Linear inequalities in two variables, whose solutions are ordered pairs filling a half-plane. Includes checking a pair by substitution, the three-step graphing procedure with dashed and solid boundaries, using the origin as a test point, and rewriting into slope-intercept form to read the shading directly.
Subject: Algebra 1 · 65 slides · symbolic lesson
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Title
Algebra 1 · Chapter 6 — Solving and Graphing Linear Inequalities
Graphing Linear Inequalities in Two Variables
Objectives
Five outcomes, each one you can test yourself on with a pencil and no answer key.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.8 Graphing Linear Inequalities in Two Variables §6.8, pp. 367-373 — the lesson these objectives are drawn from
Warm-up
Lesson 4.2 checked whether an ordered pair solved a two-variable equation. This lesson asks the same question with an inequality.
Discussion prompt
Is (0, 0) a solution of 2x plus 3y at most 2? What about (0, 1)? Substitute both coordinates each time.
Hint: The routine is the same; only the final comparison changes.
Answer:
\[ (0, 0): \; 0 \le 2 \;\checkmark \qquad (0, 1): \; 3 \le 2 \;\times \]
The substitution is exactly as in Lesson 4.2 and the verdict comes from comparing rather than from matching. Because many pairs make the statement true, the solutions fill a region rather than lying on a line.
Concept
A linear inequality in two variables is one that can be written as ax plus by compared with c. An ordered pair is a solution if substituting its two numbers makes the inequality true, and the set of all solutions is a half-plane.
linear inequality in two variables — An inequality of the form ax plus by compared with c. Its solutions are ordered pairs, and they form a half-plane bounded by the line ax plus by equals c.
A line divides the coordinate plane into two half-planes.
Figure (svg): A line dividing the plane, with one half shaded as the solution set
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.8 Graphing Linear Inequalities in Two Variables §6.8, pp. 367-367
Section
Section 1
Concept
An ordered pair is a solution of a linear inequality when substituting its coordinates produces a true statement. The routine is Lesson 4.2's with a comparison at the end instead of a match.
Unlike an equation, many different pairs give true statements.
Figure (svg): Three ordered pairs tested against a two-variable inequality
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.8 Graphing Linear Inequalities in Two Variables §6.8, pp. 367-367 — Example 1, Check Solutions of a Linear Inequality
Picture it
Two pass and one fails.
Figure (svg): Three ordered pairs tested against a two-variable inequality
The two passing pairs are on the same side of the boundary and the failing one is on the other, which is the first hint that the solutions form a region rather than a scatter.
Worked example
This is Example 1 from the textbook.
\[ \text{Is } (0, 0), \; (0, 1) \text{ or } (2, -1) \text{ a solution of } \; 2x + 3y \le 2? \]
Test the origin
Why: Two times zero plus three times zero is zero.
\[ 0 \le 2,\text{ true} \]
Test (0, 1)
Why: Zero plus three is three.
\[ 3 \le 2,\text{ false} \]
Test (2, -1)
Why: Four plus negative three is one.
\[ 1 \le 2,\text{ true} \]
State the verdicts
Why: The first and third are solutions; the second is not.
Figure (svg): Three ordered pairs tested against a two-variable inequality
\[ (0, 0) \;\checkmark \quad (0, 1) \;\times \quad (2, -1) \;\checkmark \]
Verify: say what each verdict means about the picture
Why: The two solutions lie on one side of the line 2x plus 3y equals 2, and the failure lies on the other. Every check of a pair is a question about which side of the boundary a point falls on, which is what makes the graph worth drawing.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.8 Graphing Linear Inequalities in Two Variables §6.8, pp. 367-367
Sorting
Substitute both coordinates into 2x plus 3y at most 2.
Sort into buckets
Sort each pair by whether it is a solution.
One of the solutions gives exactly two, which the or-equal-to admits — that point lies on the boundary itself, which is why the line will be drawn solid.
Worked example
Producing solutions is as easy as checking them.
\[ \text{Find three solutions of } \; 2x + 3y \le 2 \; \text{ with } x = 1. \]
Substitute x equal to 1
Why: Two plus three y is at most two.
\[ 3 y \le 0 \]
Solve for y
Why: Divide by three.
\[ y \le 0 \]
Choose three values of y
Why: Zero, negative one and negative five all qualify.
Write the pairs
Why: Each is a solution.
\[ (1, 0), (1, -1), (1, -5) \]
Figure (svg): The solution to Worked example find a solution deliberately shown as a ladder of expressions, one row per algebraic move
\[ (1, 0), \; (1, -1), \; (1, -5) \]
Verify: notice how many solutions there are at one x-value
Why: Fixing x at one leaves a whole ray of permitted y-values rather than a single one. So the solution set is not just infinite but two-dimensional, which is why it is drawn as a shaded region rather than a curve.
Trap
\[ \text{Is } (2, -1) \text{ a solution of } 2x + 3y \le 2? \]
Put -1 in for x and 2 in for y: 2(-1) + 3(2) = 4
Why: Both numbers are present and nothing in the substitution announces which is which.
Four is not at most two, so the pair would be rejected — but (2, -1) really is a solution. The first coordinate is always x, by the convention from Lesson 4.1.
\[ (2, -1): \; x = 2, \; y = -1 \;\Longrightarrow\; 4 - 3 = 1 \le 2 \;\checkmark \]
Write down which value goes with which letter before substituting
Why: The order is fixed and the substitution becomes mechanical.
This is exactly the check from Lesson 4.2, and the same discipline prevents the same error.
Faded example
Both coordinates, then compare.
Fill in the blanks
(2, -1) \text-1 2x + 3y \le 2: \quad 2(2) + 3(1) = ___
Why: The y-coordinate of negative one gives three times negative one, which is negative three, and four minus three is one. Since one is at most two, the pair is a solution and the point lies in the shaded half-plane.
Elimination
Testing whether (3, 1) solves x minus 2y greater than 0.
Eliminate the wrong options
Which is right?
Survives elimination: A
Why: Three goes in for x and one for y, giving three minus two, which is one — greater than zero, so the pair is a solution. Option B is the standard error and it gives negative five here, producing exactly the wrong verdict.
Socratic
A two-variable equation gave a line.
Discussion prompt
Explain why a two-variable inequality has a solution set filling a region while the corresponding equation gives only a line. Then say how many solutions each has at a fixed value of x.
Hint: Fix x and count the permitted y-values.
Answer:
Fixing x in the equation leaves a one-variable equation with a single solution, so each x-value contributes exactly one point and the points form a curve. Fixing x in the inequality leaves a one-variable inequality with a whole ray of solutions, so each x-value contributes a ray and the rays sweep out a region.
So the equation has one solution per x-value and the inequality has infinitely many. That difference of dimension is why one is drawn as a line and the other as a shaded area, and it is the same jump as from a point to a ray in one variable.
Section
Section 2
Concept
The line ax plus by equals c divides the coordinate plane into two half-planes. The solutions of the inequality are one of them, together with the line itself when the symbol includes or equal to.
The line plays exactly the role the endpoint dot played in one variable.
Figure (svg): Two columns contrasting a dashed boundary with a solid one
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.8 Graphing Linear Inequalities in Two Variables §6.8, pp. 367-368 — the Vocabulary Tip on half-planes and the Study Tip on dashed and solid lines
Picture it
The boundary excluded or included.
Figure (svg): Two columns contrasting a dashed boundary with a solid one
The same decision as an open or solid dot, made the same way: substitute a point of the boundary and see whether the statement holds.
Worked example
The symbol decides, exactly as the dot did in Lesson 6.1.
\[ \text{Which boundary lines are dashed: } \; x < -2, \; y \le 1, \; x + y > 3, \; 2x - y \le 2? \]
Take the first
Why: Strictly less than, so the boundary is excluded.
Take the second
Why: At most, so the boundary is included.
Take the third
Why: Strictly greater than.
Take the fourth
Why: At most again.
Figure (svg): The solution to Worked example dashed or solid shown as a ladder of expressions, one row per algebraic move
\[ \text{dashed: } x < -2, \; x + y > 3 \]
Verify: test a point on one of the boundaries
Why: The point (0, 1) is on the line y equals 1, and substituting into y at most one gives a true statement — so it is a solution and the line is solid. Testing a boundary point settles the question independently of remembering which symbol is which.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.8 Graphing Linear Inequalities in Two Variables §6.8, pp. 368-368
Sorting
Look at whether the symbol includes or equal to.
Sort into buckets
Sort each inequality by the kind of boundary line it needs.
The direction of the symbol plays no part in this decision, and three of these shade one way while three shade the other. Dashed-or-solid and which-side are independent questions.
Worked example
The line is where the two sides change verdict.
\[ \text{For } 2x + 3y \le 2, \text{ what happens at points on the line } 2x + 3y = 2? \]
Substitute a boundary point
Why: Take (1, 0), which gives exactly two.
\[ 2 \le 2 \]
Judge it
Why: Two is at most two, so it is a solution.
Consider a point just above the line
Why: The expression exceeds two.
Consider a point just below
Why: The expression is less than two.
Figure (svg): A line dividing the plane, with one half shaded as the solution set
\[ 2x + 3y = 2 \text{ separates the two verdicts} \]
Verify: explain why the verdict cannot change anywhere else
Why: The expression takes every value on one side of the line above two and every value on the other side below two, changing only by passing through the line. So the boundary is the only place a verdict can flip, which is why testing a single point on one side settles the whole half-plane.
Error analysis
The student graphed the inequality y at most 1.
Annotate
On: \( \begin{aligned} &\text{draw the line } y = 1 \text{ dashed} \\ &\text{test } (0, 0): \; 0 \le 1 \text{ is true} \\ &\text{shade below the line} \end{aligned} \)
The dashed-or-solid decision is independent of the shading decision, and both have to be made. Testing a point on the line settles the first as reliably as testing a point off it settles the second.
Faded example
The symbol decides.
Fill in the blanks
For y <= 1 the boundary line is solid, and for x + y > 3 it is dashed.
Why: An or-equal-to symbol makes the boundary points solutions, drawn as a solid line, and a strict symbol excludes them, drawn dashed. It is the same decision as an open or filled endpoint dot in one variable.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| One variable (6.1) | Two variables (6.8) | |
|---|---|---|
| The solutions form | a ray on a line | a half-plane |
| The boundary is | a point | a line |
| Excluded boundary shown by | an open dot | a dashed line |
Every feature has a counterpart one dimension up. Recognising that makes this lesson a translation of Lesson 6.1 rather than a new subject.
Socratic
A line divides the plane into exactly two pieces.
Discussion prompt
Explain why the solutions of a linear inequality are always exactly one of the two half-planes, rather than some more complicated region. Then say what would change if the boundary were a curve.
Hint: Think about how the expression's value changes across the plane.
Answer:
The expression ax plus by takes larger values as you move in one direction perpendicular to the line and smaller values as you move the other way, changing steadily and passing through c exactly on the line. So everything on one side exceeds c and everything on the other falls short, and the inequality selects one side entirely.
With a curved boundary the plane can be divided into more than two regions — a circle divides it into an inside and an outside, and a parabola into two regions of very different shapes. The half-plane result is specific to linear inequalities, and it is what makes the test-one-point method reliable here.
Section
Section 3
Concept
To graph a linear inequality, draw the corresponding equation with the right kind of line, test a point not on the line, and shade the half-plane containing it if it is a solution — or the other one if it is not.
The origin is the convenient test point when the line does not pass through it.
Figure (svg): The three steps of graphing a two-variable inequality
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.8 Graphing Linear Inequalities in Two Variables §6.8, pp. 368-368 — the Graphing a Linear Inequality summary and its Study Tips
Picture it
Draw, substitute, shade.
Figure (svg): The three steps of graphing a two-variable inequality
The second step is what makes the third automatic. Without a test point the shading is a guess, and with one it is a consequence.
Worked example
This is Example 2 from the textbook.
\[ \text{Graph the inequality } \; x < -2. \]
Graph the corresponding equation
Why: x equals negative two is a vertical line.
Choose the line style
Why: The symbol is strict, so dash it.
Test the origin
Why: Zero is not less than negative two.
Shade the other side
Why: The origin is to the right, so shade to the left.
Figure (svg): A vertical dashed boundary with the left half-plane shaded
\[ x < -2: \text{ left of } x = -2 \]
Verify: test a point in the shaded region
Why: The point (-5, 0) gives negative five, which is less than negative two, so it is a solution and lies where the shading is. Testing a point in the region you shaded confirms the decision rather than repeating it.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.8 Graphing Linear Inequalities in Two Variables §6.8, pp. 368-368
Faded example
Pass shades this side; fail shades the other.
Fill in the blanks
For x < -2 the origin gives 0 < -2, which is false, so shade the side the origin is not on.
Why: The origin fails, so it is not in the solution set and the shading goes on the other side of the line — to the left. A failed test is decisive in exactly the same way a successful one is.
Worked example
This is Example 3 from the textbook.
\[ \text{Graph the inequality } \; y \le 1. \]
Graph the corresponding equation
Why: y equals one is a horizontal line.
Choose the line style
Why: The symbol includes or equal to, so draw it solid.
Test the origin
Why: Zero is at most one.
Shade its side
Why: The origin is below the line, so shade below.
Figure (svg): A horizontal solid boundary with the lower half-plane shaded
\[ y \le 1: \text{ on or below } y = 1 \]
Verify: test a point above the line
Why: The point (0, 3) gives three, which is not at most one, so it is correctly outside the shading. Testing on both sides is worth doing once, because two agreeing verdicts confirm the boundary as well as the direction.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.8 Graphing Linear Inequalities in Two Variables §6.8, pp. 368-368
Trap
\[ x < -2 \text{, testing } (0, 0) \]
The origin is to the right, so shade to the right
Why: The test point is where attention is, so its side is the one that gets shaded.
The origin failed the test, so it is not a solution and its side is not the answer. A failed test tells you to shade the other half-plane.
Read the verdict before choosing the side
Why: A pass means shade this side; a fail means shade the other.
A failed test is exactly as useful as a successful one, and half the problems will produce one.
Elimination
The boundary is the line y = 2x.
Eliminate the wrong options
Which point is a suitable test point?
Survives elimination: A
Why: The test point must lie in one of the half-planes, so it cannot be on the line. The origin is the convenient choice whenever the boundary misses it, and a line through the origin is exactly when another point has to be found.
Sorting
Three steps in a fixed order.
Sort into buckets
Sort each action by the step it belongs to.
The order matters: shading before testing turns a decision into a guess, and choosing the line style after shading invites forgetting it altogether.
Socratic
The half-plane contains infinitely many points.
Discussion prompt
Explain why testing a single point decides the shading for a whole half-plane. Then say what property of linear inequalities makes this reliable.
Hint: Ask whether two points on the same side can disagree.
Answer:
Every point on one side of the line gives a value of ax plus by on the same side of c, because the expression changes steadily and can only pass through c by crossing the line. So all the points of one half-plane give the same verdict, and testing one of them determines all of them.
It is the linearity that guarantees this. For a non-linear inequality the region can be more complicated and a single test point may not represent its whole side, which is why this method is stated for linear inequalities specifically.
Section
Section 4
Concept
Rewriting the corresponding equation in slope-intercept form makes the boundary easy to draw. Once y is isolated, a greater-than symbol shades above the line and a less-than symbol shades below.
The rule works only after y has been isolated.
Figure (svg): The rule that greater than shades above and less than shades below
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.8 Graphing Linear Inequalities in Two Variables §6.8, pp. 369-369 — Examples 4 and 5 and the Study Tip on above and below
Picture it
Once y is alone, the symbol names the side.
Figure (svg): The rule that greater than shades above and less than shades below
Applying this to an inequality that has not been rewritten is the standard error, since isolating y sometimes requires a reversal that changes which side is meant.
Worked example
This is Example 4 from the textbook.
\[ \text{Graph } \; x + y > 3 \; \text{ using slope-intercept form.} \]
Rewrite the corresponding equation
Why: Subtract x from each side.
\[ y = -x + 3 \]
Draw the line
Why: Slope negative one, intercept three, dashed for the strict symbol.
Test the origin
Why: Zero plus zero is zero, which is not greater than three.
Shade the other side
Why: The origin lies below, so shade above.
Figure (svg): A slanted boundary with the half-plane above it shaded
\[ x + y > 3: \text{ above } y = -x + 3 \]
Verify: check the shading against the isolated form
Why: Isolating y gives y greater than negative x plus three, and greater than means above — the same answer the origin test gave. Two independent routes agreeing is a genuine check on the shading.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.8 Graphing Linear Inequalities in Two Variables §6.8, pp. 369-369
Translation
Isolate y, then apply the rule.
Match the pairs
Why: The second and fourth both required dividing by negative one, which reversed their symbols — so both ended up shading the opposite side from what the original symbol suggested. That is exactly why the rule applies only after isolating.
Worked example
This is Example 5 from the textbook, where isolating y needs a reversal.
\[ \text{Graph } \; 2x - y \le 2 \; \text{ using slope-intercept form.} \]
Move the x-term
Why: Subtract 2x from each side.
\[ -y \le - 2 x + 2 \]
Divide by -1 and reverse
Why: The symbol turns from at most to at least.
\[ y \ge 2 x - 2 \]
Draw the line
Why: Slope two, intercept negative two, solid for the or-equal-to.
Read the side
Why: Greater than or equal to means shade above.
Figure (svg): The solution to Worked example an isolation that reverses shown as a ladder of expressions, one row per algebraic move
\[ 2x - y \le 2 \;\Longleftrightarrow\; y \ge 2x - 2 \]
Verify: confirm with the origin
Why: The origin gives zero minus zero, which is zero, and zero is at most two — so the origin is a solution. It lies above the line y equals 2x minus 2, since at x equal to zero the line is at negative two. Both methods agree on shading above.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.8 Graphing Linear Inequalities in Two Variables §6.8, pp. 369-369
Trap
\[ 2x - y \le 2 \]
The symbol is at most, so shade below the line
Why: The above-or-below rule is applied to the inequality as written.
Isolating y requires dividing by negative one, which reverses the symbol to at least — so the correct shading is above. Testing the origin confirms it: the origin satisfies the inequality and lies above the line.
\[ 2x - y \le 2 \;\Longrightarrow\; y \ge 2x - 2 \;\Longrightarrow\; \text{shade above} \]
Isolate y first, then apply the rule
Why: The rule describes the relationship between y and the expression, so y has to be alone for it to mean anything.
Testing the origin is the safe method in every case, and it never depends on remembering which reversals happened.
Faded example
Dividing by negative one turns the symbol.
Fill in the blanks
From 2x - y <= 2 you get -y <= -2x + 2, then y >= 2x - 2, so shade above the line.
Why: Dividing by negative one reverses at most into at least, and at least shades above. The original symbol suggested below, which is why applying the rule before isolating gives the wrong half-plane here.
Elimination
The inequality is 3x minus y at least 4.
Eliminate the wrong options
Which half-plane is the solution set?
Survives elimination: A
Why: Isolating y gives y at most 3x minus 4, which shades below. The origin confirms it: zero minus zero is zero, which is not at least four, so the origin fails — and the origin lies above the line, so the answer is the other side.
Socratic
The rule is stated for the isolated form.
Discussion prompt
Explain why y greater than an expression shades the region above the line, in terms of what the two sides of the inequality mean. Then say why the rule fails if applied before y is isolated.
Hint: Ask what a point above the line has in common with the line below it.
Answer:
At any given x-value, the line's height is the value of the expression. A point with the same x and a larger y sits directly above that point of the line, so y greater than the expression describes exactly the points lying above it — the vertical comparison is what above means.
Before isolating, the inequality compares some combination of x and y with a constant rather than comparing y with a height, so there is no vertical comparison to read. And if isolating would require dividing by a negative, the symbol reverses, so the unrewritten symbol names the wrong side — which is why the rule has to wait until y stands alone.
Section
Section 5
Concept
In a real situation the shaded half-plane is the set of combinations that satisfy a condition. Often only part of it is meaningful, because negative quantities have no interpretation.
The boundary line is where the condition is met exactly.
Figure (svg): A line dividing the plane, with one half shaded as the solution set
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.8 Graphing Linear Inequalities in Two Variables §6.8, pp. 367-373 — the chapter opener on meal planning and Exercises 51 and 52
Picture it
Every shaded point is a valid combination.
Figure (svg): A line dividing the plane, with one half shaded as the solution set
Reading a point out of the region gives one concrete answer, and the whole region shows how much freedom there is — which is what an inequality provides that an equation does not.
Worked example
This is the situation behind Exercises 51 and 52.
\[ \text{One food has } 100 \text{ calories a serving and another } 150. \text{ A meal must have at most } 600 \text{ calories.} \]
Name the variables
Why: Servings of each food.
Write the total
Why: One hundred x plus one hundred and fifty y calories.
\[ 100 x + 150 y \]
Write the condition
Why: At most six hundred.
\[ 100 x + 150 y \le 600 \]
Describe the graph
Why: A solid boundary with the region below it shaded.
Figure (svg): The solution to Worked example a nutrition constraint shown as a ladder of expressions, one row per algebraic move
\[ 100x + 150y \le 600 \]
Verify: test the origin and read what it means
Why: The origin gives zero calories, which is at most six hundred, so it is a solution — eating nothing satisfies a calorie ceiling, which is true and not very useful. The origin passing tells you to shade its side, and the situation then restricts attention to the part with both variables at least zero.
Elimination
The constraint is 100x plus 150y at most 600, with x and y servings.
Eliminate the wrong options
Which pair is not a meaningful solution?
Survives elimination: A
Why: The pair satisfies the inequality — eight hundred minus one hundred and fifty is six hundred and fifty, which is not at most six hundred, so in fact it fails that too — but the decisive objection is that negative one serving is meaningless. The algebra's solution set includes points the situation excludes, which is why the restriction has to be stated.
Worked example
The region contains every valid combination.
\[ \text{Which servings satisfy } \; 100x + 150y \le 600 \; \text{ with whole numbers?} \]
Try no servings of the second food
Why: One hundred x at most six hundred, so up to six servings.
\[ x\text{ from } 0\text{ to } 6 \]
Try two servings of the second
Why: Three hundred used, so up to three of the first.
\[ x\text{ from } 0\text{ to } 3 \]
Try four servings of the second
Why: Six hundred used, so none of the first.
\[ x = 0 \]
Note the pattern
Why: More of one leaves room for less of the other.
Figure (svg): A line dividing the plane, with one half shaded as the solution set
\[ (6, 0), \; (3, 2), \; (0, 4), \text{ among others} \]
Verify: check a point on the boundary
Why: Six servings of the first gives exactly six hundred calories, which the at-most symbol admits — so it lies on the solid boundary and is a valid meal. Boundary points are the combinations that use the whole allowance exactly.
Trap
\[ 100x + 150y \le 600 \]
Report the whole shaded half-plane as the answer
Why: The algebra shades everything below the line, so all of it looks like the answer.
The region extends into negative servings, and eating negative two portions of something is meaningless. The mathematical solution set is larger than the set of real answers.
Restrict to the part with both variables at least zero.
State the restriction alongside the graph
Why: The situation adds conditions the algebra does not know about.
This is the same restriction as the taxi fare in Lesson 4.2 and the budget line in Lesson 4.4, and it should be stated rather than assumed.
Faded example
Calories per serving times servings, added.
Fill in the blanks
\text100: \quad 150x + ___y \le 600
Why: Each coefficient is calories per serving and each variable a number of servings, so every term is in calories and can be compared with the six hundred. The units check confirms the model, exactly as in Lesson 5.5.
Hypothesis
Predict before you decide.
Predict first
What do the points on the boundary line of 100x plus 150y at most 600 represent?
Correct: Meals using exactly the whole 600-calorie allowance.
\[ 100(6) + 150(0) = 600 \qquad 100(0) + 150(4) = 600 \]
Why: On the line the total is exactly six hundred, so those combinations use the full allowance. The at-most symbol admits them, which is why the line is drawn solid — had the condition been strictly under six hundred, the line would be dashed and those meals excluded. The boundary is where the condition is met exactly, and that is usually the interesting case.
Socratic
Lesson 4.4 modelled a budget with an equation.
Discussion prompt
Compare modelling a budget with an equation against modelling it with an inequality, and say what each is claiming. Then say which is usually the more realistic model.
Hint: Ask whether the whole budget must be spent.
Answer:
The equation says the total is exactly the budget, so its solutions form a line and every combination spends every last unit. The inequality says the total is at most the budget, so its solutions fill the region below that line and include every combination that spends less as well.
The inequality is usually more realistic, since most constraints are ceilings rather than requirements — you may spend up to your budget, eat up to a calorie limit, carry up to a weight rating. The equation describes only the extreme cases, which are the boundary of the region and often the most efficient options, which is why they still matter.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| Inequality in one variable | Inequality in two variables | |
|---|---|---|
| A solution is | a number | an ordered pair |
| The solutions form | a ray on a number line | a half-plane |
| The boundary is drawn as | an open or solid dot | a dashed or solid line |
Every idea from Lesson 6.1 has a counterpart here, one dimension up. The boundary grows from a point to a line and the solution set from a ray to a region.
Pattern
Whether the boundary is vertical, horizontal or slanted, the same five moves cover it.
Rewriting into slope-intercept form is optional and makes the line easier to draw. If you use it to decide the shading, isolate y first — otherwise a reversal will send you to the wrong side.
OpenStax Elementary Algebra 2e, §4.7 Graphs of Linear Inequalities §4.7
Check
Substitute both coordinates.
Check your understanding
Is (1, 2) a solution of 3x - y >= 1?
Answer: A
Why: Substituting gives three minus two, which is one, and one is greater than or equal to one. The pair lies exactly on the boundary line, which the or-equal-to symbol includes.
Check
The symbol chooses the line style.
Check your understanding
What kind of boundary line does x + 2y < 6 need?
Answer: A
Why: A strict symbol means the points on the line are not solutions, which a dashed line records. Testing a boundary point such as (6, 0) gives exactly six, which is not less than six, confirming the exclusion.
Check
Read the verdict, then choose the side.
Check your understanding
Graphing 2x + y > 4, the origin gives 0 > 4, which is false. What do you shade?
Answer: A
Why: A failed test means the origin is not a solution, so its half-plane is not the answer and the other one is. The failed verdict is exactly as decisive as a successful one.
Real world
You are planning a meal from two foods. One has 100 calories a serving and 5 grams of protein; the other has 150 calories and 12 grams of protein. The meal must have at most 600 calories.
Discussion prompt
Write the calorie constraint as an inequality, graph it, and say which part of the region describes a real meal. Then say what a second condition — at least 30 grams of protein — would add to the picture.
Hint: Each coefficient is calories per serving.
Answer:
\[ 100x + 150y \le 600 \;\Longleftrightarrow\; 2x + 3y \le 12 \]
The graph is a solid line with the region below it shaded, and only the part with both variables at least zero describes a real meal — you cannot eat a negative number of servings. That restricts the answer to a triangle with corners at the origin, six servings of the first food, and four of the second.
The protein condition adds a second inequality, 5x plus 12y at least 30, whose solutions are the region above another line. A meal must satisfy both, so the answer becomes the overlap of two half-planes — which is exactly the and idea from Lesson 6.4, moved into two dimensions. Chapter 7 takes this up as a system of inequalities.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
For 2x - y <= 2, which half-plane should you shade?
Correct: Above the line, since isolating y reverses the symbol.
\[ 2x - y \le 2 \;\Longrightarrow\; -y \le -2x + 2 \;\Longrightarrow\; y \ge 2x - 2 \]
Why: Isolating y requires dividing by negative one, which turns at most into at least, so the answer is y at least 2x minus 2 and the shading goes above. The origin confirms it: zero minus zero is zero, which is at most two, so the origin is a solution — and the origin lies above the line, since at x equal to zero the line is at negative two. The first option applies the above-or-below rule before isolating, which is exactly when it fails.
Explain it
They can graph a line and have never shaded a region.
Discussion prompt
In no more than four sentences, explain how to graph a two-variable inequality without asking them to remember which side goes with which symbol. Then tell them the one point that makes the decision easiest.
Hint: Draw the line, then test.
Answer:
A usable answer: start by graphing the line you would get if the symbol were an equals sign, drawing it dashed if the symbol is strict and solid if it includes or equal to. Then pick any point that is not on the line and substitute it — if it makes the inequality true, shade its side, and if it makes it false, shade the other one.
The easiest test point is the origin, because putting zero in for both letters makes every product vanish. The only time you cannot use it is when the line passes through it, and then any other convenient point will do.
Exit ticket
Name the weakest spot before you close the deck. That is the one worth ten minutes tonight.
Predict first
Which of these would you least want to be handed cold on a quiz tomorrow?
Correct: Whichever you picked is the right answer — and each one has a specific fix.
Why: The line style is fixed by testing a point on the boundary itself. The shading is fixed by reading the verdict before choosing a side — pass means this side, fail means the other. The above-or-below rule is fixed by isolating y first, or by ignoring the rule entirely and testing the origin. Real restrictions are fixed by asking which quantities cannot be negative and saying so. Pick yours and do five of that kind tonight rather than twenty mixed ones.
Connect it up
Do this on paper. It is worth more than rereading the slides.
Draw it
Draw four coordinate planes across a page and graph one inequality on each: one with a vertical boundary, one horizontal, one slanted with a strict symbol, and one slanted whose isolation of y requires a reversal. For each, write the corresponding equation beside the plane, mark the line dashed or solid with a note saying which symbol caused it, write your test point with its substitution and verdict, and shade accordingly. Underneath the fourth, write the inequality in both its original and its isolated form and circle the symbol that changed. In the lower half, invent a real constraint of your own with two quantities, write it as an inequality, graph it, and shade only the part that describes something real, writing the restriction as a sentence. Finally, in the margin, write the three steps of the procedure in order.
For the fourth graph, the side you shaded from the test point must match the side the isolated form names. If they disagree, the reversal during isolation was missed, and the test point is the one to trust.
Recap
Five things, and the third is what turns a guess into a decision.
| If the question says | Your first move is |
|---|---|
| Is this pair a solution | Substitute both coordinates |
| Graph the inequality | Draw the corresponding equation first |
| The symbol is strict | Dash the boundary line |
| Which side do I shade | Test the origin, if it is off the line |
| It models a real quantity | Restrict to non-negative values |
That completes Chapter 6. Chapter 7 puts two equations together and asks where their graphs meet, which turns the overlap idea from this chapter into a method for finding a single point that satisfies two conditions at once.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.8 Graphing Linear Inequalities in Two Variables §6.8, pp. 367-373 — everything on these slides traces back here
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