutions are every point whose x coordinate is under 2 — the whole region left of a vertical line at x equal to 2.
The open dot becomes a dashed line. The idea is identical: the boundary is excluded, and the only change is that a point has grown into a line.
Concept
A linear inequality in two variables is satisfied by whole areas of the plane rather than by isolated points or a single line. Its graph is one of the two halves the boundary line divides the plane into.
half-plane — One of the two regions a line divides a coordinate plane into. The graph of a linear inequality in two variables is a half-plane, with or without its boundary.
\[ Ax + By < C, \;\; Ax + By \leq C, \;\; Ax + By > C, \;\; Ax + By \geq C \]
An ordered pair is a solution when substituting both coordinates makes the inequality true — the same definition as for an equation, with equals replaced by an inequality symbol.
Figure (svg): A boundary line dividing the plane into two half-planes with one of them shaded and a test point marked
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.8 Graph Linear Inequalities in Two Variables §2.8, pp. 132-132
Section
Section 1
Concept
An ordered pair is a solution of a linear inequality in two variables exactly when substituting the two values makes the inequality a true statement. There is nothing more to it, and no graphing is needed.
solution of a linear inequality — An ordered pair that makes the inequality true when its coordinates are substituted for the variables.
\[ 2x + 5y > 9 \quad \text{at } (-2, 3): \; -4 + 15 = 11 > 9 \; \checkmark \]
This is the fastest way to answer a multiple-choice question about inequalities: test the options rather than graphing anything.
Figure (svg): A boundary line dividing the plane into two half-planes with one of them shaded and a test point marked
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.8 Graph Linear Inequalities in Two Variables §2.8, pp. 132-132
Picture it
One boundary line, and a point on each side of it.
Figure (svg): A boundary line dividing the plane into two half-planes with one of them shaded and a test point marked
Every point on one side gives a true statement and every point on the other gives a false one. That is what makes a single test point enough to decide the whole region.
Worked example
Example 1. Four candidates, tested one at a time.
\[ \text{Which of } (-4,-1), (-2,3), (2,-4), (6,-1) \text{ satisfies } 2x + 5y > 9? \]
Test the first pair
Why: Two times negative four is negative eight; five times negative one is negative five. The sum is negative thirteen, which is not greater than nine.
\[ -13,\text{ fails} \]
Test the second pair
Why: Two times negative two is negative four; five times three is fifteen. The sum is eleven, which is greater than nine.
\[ 11,\text{ works} \]
Test the third pair
Why: Four minus twenty is negative sixteen, which fails.
\[ -16,\text{ fails} \]
Test the fourth pair
Why: Twelve minus five is seven, which is not greater than nine.
\[ 7,\text{ fails} \]
Figure (svg): The solution to Worked example which pair is a solution shown as a ladder of expressions, one row per algebraic move
\[ (-2, 3) \quad \text{is the solution} \]
Verify: check the near miss
Why: The fourth pair gave 7, which is close to 9 but still short — a reminder that being near the boundary is not the same as being on the right side of it. Only one of the four produced a true statement, which is what a well-written multiple-choice question guarantees.
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.8 Graph Linear Inequalities in Two Variables §2.8, pp. 132-132
Sorting
Test each pair against 5x minus 2y at most 6.
Sort into buckets
Sort each ordered pair.
The origin is worth testing on any inequality: it is the easiest substitution there is, and it is the standard test point for the graphing procedure.
Worked example
Guided Practice 1 through 4, including a pair that lands exactly on the boundary.
\[ \text{Which of } (0,-4), (2,2), (-3,8), (-1,-7) \text{ satisfy } 5x - 2y \leq 6? \]
Test the first pair
Why: Zero minus two times negative four is positive eight, which is not at most six.
\[ 8,\text{ fails} \]
Test the second pair
Why: Ten minus four is exactly six, and six is at most six because the symbol is inclusive.
\[ 6,\text{ works} \]
Test the third pair
Why: Negative fifteen minus sixteen is negative thirty-one, comfortably at most six.
\[ -31,\text{ works} \]
Test the fourth pair
Why: Negative five plus fourteen is nine, which is not at most six.
\[ 9,\text{ fails} \]
Figure (svg): The solution to Worked example four pairs, one inequality shown as a ladder of expressions, one row per algebraic move
\[ (2, 2) \;\text{ and }\; (-3, 8) \quad \text{are solutions} \]
Verify: look closely at the pair that gave exactly six
Why: The pair (2,2) lands exactly on the boundary, and it counts as a solution because the symbol allows equality. Had the inequality been strictly less than, that same pair would have failed — which is precisely what the dashed-versus-solid distinction records on a graph.
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.8 Graph Linear Inequalities in Two Variables §2.8, pp. 132-132
Trap
\[ \text{Is } (6, -1) \text{ a solution of } 2x + 5y > 9? \]
Substitute the x value and stop
Why: The pair is treated as a single value rather than as two.
\[ 2(6) = 12 > 9 \;\Longrightarrow\; \text{yes} \quad \text{(wrong)} \]
The y term was never included. With it, the value is 12 minus 5, which is 7 — and 7 is not greater than 9.
\[ \text{Is } (6, -1) \text{ a solution of } 2x + 5y > 9? \]
Substitute BOTH coordinates and evaluate the whole left side
Why: An ordered pair carries two values, and the inequality has a term for each.
\[ 2(6) + 5(-1) = 12 - 5 = 7 \]
\[ 7 \ngtr 9 \;\Longrightarrow\; \text{not a solution} \]
Prediction
Commit before testing anything.
Predict first
For 2x plus 5y greater than 9, will the origin be a solution?
Correct: No — substituting gives 0, and 0 is not greater than 9.
\[ 2(0) + 5(0) = 0 \ngtr 9 \]
Why: Substituting the origin into any expression of the form Ax plus By gives zero, so the verdict depends entirely on how zero compares with the constant on the right. Here zero is not greater than nine, so the origin fails and the shaded region is the half-plane on the other side of the boundary. That single observation is what makes the origin the natural test point.
Fill the middle
Guided Practice 3.
Fill in the blanks
5(-3) - 2(8) = -31 \leq 6 \quad \checkmark
Why: Five times negative three is negative fifteen, and minus two times eight is minus sixteen, giving negative thirty-one. That is comfortably at most six, so the pair is a solution. Note that both terms came out negative, which is why the result is so far below the boundary — this point sits well inside the shaded region rather than near its edge.
Explain it to yourself
The graphing procedure tests a single point and shades an entire half-plane.
\[ Ax + By < C \]
Discussion prompt
Explain why testing one point is enough to decide which whole half-plane to shade. What would have to be true for a half-plane to contain both solutions and non-solutions, and why can that not happen?
Hint: Think about what would have to happen between a solution and a non-solution.
Answer:
Moving continuously from a solution to a non-solution, the value of Ax plus By changes continuously, so at some moment it must equal C exactly — and that moment is a point ON the boundary line.
So any path from a solution to a non-solution has to cross the boundary. Within one half-plane there is no boundary to cross, so every point there gives the same verdict. That is why one test settles the whole region.
Section
Section 2
Concept
Replacing the inequality symbol with an equals sign gives the boundary line. It splits the plane into two half-planes, exactly one of which is the graph. Whether the line itself is included depends on the symbol.
boundary line — The line obtained by replacing the inequality symbol with an equals sign. It separates the solutions from the non-solutions.
\[ 3x - 2y > 2 \;\Longrightarrow\; \text{boundary } 3x - 2y = 2 \]
Dashed for strict inequalities, solid for inclusive ones. It is the open and solid dot rule from Lesson 1.6, promoted from a point to a line.
Figure (svg): Two columns contrasting a dashed boundary for strict inequalities with a solid boundary for inclusive ones
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.8 Graph Linear Inequalities in Two Variables §2.8, pp. 132-133 — the Interpret Graphs note and Graphing a Linear Inequality
Picture it
The two kinds of boundary, and what each claims.
Figure (svg): Two columns contrasting a dashed boundary for strict inequalities with a solid boundary for inclusive ones
A dashed line says the points on it fail; a solid line says they succeed. Getting this wrong changes the answer for infinitely many points, all of them on the line.
Worked example
Four inequalities, four boundaries.
\[ \text{For } y > -2x, \; 5x - 2y \leq -4, \; y \leq -3, \; x < 2: \text{ name each boundary and its style.} \]
Replace each symbol with an equals sign
Why: That is the whole of finding the boundary; the algebra is unchanged.
\[ y = -2 x, 5 x - 2 y = -4, y = -3, x = 2 \]
Classify the first two symbols
Why: Greater than is strict, so dashed; at most is inclusive, so solid.
Classify the last two symbols
Why: At most is inclusive, so solid; less than is strict, so dashed.
Note what kind of line each boundary is
Why: A line through the origin, a slanted line, a horizontal line, and a vertical line.
Figure (svg): The solution to Worked example identify the boundary and its style shown as a ladder of expressions, one row per algebraic move
\[ \text{dashed, solid, solid, dashed} \]
Verify: check one boundary point against its inequality
Why: For the third, the point (0,-3) is on the boundary, and substituting gives negative three at most negative three, which is true — so the line belongs and must be solid. For the fourth, (2,0) gives 2 less than 2, which is false, so the line is excluded and must be dashed. The substitution confirms the symbol rule.
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.8 Graph Linear Inequalities in Two Variables §2.8, pp. 133-133
Sorting
Decide from the symbol alone.
Sort into buckets
Sort each inequality by the boundary style its graph needs.
Only the symbol matters. The direction of the inequality decides which side to shade, and that is a separate question answered by the test point.
Worked example
The boundary drawn, and the shading decided.
\[ \text{For } 3x - 2y > 2, \text{ decide which half-plane to shade using the origin.} \]
Check that the origin is not on the boundary
Why: Substituting gives 0, and the boundary requires 2, so the origin is safely off the line.
Substitute the origin into the inequality
Why: Three times zero minus two times zero is zero.
Read the verdict
Why: Zero is not greater than two, so the origin is not a solution.
Shade the OTHER half-plane
Why: Since the test point fails, the solutions lie on the side that does not contain it.
Figure (svg): The solution to Worked example which half-plane, from a single point shown as a ladder of expressions, one row per algebraic move
\[ \text{origin fails} \;\Longrightarrow\; \text{shade the other half-plane} \]
Verify: test a second point in the shaded region
Why: Take (3,-3): three times three minus two times negative three is nine plus six, which is 15, comfortably greater than 2. A point in the region you shaded must satisfy the inequality, and it does — so the shading is right.
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.8 Graph Linear Inequalities in Two Variables §2.8, pp. 132-132
Trap
\[ y > -2x \]
Draw the boundary as a solid line
Why: The boundary is drawn the way a line is normally drawn, without asking whether it belongs.
The graph now claims every point on the line is a solution. But at (1,-2) the statement reads negative two greater than negative two, which is false.
\[ y > -2x \]
Read the symbol before drawing the line
Why: Greater than is strict, so the boundary is excluded and must be dashed.
Points on the line give equality, not strict inequality, so they fail. A dashed line is how a graph says so.
\[ \text{at } (1,-2): \; -2 > -2 \text{ is false} \]
Matching
Replace the symbol with an equals sign.
Match the pairs
Why: Finding the boundary is purely mechanical: the symbol becomes an equals sign and nothing else changes. What varies is the KIND of line that results, and that determines how you draw it — the first passes through the origin, which matters because the origin is then unavailable as a test point.
Prediction
Commit before reasoning.
Predict first
For which of these inequalities is the origin an unsuitable test point?
Correct: y greater than negative 2x — the origin lies on its boundary.
The tell is quick: if the inequality has no constant term, its boundary passes through the origin and another test point is needed.
Why: The boundary of that inequality is y equals negative two x, which passes through the origin. A test point must be OFF the boundary, because a point on it gives equality and therefore cannot distinguish the two sides. The textbook flags exactly this in an Avoid Errors note. For that inequality a point like (1,1) is used instead.
Explain it
A classmate keeps confusing which symbol gives a dashed line with which side to shade.
Discussion prompt
In three sentences, explain that these are two separate questions, say what answers each, and give them a way to remember which is which.
Hint: One is answered by the symbol, the other by a substitution.
Answer:
The style of the line and the side to shade are decided by different things. The SYMBOL alone decides dashed or solid — strict is dashed, inclusive is solid — and it says nothing about which side.
The side is decided by substituting a test point off the line: if it works, shade its side; if not, shade the other. Symbol for the line, substitution for the side.
Section
Section 3
Concept
Graph the boundary line, dashed or solid according to the symbol. Then substitute any point not on the line: if it satisfies the inequality, shade its half-plane; if not, shade the other.
\[ y > -2x \quad \text{test } (1, 1): \; 1 > -2 \; \checkmark \]
The origin is usually the easiest test point, but it can only be used when it does not lie on the boundary.
Figure (svg): The two-step procedure: draw the boundary, then test a point off it to decide which half-plane to shade
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.8 Graph Linear Inequalities in Two Variables §2.8, pp. 133-133 — Graphing a Linear Inequality
Picture it
Example 3a: y greater than negative two x.
Figure (svg): The two-step procedure: draw the boundary, then test a point off it to decide which half-plane to shade
Because the boundary passes through the origin, the test point had to be somewhere else — (1,1) is the textbook's choice, and any point off the line would do.
Worked example
Example 3a. The origin is unavailable, so another point is chosen.
\[ \text{Graph } y > -2x. \]
Graph the boundary line, dashed
Why: The boundary is y equals negative two x, and the symbol is strict, so the line is dashed.
Notice the origin lies on the boundary
Why: Substituting gives 0 equals 0, so the origin cannot distinguish the two sides.
Choose another test point and substitute
Why: Take (1,1): the inequality reads 1 greater than negative 2, which is true.
\[ (1, 1)\text{ works} \]
Shade the half-plane containing that point
Why: Since the test point is a solution, its side is the graph.
\[ \text{shade toward } (1, 1) \]
Figure (svg): The solution to Worked example a boundary through the origin shown as a ladder of expressions, one row per algebraic move
\[ y > -2x \]
Verify: test a point on the other side
Why: Take (-1,-1): the inequality reads negative one greater than positive two, which is false. So the unshaded side really does fail, confirming the shading rather than merely repeating the first test.
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.8 Graph Linear Inequalities in Two Variables §2.8, pp. 133-133
Ranking
Graphing a linear inequality in two variables.
Put in order
Why: The boundary has to exist before its style can be chosen or a point can be checked against it — which is why finding it comes first. Choosing the test point before substituting matters because the choice has a constraint: it must be off the line. And the shading rule depends on the verdict, so it comes last.
Worked example
Example 3b. This time the origin can be used, and it does not work.
\[ \text{Graph } 5x - 2y \leq -4. \]
Graph the boundary line, solid
Why: The boundary is 5x minus 2y equals negative 4, and at most is inclusive, so the line is solid.
Find two points to draw the boundary
Why: Setting y to zero gives x equal to negative four fifths; setting x to zero gives y equal to 2.
\[ (-0.8, 0)\text{ and } (0, 2) \]
Test the origin
Why: Zero minus zero is zero, and zero is not at most negative four.
Shade the other half-plane
Why: Since the origin is not a solution, the graph is the side that does not contain it.
Figure (svg): The solution to Worked example a solid boundary, origin fails shown as a ladder of expressions, one row per algebraic move
\[ 5x - 2y \leq -4 \]
Verify: test a point in the shaded region
Why: Take (-2, 0): five times negative two minus zero is negative ten, which is at most negative four — true. And the boundary point (0,2) gives negative four, which is at most negative four, confirming the solid line belongs.
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.8 Graph Linear Inequalities in Two Variables §2.8, pp. 133-133
Error analysis
A student graphs 5x minus 2y at most negative 4 and shades the half-plane containing the origin.
Annotate
On: \( \text{test } (0,0): \; 5(0) - 2(0) = 0 \leq -4 \;\Longrightarrow\; \text{shade toward the origin} \)
After shading, test one more point inside the shaded region. It costs ten seconds and it catches this error every time.
Discrimination
Say whether the origin may be used for each.
Sort into buckets
Sort each inequality by whether the origin is a legal test point.
Fill the middle
Guided Practice 10: graph 2x minus 6y greater than 12.
Fill in the blanks
\textfalse (0,0): \; 2(0) - 6(0) = 0, \text___ 0 > 12 \text___ ___
Why: Zero is not greater than twelve, so the origin fails and the shaded region is the half-plane away from it. The boundary is 2x minus 6y equals 12, which crosses at (6,0) and (0,-2), and the shading is below and right of it. Testing (10,0) confirms: 20 is greater than 12.
Commit first
Answer, then rate your confidence honestly.
Predict first
Does solving a two-variable inequality for y ever require reversing the symbol?
Correct: Yes, whenever you divide by a negative coefficient of y.
\[ 5x - 2y \leq -4 \;\Longrightarrow\; -2y \leq -5x - 4 \;\Longrightarrow\; y \geq \tfrac{5}{2}x + 2 \]
Why: The reversal rule from Lesson 1.6 is unchanged: dividing both sides by a negative number reverses the symbol, however many variables are present. So 5x minus 2y at most negative 4 becomes, after subtracting 5x and dividing by negative 2, y at least five halves x plus 2 — an at-least where there was an at-most. Missing that flip shades the wrong half-plane, which is why the test-point method is safer: it never requires solving for y at all.
Section
Section 4
Concept
An inequality naming only one variable still graphs as a half-plane, because the other variable may take any value at all. The boundary is a horizontal or vertical line, and the shading covers everything on one side of it.
\[ y \leq -3 \text{ and } x < 2 \]
This is the same fact as in Lesson 2.3, where y equals 2 was a whole horizontal line rather than a single point.
Figure (svg): Two graphs: y at most negative three shaded below a solid horizontal line, and x less than two shaded left of a dashed vertical line
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.8 Graph Linear Inequalities in Two Variables §2.8, pp. 133-133 — Graph linear inequalities with one variable
Picture it
Example 2: y at most negative three, and x less than two.
Figure (svg): Two graphs: y at most negative three shaded below a solid horizontal line, and x less than two shaded left of a dashed vertical line
Note that the origin fails the first and satisfies the second, which is why one is shaded away from it and the other toward it.
Worked example
Example 2a. Solid, and shaded away from the origin.
\[ \text{Graph } y \leq -3. \]
Find the boundary and its style
Why: The boundary is y equals negative three, a horizontal line, and at most is inclusive, so it is solid.
Test the origin
Why: Substituting gives 0 at most negative 3, which is false.
Shade the other half-plane
Why: The solutions lie on the side away from the origin, which is below the line.
Say what the region contains
Why: Every point whose y coordinate is negative three or less, whatever its x coordinate.
\[ \text{any } x, y\text{ at most } -3 \]
Figure (svg): The solution to Worked example a horizontal boundary shown as a ladder of expressions, one row per algebraic move
\[ y \leq -3 \]
Verify: test two points in the region with very different x values
Why: Both (100, -5) and (-100, -5) have y equal to negative five, which is at most negative three, so both are solutions. That the x coordinate can be anything at all is exactly what makes the graph a half-plane rather than a ray.
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.8 Graph Linear Inequalities in Two Variables §2.8, pp. 133-133
Comparison
Fill the blanks. The same inequality, in two different spaces.
Comparison matrix
| Feature | On a number line | In the coordinate plane |
|---|---|---|
| What x < 2 graphs as | a ray | a half-plane |
| The boundary | a point at 2 | a vertical line at x = 2 |
| Excluded boundary shown by | an open dot | a dashed line |
| Included boundary shown by | a solid dot | a solid line |
| What the other variable does | there is none | takes any value |
Every row is the same idea one dimension up. Nothing new was invented; a point grew into a line and a ray grew into a region.
Worked example
Example 2b. Dashed, and shaded toward the origin.
\[ \text{Graph } x < 2. \]
Find the boundary and its style
Why: The boundary is x equals two, a vertical line, and less than is strict, so it is dashed.
Test the origin
Why: Substituting gives 0 less than 2, which is true.
Shade the half-plane containing it
Why: The solutions lie on the same side as the origin, which is to the left.
Say what the region contains
Why: Every point whose x coordinate is less than two, whatever its y coordinate.
\[ x < 2,\text{ any } y \]
Figure (svg): The solution to Worked example a vertical boundary shown as a ladder of expressions, one row per algebraic move
\[ x < 2 \]
Verify: test a point on the boundary and one just past it
Why: The point (2, 5) gives 2 less than 2, which is false — so the dashed line correctly excludes it. And (1.9, 5) gives a true statement, so the region really does come right up to the line without including it.
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.8 Graph Linear Inequalities in Two Variables §2.8, pp. 133-133
Error analysis
A student is asked to graph x less than 2 in the coordinate plane and draws a shaded ray along the horizontal axis.
Annotate
On: \( x < 2 \;\Longrightarrow\; \text{an open dot at } 2 \text{ on the } x\text{-axis, shaded left along the axis} \)
The question is always which SPACE you are graphing in. The same inequality is a ray on a number line and a half-plane in the coordinate plane.
Sorting
Decide from which variable is named.
Sort into buckets
Sort each inequality by the direction of its boundary line.
The named variable is the one that is pinned; the line runs along the direction of the free one. That is exactly the rule from Lesson 2.3.
Prediction
Commit before reasoning.
Predict first
For y at least negative one, which half-plane is shaded?
Correct: Above the line y equals negative one.
\[ y \geq -1: \; \text{solid line, shaded above} \]
Why: At least negative one means the y coordinate is negative one or larger, and larger y values are higher on the plane, so the region is above the boundary. For a one-variable inequality the direction can be read straight off the symbol without a test point — greater means above for y and right for x. The test point method still works, and testing the origin here gives 0 at least -1, which is true, confirming the region containing the origin.
Counterexample
A classmate offers a general rule.
\[ \text{a greater-than inequality is always shaded above the boundary} \]
Discussion prompt
Find an inequality where greater than does not mean shaded above, and say when the rule does hold and why.
Hint: Try one that names x rather than y.
Answer:
\[ x > 3 \quad \text{is shaded to the RIGHT, not above} \]
The rule holds only when the inequality is solved for y — then y greater than something really does mean above, because larger y is higher up. For x the same reasoning gives right rather than above.
And for a general inequality like 5x minus 2y at most negative 4, the rule cannot be applied at all until it is solved for y, which risks the reversal from Lesson 1.6. The test point method avoids the whole question.
Section
Section 5
Concept
Real constraints say at most or at least, so they model as inequalities rather than equations. The graph is then the set of every allowable combination, and the boundary is the case that uses the resource exactly.
\[ 0.4x + 1.2y \leq 300 \]
Quantities that cannot be negative restrict the picture further: only the part of the half-plane in the first quadrant makes physical sense.
Figure (svg): A shaded region in the first quadrant showing the possible combinations of standard and high-quality recording time
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.8 Graph Linear Inequalities in Two Variables §2.8, pp. 134-134 — the DVD recording example
Picture it
Example 4: up to 300 megabytes of DVD space, at 0.4 MB per second of standard video and 1.2 for high quality.
Figure (svg): A shaded region in the first quadrant showing the possible combinations of standard and high-quality recording time
Points on the boundary use every megabyte; points inside leave space spare. Both are allowed, which is what an at-most constraint means.
Worked example
Example 4, the first two parts.
\[ \text{Up to } 300 \text{ MB, at } 0.4 \text{ MB/sec standard and } 1.2 \text{ MB/sec high quality. Write and graph the inequality.} \]
Write the verbal model with units
Why: Standard rate times standard time, plus high-quality rate times high-quality time, is at most the total space. Megabytes per second times seconds gives megabytes on both terms.
\[ MB / \sec \times \sec = MB \]
Write the inequality
Why: Letting x be standard seconds and y high-quality seconds.
\[ 0.4 x + 1.2 y \le 300 \]
Graph the boundary, solid
Why: At most is inclusive. The boundary crosses at (750, 0) and (0, 250).
Test the origin and shade
Why: Zero is at most 300, so the origin is a solution and its side is shaded — but only the part in the first quadrant, since neither time can be negative.
Figure (svg): The solution to Worked example write and graph the constraint shown as a ladder of expressions, one row per algebraic move
\[ 0.4x + 1.2y \leq 300 \]
Verify: check the two intercepts against the situation
Why: At (750, 0) the recording is all standard: 750 seconds at 0.4 MB per second is exactly 300 MB. At (0, 250) it is all high quality: 250 times 1.2 is also 300 MB. Both use the whole disc, which is what being on the boundary should mean.
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.8 Graph Linear Inequalities in Two Variables §2.8, pp. 134-134
Real world
You have 40 dollars for a party. Pizzas cost 8 dollars and drinks cost 2 dollars.
Discussion prompt
Write the inequality, say what the boundary and the interior of the region mean, and give two combinations that work. What extra restriction does the situation impose that the algebra does not?
Hint: Think about what a fractional or negative pizza would mean.
Answer:
\[ 8p + 2d \leq 40 \]
The boundary is the combinations that spend exactly 40 dollars; the interior is those that leave change. Two that work: three pizzas and five drinks, costing 34, or two pizzas and twelve drinks, costing 40 exactly.
The algebra allows any real values, but the situation requires p and d to be non-negative whole numbers — you cannot buy 2.4 pizzas or negative drinks. So the real answer is a set of lattice points inside the triangle, not the whole shaded region.
Worked example
Example 4's third part. Some use everything and some do not.
\[ \text{Check whether } (150, 200), (300, 120) \text{ and } (600, 25) \text{ satisfy } 0.4x + 1.2y \leq 300. \]
Test the first combination
Why: Nought point four times 150 is 60, and 1.2 times 200 is 240. The total is exactly 300.
\[ 300,\text{ on the boundary} \]
Test the second
Why: Nought point four times 300 is 120, and 1.2 times 120 is 144. The total is 264.
\[ 264,\text{ inside} \]
Test the third
Why: Nought point four times 600 is 240, and 1.2 times 25 is 30. The total is 270.
\[ 270,\text{ inside} \]
Interpret the difference
Why: The first uses every megabyte; the other two leave 36 and 30 megabytes unused.
Figure (svg): The solution to Worked example identify three solutions shown as a ladder of expressions, one row per algebraic move
\[ 300, \; 264, \; 270 \quad \text{all at most } 300 \]
Verify: confirm each lies in the first quadrant
Why: All three have positive coordinates, so all three describe real recordings with non-negative times. A pair like (800, -50) would satisfy the arithmetic — giving 260 — while describing negative fifty seconds of video, which is why the physical restriction to the first quadrant is part of the model rather than an afterthought.
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.8 Graph Linear Inequalities in Two Variables §2.8, pp. 134-134
Error analysis
A student models the DVD problem and writes an equation instead.
Annotate
On: \( 0.4x + 1.2y = 300 \)
Read the wording for at most, at least, up to and no more than. Those phrases are inequalities, and modelling them as equations throws away most of the valid answers.
Matching
Each phrase is a constraint in ordinary English.
Match the pairs
Why: Three of these are inclusive and one is strict, and the everyday wording is what decides it. Up to, at least and no more than all admit the boundary case; fewer than excludes it. That distinction decides whether a group using exactly 300 megabytes is within their allowance, which is a real question rather than a technicality.
Estimation
The DVD constraint: 0.4 MB per second standard, 1.2 high quality, 300 MB total.
Predict first
Roughly how many seconds of pure high-quality video fit on the disc?
Correct: About 250 seconds.
\[ \frac{300}{1.2} = 250 \qquad \frac{300}{0.4} = 750 \]
Why: At 1.2 megabytes a second, 300 megabytes buys 300 divided by 1.2, which is 250 seconds — a little over four minutes. Estimating first catches the common slip of dividing by 0.4 instead, which gives 750 and is the answer for pure standard video. Note that high quality costs three times as much per second, so it buys exactly one third the time, which is a quicker route to the same check.
Socratic
One question, and nothing else on this slide.
\[ 0.4x + 1.2y = 300 \]
Discussion prompt
Every point on this boundary uses the disc exactly. Is a point on the boundary a better answer than a point inside the region, or a worse one? What would make you prefer one over the other, and what does that tell you about what an inequality model does and does not decide?
Hint: Ask what the film class is actually trying to achieve.
Answer:
Neither is better as a matter of mathematics. The inequality says which combinations are ALLOWED; it says nothing about which is desirable.
A group wanting the longest possible film would prefer the boundary, since anything inside wastes space. A group wanting a safety margin against a miscalculated encoding rate would prefer the interior.
So the model divides the plane into permitted and forbidden, and stops. Choosing the best permitted point is a different question — the one linear programming answers, and one you will meet in Chapter 3.
Comparison
Fill the blanks. The style and the side are independent.
Comparison matrix
| Decision | Decided by | Rule |
|---|---|---|
| Dashed or solid | the inequality symbol | strict dashed, inclusive solid |
| Which side to shade | a test point off the boundary | works: its side. fails: the other |
| Which test point | convenience | the origin, unless it is on the boundary |
| Direction of the boundary | which variables appear | one variable gives a horizontal or vertical line |
| How much of the region is real | the situation being modelled | often only the first quadrant |
Confusing the first two rows is the single most common error in this lesson, and keeping them as separate questions prevents it.
Pattern
One routine graphs any linear inequality in two variables.
Step five is worth the ten seconds. It catches a mis-read verdict in step four, which is the error that shades an entire half-plane wrongly.
Check
Testing an ordered pair. Substitute both coordinates.
Check your understanding
Which ordered pair is a solution of 2x + 5y > 9?
Answer: A
Why: Substituting gives -4 + 15, which is 11, and 11 is greater than 9. The other three give -13, -16 and 7, none of which exceeds 9.
Check
Boundary style and shading. Two separate questions.
Check your understanding
For the graph of 5x - 2y <= -4, which is correct?
Answer: A
Why: At most is inclusive, so the boundary is solid. Testing the origin gives 0, which is not at most -4, so the origin fails and the other half-plane is shaded.
Check
Modelling a constraint. Read the wording.
Check your understanding
A group has up to 300 MB of space, using 0.4 MB per second of standard video and 1.2 per second of high quality. Which inequality models this?
Answer: A
Why: Up to 300 means at most 300, so the symbol is inclusive and points the right way. Each term is a rate in MB per second times a time in seconds, giving megabytes.
Real world
A delivery van can carry at most 1200 kg. Boxes of type A weigh 30 kg and type B weigh 45 kg.
Discussion prompt
Write the constraint, graph it, and describe the region in words. Then say what happens to the region if a second constraint is added — that the van can hold at most 32 boxes in total — and what the two together describe.
Hint: A second constraint is a second half-plane.
Answer:
\[ 30a + 45b \leq 1200 \qquad \text{and} \qquad a + b \leq 32 \]
Each inequality is a half-plane, and a load must satisfy both, so the allowable region is where the two overlap — plus the first quadrant, since box counts cannot be negative. The result is a four-sided region with corners.
That overlap is a system of inequalities, which is Lesson 3.3, and finding the best point in it — the most boxes, or the most value — is linear programming. This lesson has just built the single half-plane those methods are made of.
Commit first
Answer, then rate your confidence honestly.
Predict first
How many solutions does a linear inequality in two variables have?
Correct: Infinitely many, filling a half-plane.
This is also why checking an inequality graph means testing a sample point rather than verifying every solution. Verification of infinitely many points is only possible by the argument from Section 1: crossing from a solution to a non-solution requires crossing the boundary.
Why: An equation in two variables has infinitely many solutions already — every point on a line. An inequality has infinitely many more: every point on one whole side of that line. This is the pattern of the whole chapter, and it is why the answer is drawn as a shaded region rather than listed: no list could contain it. The coefficients change where the region is, never how many points it holds.
Explain it
They can graph a line and can shade a number line, but have never combined the two.
Discussion prompt
In four sentences or fewer, give them the whole procedure, and tell them the one thing to check before using the origin as their test point.
Hint: The check is one substitution.
Answer:
Change the inequality sign to an equals sign and graph that line, drawing it dashed if the symbol is strict and solid if it allows equality. Then pick a point that is not on the line, substitute it, and shade the point's side if it makes the inequality true and the other side if it does not.
Before using the origin, substitute it into the BOUNDARY equation. If it satisfies the boundary the origin is on the line and cannot be used, so pick something like (1,1) instead.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For the style, read the symbol and nothing else: strict is dashed. For the shading, remember that a failing test point sends you to the OTHER side. For the origin, glance at the constant term — a zero constant means the boundary passes through it. For worded constraints, look for at most, at least, up to and fewer than, and match the symbol to the phrase. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
Draw one coordinate plane and graph a linear inequality of your own with a slanted boundary, showing the dashed or solid line, marking your test point, writing the substitution beside it, and shading the correct half-plane. Beneath it, draw two smaller planes: one for a one-variable inequality naming y and one naming x, labelling which variable is pinned and which is free in each. In the margin write the two decisions — style and side — with what answers each, and box the fact that they are independent. At the bottom of the page, invent a real constraint with two costs and a budget, write the inequality, graph it, shade only the part that makes physical sense, and mark one point on the boundary and one inside, writing what each means in the situation.
If the boxed note in your margin is one sentence rather than two, look again: the symbol decides the line, and a substitution decides the side, and neither has anything to say about the other.
Recap
Five things, and the whole method is only two steps of them.
| If you see | Then |
|---|---|
| A strict symbol | Dashed boundary |
| An inclusive symbol | Solid boundary |
| A test point that works | Shade its side |
| A test point that fails | Shade the other side |
| No constant term | The origin is on the boundary; test elsewhere |
That completes Chapter 2. Chapter 3 puts two of these conditions together at once and asks where they are both satisfied — which is where lines start intersecting and regions start overlapping.
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.8 Graph Linear Inequalities in Two Variables §2.8, pp. 132-137 — everything on these slides traces back here
Want this taught 1-on-1? Alexander tutors Algebra 2 — $55/session, free consultation.