Systems of linear inequalities, the graph as the overlap of half-planes, systems whose regions do not meet, systems of three or more inequalities and their bounded regions, absolute value inequalities inside a system, and modelling a pair of constraints.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 3 — Linear Systems and Matrices
Graph Systems of Linear Inequalities
Objectives
Five outcomes. The first is the whole method, and the rest is what it produces.
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.3 Graph Systems of Linear Inequalities §3.3, pp. 168-171 — the lesson these objectives are drawn from
Warm-up
Lesson 2.8 graphed one inequality as a half-plane. This lesson asks for two of them at once.
Discussion prompt
The point (5, -2) satisfies x plus y at most 8. Does that make it a solution of a system containing that inequality? What else would you have to check?
Hint: Compare with how you checked a solution of a system of equations in Lesson 3.1.
Answer:
\[ 5 + (-2) = 3 \leq 8 \quad \checkmark \]
Not yet. A solution of a system must satisfy every inequality in it, so the other one has to be checked too. If the system also contains 4x minus y greater than 6, then 20 plus 2 is 22, which is greater than 6 — so this pair does solve that system.
Concept
Each inequality shades a half-plane. A solution of the system must satisfy all of them, so the graph is the region shaded by every inequality at once — their intersection.
solution of a system of inequalities — An ordered pair that is a solution of each inequality in the system. The graph of the system is the set of all of them.
\[ \begin{cases} x + y \leq 8 \\ 4x - y > 6 \end{cases} \]
Coloured pencils are worth using: shade each inequality in its own colour, and the graph of the system is whatever ends up with every colour on it.
Figure (svg): Two shaded half-planes overlapping, with the region belonging to both marked as the graph of the system
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.3 Graph Systems of Linear Inequalities §3.3, pp. 168-168 — Graphing a System of Linear Inequalities
Section
Section 1
Concept
Graph the first inequality exactly as in Lesson 2.8, then the second on the same axes. The graph of the system is the part of the plane lying in both shaded regions.
system of linear inequalities — Two or more linear inequalities in the same variables, considered together. Its graph is the intersection of their individual graphs.
Every boundary keeps its own style: a dashed line stays dashed and a solid one stays solid, even where they bound the same region.
Figure (svg): Two shaded half-planes overlapping, with the region belonging to both marked as the graph of the system
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.3 Graph Systems of Linear Inequalities §3.3, pp. 168-168 — Graph a system of two inequalities
Picture it
Example 1: y greater than negative 2x minus 5, and y at most x plus 3.
Figure (svg): Two shaded half-planes overlapping, with the region belonging to both marked as the graph of the system
Only the doubly shaded region is the answer. A point in the red-only part satisfies the first inequality and fails the second, which makes it no more a solution than a point outside both.
Worked example
Example 1, both steps.
\[ \begin{cases} y > -2x - 5 \\ y \leq x + 3 \end{cases} \]
Graph the first inequality
Why: Boundary y equals negative 2x minus 5, dashed because the symbol is strict. Testing the origin gives 0 greater than negative 5, which is true, so shade toward the origin.
Graph the second inequality
Why: Boundary y equals x plus 3, solid because the symbol is inclusive. Testing the origin gives 0 at most 3, true, so shade toward the origin again.
Find the region shaded by both
Why: The two shadings overlap in a wedge opening to the right.
Check a point in the overlap
Why: Take (0, 0): it satisfies both, as the two tests already showed.
\[ (0, 0)\text{ works} \]
Figure (svg): The solution to Worked example graph a system of two shown as a ladder of expressions, one row per algebraic move
\[ \text{the region above } y = -2x - 5 \text{ and below } y = x + 3 \]
Verify: test one point in each single-shaded region
Why: Take (-4, 0), which is above the first boundary but above the second too: 0 is greater than 3 is false, so it fails the second inequality and correctly lies outside the overlap. Take (0, -6): it is below both, failing the first. Points in the singly shaded parts really do fail one inequality each.
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.3 Graph Systems of Linear Inequalities §3.3, pp. 168-168
Sorting
Test each pair against BOTH y greater than negative 2x minus 5 and y at most x plus 3.
Sort into buckets
Sort each ordered pair.
The point (1,4) lies exactly on the solid boundary, and an inclusive symbol means boundary points count. Check the boundary cases deliberately rather than assuming they fail.
Worked example
Guided Practice 2. Both need rearranging or a test point before shading.
\[ \begin{cases} 2x - \tfrac{1}{2}y \geq 4 \\ 4x - y \leq 5 \end{cases} \]
Graph the first boundary
Why: Two x minus half y equals 4 crosses at (2, 0) and (0, -8). The symbol is inclusive, so the line is solid.
\[ \text{solid through } (2, 0), (0, -8) \]
Test the origin in the first
Why: Zero is not at least 4, so shade away from the origin.
Graph the second boundary
Why: Four x minus y equals 5 crosses at (1.25, 0) and (0, -5). Inclusive again, so solid.
\[ \text{solid through } (1.25, 0), (0, -5) \]
Test the origin in the second
Why: Zero is at most 5, so shade toward the origin.
Identify the overlap
Why: The two shadings meet in a wedge below and right of both lines.
Figure (svg): The solution to Worked example two inequalities in standard form shown as a ladder of expressions, one row per algebraic move
\[ \text{the region right of the first line and left of the second} \]
Verify: test a point in the claimed overlap
Why: Take (4, 0): the first gives 8, which is at least 4 — true. The second gives 16, which is not at most 5 — false. So (4,0) is NOT in the overlap, and the region must be narrower than a first glance suggests. Testing candidates rather than trusting the sketch is what catches this.
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.3 Graph Systems of Linear Inequalities §3.3, pp. 169-169
Trap
\[ \begin{cases} y > -2x - 5 \\ y \leq x + 3 \end{cases} \]
Shade everything either inequality covers
Why: The two regions are combined rather than intersected.
That region includes (0, -6), which satisfies the second inequality but not the first: -6 is not greater than -5.
\[ \begin{cases} y > -2x - 5 \\ y \leq x + 3 \end{cases} \]
Keep only the region covered by BOTH
Why: A solution must satisfy every inequality in the system, so satisfying one is not enough.
This is the same distinction as and against or in Lesson 1.6. A system is an and, so the answer is the overlap, never the union.
Prediction
Commit before graphing.
Predict first
Two half-planes with non-parallel boundaries overlap. What shape is the overlap?
Correct: A wedge, unbounded in one direction.
This is worth predicting before drawing: if your two-inequality answer came out as a closed triangle, one of the shadings is wrong.
Why: Two non-parallel lines cross once and divide the plane into four wedges. Each inequality selects one side of each line, so together they select exactly one of the four wedges — which is unbounded because nothing closes it off. Getting a bounded shape needs at least three inequalities, which is Section 4.
Fill the middle
Checking the origin against the second inequality of Example 1.
Fill in the blanks
\texttrue (0,0): \; 0 \leq 0 + 3 \text___ ___ \;\Longrightarrow\; \text___
Why: Zero is at most three, so the origin satisfies the second inequality and its side is shaded. Doing this test for each inequality separately, and only then looking for the overlap, is what keeps a two-inequality system from becoming confusing — each half is just Lesson 2.8.
Explain it to yourself
The word system is doing specific work here.
\[ \begin{cases} x + y \leq 8 \\ 4x - y > 6 \end{cases} \]
Discussion prompt
Explain why the graph of a system of inequalities is the overlap rather than the combination of the two regions. What would the combination represent instead?
Hint: Compare with and against or from Lesson 1.6.
Answer:
A system asks for pairs satisfying every inequality, so a point must be in every region. That is exactly what intersection means.
The combination — the union — would represent an OR condition: satisfy the first or the second or both. That is a perfectly good thing to graph, but it is not what a system asks for, and the distinction is the same one that separated and-inequalities from or-inequalities on the number line.
Section
Section 2
Concept
If the two shaded regions have no point in common, no ordered pair satisfies both inequalities and the system has no solution. The commonest case is parallel boundaries with the shadings facing apart.
\[ \begin{cases} 2x + 3y < 6 \\ y \geq -\tfrac{2}{3}x + 4 \end{cases} \]
Rewriting both in slope-intercept form makes it obvious: one region is everything below a line and the other is everything above a parallel line two units higher.
Figure (svg): Two shaded regions with parallel boundaries that never overlap, so the system has no solution
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.3 Graph Systems of Linear Inequalities §3.3, pp. 169-169 — Graph a system with no solution
Picture it
Example 2: 2x plus 3y less than 6, and y at least negative two thirds x plus 4.
Figure (svg): Two shaded regions with parallel boundaries that never overlap, so the system has no solution
The first region lies below the lower line and the second above the upper one, so there is a permanent gap between them. No point is in both.
Worked example
Example 2. Rewriting the first inequality is what makes the verdict visible.
\[ \begin{cases} 2x + 3y < 6 \\ y \geq -\tfrac{2}{3}x + 4 \end{cases} \]
Rewrite the first in slope-intercept form
Why: Subtracting 2x and dividing by 3 gives y less than negative two thirds x plus 2. The division was by a positive, so no symbol flips.
\[ y < -(\frac{2}{3}) x + 2 \]
Compare the two boundaries
Why: Both have slope negative two thirds, so the boundaries are parallel, with intercepts 2 and 4.
Read the two shadings
Why: The first is everything strictly below its line; the second is everything on or above the higher line.
Conclude
Why: A point below the line at height 2 cannot also be above the line at height 4, so the regions never meet.
Figure (svg): The solution to Worked example no solution shown as a ladder of expressions, one row per algebraic move
\[ \text{the regions do not intersect: no solution} \]
Verify: test a point from each region
Why: Take (0, 0): it satisfies the first, since 0 is less than 6, but fails the second, since 0 is not at least 4. Take (0, 5): it satisfies the second but gives 15 in the first, which is not less than 6. Every point is in at most one region, which is what no overlap means.
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.3 Graph Systems of Linear Inequalities §3.3, pp. 169-169
Sorting
Both boundaries are y equals negative two thirds x plus something, so all are parallel.
Sort into buckets
Sort each system by what its overlap looks like.
Parallel boundaries alone decide nothing. The pair of directions decides everything, and there are exactly three outcomes.
Worked example
The same boundaries with one shading reversed.
\[ \begin{cases} 2x + 3y < 6 \\ y \leq -\tfrac{2}{3}x + 4 \end{cases} \]
Rewrite the first as before
Why: The same region: below the line at intercept 2.
\[ y < -(\frac{2}{3}) x + 2 \]
Read the second shading
Why: Now it is everything on or below the line at intercept 4.
Compare the regions
Why: Everything below the lower line is automatically below the upper one, so the first region sits entirely inside the second.
Conclude
Why: The overlap is the whole of the first region, so the second inequality adds nothing.
Figure (svg): The solution to Worked example a system that does overlap, for contrast shown as a ladder of expressions, one row per algebraic move
\[ \text{the first region entirely: the second is redundant} \]
Verify: test a point in the strip between the two lines
Why: Take (0, 3): the second inequality gives 3 at most 4, true, but the first gives 9, which is not less than 6, so it fails. The strip between the lines belongs to the second region only, confirming that the overlap stops at the lower line.
Error analysis
A student sees parallel boundaries and concludes there is no solution.
Annotate
On: \( \begin{cases} y < -\tfrac{2}{3}x + 2 \\ y \leq -\tfrac{2}{3}x + 4 \end{cases} \;\Longrightarrow\; \text{parallel, so no solution} \)
Parallel boundaries give three possibilities: an empty overlap, one region inside the other, or a strip between them. Always test a point before deciding.
Prediction
Commit before drawing.
Predict first
Two inequalities have parallel boundaries and both are shaded upward. What is the overlap?
Correct: The region above the higher line.
\[ y > 2 \;\text{ and }\; y > 5 \;\Longrightarrow\; y > 5 \]
Why: Everything above the higher line is automatically above the lower one, so the higher constraint is the binding one and the lower adds nothing. This is the nested case, and the overlap is the smaller of the two regions. Reasoning about which constraint is binding is exactly what linear programming does with much larger systems.
Counterexample
A classmate offers a shortcut.
\[ \text{a system of inequalities always has infinitely many solutions} \]
Discussion prompt
Find a system of two linear inequalities with no solution at all, and then find one whose solution set is a single point. What has to be true in each case?
Hint: For a single point, think about combining inclusive inequalities that just touch.
Answer:
\[ y < 2 \;\text{ and }\; y > 5 \;\Longrightarrow\; \text{no solution} \]
\[ y \geq 2 \;\text{ and }\; y \leq 2 \;\text{ and }\; x \geq 0 \;\text{ and }\; x \leq 0 \;\Longrightarrow\; \text{only } (0,2) \]
An empty solution set needs shadings facing apart. A single-point solution needs inclusive inequalities that pin each variable from both sides. Both are unusual, but a system is not guaranteed to have a region — or even more than one point.
Explain it
A classmate thinks parallel boundaries always mean no solution.
Discussion prompt
In three sentences, tell them the three possible outcomes when the boundaries are parallel, what decides between them, and the one test that settles any case in five seconds.
Hint: The test is a substitution.
Answer:
Parallel boundaries can give an empty overlap, one region nested inside the other, or a strip between the lines. What decides is which way each region is shaded — apart, the same way, or toward each other.
The five-second test is to substitute one convenient point, usually the origin, into both inequalities. If it satisfies both, the overlap is certainly not empty, and you can go on to work out its shape.
Section
Section 3
Concept
Adding a third inequality removes more of the plane. With three or more, the region often closes into a bounded shape with straight edges and corners.
\[ \begin{cases} x < 6 \\ y > -1 \\ y < x \end{cases} \]
The corners are where two boundaries cross. They matter enormously in applications, because the best point in a region of this kind is always at a corner.
Figure (svg): Three half-planes overlapping in a bounded triangular region
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.3 Graph Systems of Linear Inequalities §3.3, pp. 171-171 — Systems of three or more inequalities
Picture it
Exercise 17: x less than 6, y greater than negative 1, and y less than x.
Figure (svg): Three half-planes overlapping in a bounded triangular region
Each boundary contributes one edge, and each pair of boundaries that meets inside contributes one corner. Three inequalities can give at most a triangle.
Worked example
Exercise 17. Graph each, then intersect all three.
\[ \begin{cases} x < 6 \\ y > -1 \\ y < x \end{cases} \]
Graph the first
Why: A dashed vertical line at x equal to 6, with everything to its left shaded.
\[ \text{left of } x = 6 \]
Graph the second
Why: A dashed horizontal line at y equal to negative 1, with everything above it shaded.
\[ \text{above } y = -1 \]
Graph the third
Why: A dashed line y equals x; testing (1, 0) gives 0 less than 1, true, so shade below it.
\[ \text{below } y = x \]
Find the region satisfying all three
Why: The three shadings close off a triangle with corners at (-1,-1), (6,-1) and (6,6).
Figure (svg): The solution to Worked example a bounded region from three inequalities shown as a ladder of expressions, one row per algebraic move
\[ \text{the open triangle bounded by } x = 6, \; y = -1, \; y = x \]
Verify: test a point inside and one just outside
Why: Take (3, 0): 3 is less than 6, 0 is greater than -1, and 0 is less than 3 — all three hold. Take (3, 4): the first two hold but 4 is not less than 3, so it fails and lies outside. And all three corners are excluded, since every boundary is dashed.
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.3 Graph Systems of Linear Inequalities §3.3, pp. 171-171
Prediction
Commit before drawing.
Predict first
What is the smallest number of linear inequalities that can enclose a bounded region in the plane?
Correct: Three.
The same fact appears in geometry: three is the fewest straight sides a closed figure can have.
Why: Two half-planes give at most a wedge, which is always unbounded because nothing closes the open end. Three can close a triangle, provided their boundaries form one — three parallel boundaries would not. Four or more can give quadrilaterals and larger polygons, but three is the minimum, which is why triangular feasible regions are so common in textbook problems.
Worked example
Exercise 25 in structure: four constraints closing a four-sided region.
\[ \begin{cases} x \leq 10 \\ x \geq -2 \\ 3x + 2y < 6 \\ y > -4 \end{cases} \]
Graph the two vertical boundaries
Why: Solid lines at x equal to negative 2 and x equal to 10, with the region between them shaded.
Graph the horizontal boundary
Why: A dashed line at y equal to negative 4, with everything above shaded, cutting the strip from below.
\[ \text{above } y = -4 \]
Graph the slanted boundary
Why: Three x plus 2y equals 6 crosses at (2,0) and (0,3). Testing the origin gives 0 less than 6, true, so shade below it.
Intersect all four
Why: The strip, cut above by the slant and below by the horizontal, gives a four-sided region.
Figure (svg): The solution to Worked example four inequalities, a quadrilateral shown as a ladder of expressions, one row per algebraic move
\[ \text{a quadrilateral bounded by all four lines} \]
Verify: test a point inside
Why: Take (0, -1): x is between -2 and 10, y is greater than -4, and 3 times 0 plus 2 times -1 is -2, which is less than 6. All four hold. Take (0, 4): the last fails, since 8 is not less than 6, and the point sits above the slanted edge as expected.
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.3 Graph Systems of Linear Inequalities §3.3, pp. 171-171
Error analysis
A student graphs three inequalities and shades the overlap of only two.
Annotate
On: \( \begin{cases} x < 6 \\ y > -1 \\ y < x \end{cases} \;\Longrightarrow\; \text{shade the quarter-plane left of } x=6 \text{ and above } y=-1 \)
With three or more inequalities, test a point against EVERY one before believing the region. A single unapplied constraint usually leaves the region unbounded, which is a visible symptom.
Definition probe
The triangle from x less than 6, y greater than negative 1, and y less than x.
Sort into buckets
Sort each feature of the region by which boundary produces it.
Fill the middle
Where does the boundary y equals x meet the boundary y equals negative 1?
Fill in the blanks
y = x \;\text-1, -1\; y = -1 \;\Longrightarrow\; \text___ (___)
Why: Substituting y equal to negative 1 into y equals x gives x equal to negative 1, so the two boundaries meet at (-1, -1). Finding a corner is solving a two-equation system, exactly as in Lesson 3.2 — the inequalities become equations because a corner lies on both boundaries.
Real world
A workshop can build at most 10 chairs and at most 8 tables a week, and its bench space limits it to at most 14 items in total.
Discussion prompt
Write the four constraints including the non-negativity ones, describe the region, and say where you would look if you wanted to maximise the number of items built. Why there?
Hint: Count the constraints, including the ones the situation implies.
Answer:
\[ c \leq 10, \; t \leq 8, \; c + t \leq 14, \; c \geq 0, \; t \geq 0 \]
Five constraints give a five-sided region in the first quadrant. To maximise the total built you would look at the corners, in particular where c plus t equals 14 meets the other boundaries.
Corners matter because a linear quantity's largest value over a region of this kind always occurs at a corner — moving along an edge changes it steadily, so the extreme is at an end. That fact is the whole basis of linear programming.
Section
Section 4
Concept
An inequality with an absolute value has a V for its boundary rather than a line. Everything else is unchanged: draw the boundary dashed or solid, test a point, shade, and intersect with the other regions.
\[ y > \lvert x + 4 \rvert \]
Greater than shades the inside of the V, above it; less than shades the outside, below. A test point settles it as reliably as for a line.
Figure (svg): A V-shaped boundary with a horizontal line above it, and the region between them shaded
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.3 Graph Systems of Linear Inequalities §3.3, pp. 169-169 — Graph a system with an absolute value inequality
Picture it
Example 3: y at most 3, and y greater than the absolute value of x plus 4.
Figure (svg): A V-shaped boundary with a horizontal line above it, and the region between them shaded
The overlap is a triangle with a pointed base at the vertex of the V. Everything about the method is the same; only the shape of one boundary changed.
Worked example
Example 3, both steps.
\[ \begin{cases} y \leq 3 \\ y > \lvert x + 4 \rvert \end{cases} \]
Graph the first inequality
Why: A solid horizontal line at y equal to 3, with everything below shaded.
Graph the V boundary
Why: The vertex is where x plus 4 is zero, so at (-4, 0), and the branches have slopes 1 and negative 1. The symbol is strict, so the V is dashed.
\[ \text{dashed } V\text{ at } (-4, 0) \]
Test a point for the V
Why: Take (-4, 5): is 5 greater than the absolute value of 0? Yes, so shade INSIDE the V, above it.
Find the overlap
Why: Inside the V and below the horizontal line: a triangle with its point at the vertex.
Figure (svg): The solution to Worked example a line and a V shown as a ladder of expressions, one row per algebraic move
\[ \text{inside the V, below } y = 3 \]
Verify: test a point inside and one outside
Why: Take (-4, 2): 2 is at most 3, and 2 is greater than 0 — both hold, so it is in the region. Take (0, 2): 2 is at most 3, but the absolute value of 4 is 4, and 2 is not greater than 4 — so it fails and lies outside the V, as the picture shows.
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.3 Graph Systems of Linear Inequalities §3.3, pp. 169-169
Sorting
For each, decide which side of the V-shaped boundary is shaded.
Sort into buckets
Sort each inequality.
Rather than memorising this, test the point directly above or below the vertex. It settles every case in one substitution.
Worked example
Guided Practice 5. The V opens downward and the shading is outside it.
\[ \begin{cases} y > -2 \\ y \leq -\lvert x + 2 \rvert \end{cases} \]
Graph the first inequality
Why: A dashed horizontal line at y equal to negative 2, with everything above shaded.
Graph the V boundary
Why: The vertex is at (-2, 0) and the negative coefficient makes it open downward. The symbol is inclusive, so the V is solid.
\[ \text{solid downward } V\text{ at } (-2, 0) \]
Test a point for the V
Why: Take (-2, -1): is negative 1 at most negative zero, which is 0? Yes, so shade inside the downward V, below it.
Find the overlap
Why: Above y equal to negative 2 and inside the downward V: a triangle with its point at the top.
Figure (svg): The solution to Worked example a downward V shown as a ladder of expressions, one row per algebraic move
\[ \text{inside the downward V, above } y = -2 \]
Verify: find the corners of the region
Why: The V meets y equal to negative 2 where the absolute value of x plus 2 equals 2, so at x equal to 0 and x equal to negative 4. The region is the triangle with vertices at (-2, 0), (0, -2) and (-4, -2), with the bottom edge excluded because that boundary is dashed.
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.3 Graph Systems of Linear Inequalities §3.3, pp. 169-169
Trap
\[ y > \lvert x + 4 \rvert \]
Shade below the V, because greater-than usually means above the line
Why: A rule learned for straight boundaries is applied to a curved one without testing.
But below the V at, say, (-4, -3), the absolute value of 0 is 0, and -3 is not greater than 0. That point fails.
\[ y > \lvert x + 4 \rvert \]
Test a point, exactly as for a line
Why: The method does not change because the boundary is bent.
\[ \text{at } (-4, 5): \; 5 > \lvert 0 \rvert = 0 \quad \checkmark \]
So the region is INSIDE the V. For a V, greater-than means inside and above, which is a genuinely different picture from the straight-line case.
Fill the middle
Guided Practice 4: y at most 4, and y at least the absolute value of x minus 5.
Fill in the blanks
y = \lvert x - 5 \rvert \;\Longrightarrow\; \text5, 0 (___)
Why: The inside is zero at x equal to 5, and there is no constant outside the bars, so the vertex is at (5, 0) — exactly the reading skill from Lesson 2.7. Combined with y at most 4, the region is a triangle with corners at (5,0), (1,4) and (9,4), since the branches reach height 4 four units either side of the vertex.
Prediction
Commit before drawing.
Predict first
An upward V shaded inside, combined with a horizontal line shaded below it, gives what shape?
Correct: A triangle, provided the horizontal line is above the vertex.
\[ y \leq 3 \;\text{ and }\; y > \lvert x + 4 \rvert \;\Longrightarrow\; \text{triangle with vertex } (-4,0) \]
Why: The inside of the V is an unbounded wedge opening upward, and the horizontal line cuts it off at a fixed height, closing it into a triangle with its point at the vertex. If the horizontal line were BELOW the vertex the overlap would be empty instead, which is why the proviso matters — a single inequality can turn a bounded region into nothing.
Comparison
Fill the blanks. Only the shape changes.
Comparison matrix
| Step | Linear inequality | Absolute value inequality |
|---|---|---|
| Find the boundary | replace the symbol with = | replace the symbol with = |
| Shape of the boundary | a straight line | a V |
| Dashed or solid | by the symbol, as always | by the symbol, as always |
| Which side to shade | test a point off it | test a point off it |
| How it joins a system | intersect the regions | intersect the regions |
Four of the five rows are identical. Only the shape of the boundary differs, which is why nothing about the method needs relearning.
Section
Section 5
Concept
When a situation imposes two conditions at once, each becomes an inequality and the allowable combinations are the overlap. Reading a vertical slice of the region answers a question about one particular input.
\[ \begin{cases} y \geq 0.4x \\ y \leq 0.6x \end{cases} \]
As in Lesson 2.8, quantities that cannot be negative restrict the picture to the first quadrant, and that restriction is part of the model.
Figure (svg): A shaded band showing sale prices between forty and sixty percent of the regular price
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.3 Graph Systems of Linear Inequalities §3.3, pp. 170-170 — the shoe sale example
Picture it
Example 4: a sale takes between 40 and 60 percent off the regular price.
Figure (svg): A shaded band showing sale prices between forty and sixty percent of the regular price
The two boundaries both pass through the origin, so the region is a wedge rather than a strip. A vertical slice at any regular price gives the range of possible sale prices.
Worked example
Example 4. Two percentages become two inequalities.
\[ \text{Sale prices are between } 40\% \text{ and } 60\% \text{ off. Find the possible sale prices for a } \$70 \text{ item.} \]
Name the variables
Why: Let x be the regular price and y the sale price, both in dollars.
Turn 40 percent off into an inequality
Why: Forty percent off means paying at most 60 percent, so y is at most 0.6x.
\[ y \le 0.6 x \]
Turn 60 percent off into an inequality
Why: Sixty percent off means paying at least 40 percent, so y is at least 0.4x.
\[ y \ge 0.4 x \]
Read the region at x equal to 70
Why: The slice runs from 0.4 times 70 to 0.6 times 70.
\[ 28\text{ to } 42\text{ dollars} \]
Figure (svg): The solution to Worked example the shoe sale shown as a ladder of expressions, one row per algebraic move
\[ 28 \leq y \leq 42 \]
Verify: check both endpoints against the advertised discounts
Why: Forty-two dollars off a 70 dollar item is a discount of 28 dollars, which is 40 percent — the smallest advertised discount. Twenty-eight dollars is a discount of 42, which is 60 percent — the largest. Both endpoints correspond exactly to the two stated limits, which confirms the inequalities were set up the right way round.
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.3 Graph Systems of Linear Inequalities §3.3, pp. 170-170
Real world
A recipe needs at least twice as much flour as sugar, and no more than 900 grams of the two combined.
Discussion prompt
Write the system including any constraints the situation implies, describe the region, and give one combination that works and one that does not. What shape is the region?
Hint: Two of the constraints are ones the problem never states.
Answer:
\[ f \geq 2s, \quad f + s \leq 900, \quad f \geq 0, \quad s \geq 0 \]
The region is a triangle in the first quadrant with corners at the origin, (900, 0) and (600, 300). A working combination is 500 grams of flour and 200 of sugar: 500 is at least 400, and the total is 700.
A failing one is 400 flour and 300 sugar, since 400 is not at least 600. Note that the non-negativity constraints are never stated in the problem and are essential — without them the region would extend absurdly into negative quantities.
Worked example
A heart-rate style problem, in the shape of Exercise 39.
\[ \text{A target zone runs from } 60\% \text{ to } 85\% \text{ of a maximum rate of } 220 \text{ minus age. Find the zone at age } 40. \]
Write the maximum rate as an expression in age
Why: Two hundred and twenty minus the age, in beats per minute.
\[ \max = 220 - a \]
Write the two constraints
Why: The target rate r is at least 0.60 of the maximum and at most 0.85 of it.
\[ 0.60(220 - a) \le r \le 0.85(220 - a) \]
Substitute the age
Why: At 40 the maximum is 180 beats per minute.
\[ 220 - 40 = 180 \]
Compute the two endpoints
Why: Sixty percent of 180 is 108, and 85 percent is 153.
\[ 108\text{ to } 153 \]
Figure (svg): The solution to Worked example a second pair of constraints shown as a ladder of expressions, one row per algebraic move
\[ 108 \leq r \leq 153 \]
Verify: check the direction and the width
Why: Both endpoints are below the maximum of 180, as they must be since both percentages are under 100. And the zone spans 45 beats, which is 25 percent of 180 — matching the gap between 60 and 85 percent. Both checks confirm the arithmetic and the setup.
Error analysis
A student models a sale of 40 to 60 percent off.
Annotate
On: \( 0.40x \leq y \leq 0.60x \quad \text{where } y \text{ is the sale price} \)
Percent off and percent paid are complements, not the same number. Write down which one the problem states before converting it into a coefficient.
Matching
Each phrase becomes one constraint.
Match the pairs
Why: At least gives an at-least symbol and at most gives an at-most one, in every case. The third is the one worth care: at least twice as much flour as sugar means f is at least 2s, not 2f at least s — reading it as a sentence about flour, with sugar doubled on the other side, gets it right.
Estimation
Sale prices between 40 and 60 percent off, on an item regularly 90 dollars.
Predict first
Roughly what is the range of possible sale prices?
Correct: About 36 to 54 dollars.
\[ 0.4(90) = 36 \qquad 0.6(90) = 54 \]
Why: Forty percent off leaves 60 percent, which is 54 dollars; 60 percent off leaves 40 percent, which is 36. The band runs from 36 to 54. The third option halves both figures, and the fourth adds rather than subtracts the discount — a sale price above the regular price should be immediately implausible, which is why estimating the direction first is worth the second it takes.
Socratic
One question, and nothing else on this slide.
\[ 0.4x \leq y \leq 0.6x \]
Discussion prompt
The shaded region shows every price pair the advertisement permits. Does it tell you what any particular item actually costs? What would you need in addition, and what does that say about the difference between a constraint and a prediction?
Hint: Ask how many points are in a vertical slice.
Answer:
No. A vertical slice at 70 dollars contains every price from 28 to 42, and the region says all of them are permitted — it cannot single one out.
You would need the actual discount applied to that item, which is information the advertisement deliberately does not give. A constraint says what is allowed; a prediction says what will happen, and no system of inequalities can produce one.
This is the same distinction as in Lesson 2.8's DVD problem: the model divides the plane into permitted and forbidden and stops there. Choosing among the permitted points is a separate question.
Comparison
Fill the blanks. The method scales without changing.
Comparison matrix
| Feature | One inequality | A system |
|---|---|---|
| What you graph | one boundary | every boundary |
| What you shade | one half-plane | the overlap of all of them |
| How to test a candidate | substitute once | substitute into every inequality |
| Possible outcome | always a half-plane | a region, a point, or nothing |
| Shape with three constraints | not applicable | often a bounded triangle |
The only structural change is the word every. Each inequality is still handled exactly as in Lesson 2.8.
Pattern
One routine graphs any system of inequalities.
Step four is what catches an unapplied constraint, which usually shows up as a region that is unbounded when it should not be.
Check
A solution must satisfy every inequality.
Check your understanding
Which point is a solution of the system y > -2x - 5 and y <= x + 3?
Answer: A
Why: At the origin, 0 is greater than -5 and 0 is at most 3, so both inequalities hold and the point lies in the overlap.
Check
Parallel boundaries. Check the directions.
Check your understanding
What is the graph of the system 2x + 3y < 6 and y >= -(2/3)x + 4?
Answer: A
Why: Rewriting the first gives y less than -(2/3)x + 2. One region is below a line at intercept 2 and the other is above a parallel line at intercept 4, so they never meet.
Check
An absolute value boundary. Test a point.
Check your understanding
For the system y <= 3 and y > |x + 4|, what shape is the graph?
Answer: A
Why: The V has vertex (-4, 0) and the greater-than shades inside it, opening upward. The line y = 3 cuts that wedge off at height 3, closing it into a triangle.
Real world
A student has at most 15 hours a week for two jobs, must work at least 4 hours at the first, and needs to earn at least 150 dollars. The first pays 12 dollars an hour and the second 9.
Discussion prompt
Write every constraint, including the ones the situation implies but does not state, and describe the region. Then say why the corners of that region are the schedules worth examining.
Hint: There are more than three constraints once you include the implied ones.
Answer:
\[ a + b \leq 15, \quad a \geq 4, \quad 12a + 9b \geq 150, \quad b \geq 0 \]
Four constraints give a bounded region — a small polygon — whose corners are where pairs of boundaries cross. One corner is where a equals 4 meets 12a plus 9b equals 150, giving a equals 4 and b equals about 11.3.
Corners matter because any linear quantity you might want to optimise — total pay, total hours, free time — takes its extreme value at one of them. Checking a handful of corners replaces checking infinitely many schedules, which is what makes the method practical.
Commit first
Answer, then rate your confidence honestly.
Predict first
Can adding one more inequality to a system ever make the solution region larger?
Correct: No — an extra condition can only remove points, never add them.
It also explains why real problems often have redundant constraints: adding one that the others already imply costs nothing and is sometimes clearer to state.
Why: A solution must satisfy every inequality, so a point excluded by the new one is excluded from the system regardless of how it fared before. The region can stay the same, if the new inequality is redundant, or shrink, possibly to nothing — but it can never grow. This monotonicity is worth knowing because it means a region that got bigger when you added a constraint is a graphing error, not a discovery.
Explain it
They can graph one inequality but do not see what changes with two.
Discussion prompt
In four sentences or fewer, explain what the graph of a system is, how to produce it, and the one check that catches a forgotten constraint.
Hint: The check involves testing a point against everything.
Answer:
Graph each inequality on its own, exactly as you would if it were the only one, shading its half-plane. The graph of the system is the part of the plane that ended up shaded by every one of them, because a solution has to satisfy all of them at once.
The check: pick a point inside your final region and substitute it into EVERY inequality. If it fails one, a constraint was drawn or shaded wrongly — and a region that came out unbounded when you expected a closed shape usually means one was left out entirely.
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 shadings, use different colours or hatch in different directions, and mark the overlap separately. For parallel regions, test one point in both inequalities rather than reasoning from the picture. For a V, test the point directly above or below its vertex. For worded constraints, match at least and at most to the symbols and name which quantity each percentage measures. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper, and two colours if you have them. Fifteen minutes.
Draw it
Draw one large coordinate plane and graph a system of two inequalities of your own, shading each in a different colour and outlining the overlap. Beside it, draw a smaller plane showing a system with parallel boundaries that has NO solution, and beneath that one with parallel boundaries whose regions are nested, writing one sentence about what distinguishes them. In the lower half of the page, graph a system of three inequalities that closes into a triangle, and find all three corners by solving the corresponding pairs of boundary equations, marking each corner with its coordinates. Finally, in a margin, write a real pair of constraints from your own life, turn them into inequalities, and sketch the region they allow.
If your triangle's corners were read off the drawing rather than solved for, go back and solve them: a corner is a two-equation system, and reading one off a sketch is exactly the estimate Lesson 3.2 was written to replace.
Recap
Five things, and the first one is the whole method.
| If you see | Then |
|---|---|
| Two inequalities | Shade both; keep the overlap |
| Parallel boundaries | Test a point; three outcomes are possible |
| Three or more inequalities | Expect a bounded region with corners |
| An absolute value inequality | Same method, V-shaped boundary |
| A corner you need exactly | Solve the two boundary equations |
Lesson 3.4 goes the other way: from two variables to three, where each equation is a plane in space and a solution is an ordered triple.
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.3 Graph Systems of Linear Inequalities §3.3, pp. 168-171 — everything on these slides traces back here
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