Two or more linear inequalities in the same variables, whose solutions are the ordered pairs satisfying every one of them. Includes graphing each half-plane as in Lesson 6.8 and reading the overlap, handling three or more conditions, testing a point against every inequality, and recognising a bounded region and its corners.
Subject: Algebra 1 · 65 slides · symbolic lesson
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Title
Algebra 1 · Chapter 7 — Systems of Linear Equations and Inequalities
Systems of Linear Inequalities
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. 7 Systems of Linear Equations and Inequalities — Lesson 7.6 Systems of Linear Inequalities §7.6, pp. 424-430 — the lesson these objectives are drawn from
Warm-up
Lesson 6.8 shaded one half-plane at a time. This lesson shades two and keeps only what they share.
Discussion prompt
Graph x plus y less than 3 and x minus 4y at most 0 on the same plane. Which points satisfy both, and what shape do they form?
Hint: Look for where the two shadings overlap.
Answer:
Each inequality shades half the plane, and the points satisfying both lie where the two shadings overlap — a wedge bounded by the two lines.
The answer is a region rather than a point, because each condition still leaves infinitely many possibilities. Requiring both narrows the plane down without pinning it to a single place.
Concept
Two or more linear inequalities in the same variables form a system of linear inequalities. A solution is an ordered pair that satisfies every inequality in the system, and the graph of the system is the overlap of the half-planes.
system of linear inequalities — Two or more linear inequalities in the same variables. A solution is an ordered pair satisfying every one of them, and the solution set is the intersection of their half-planes.
This is the and condition of Lesson 6.4, one dimension up.
Figure (svg): Two half-planes overlapping, with the overlap shaded darker
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.6 Systems of Linear Inequalities §7.6, pp. 424-424
Section
Section 1
Concept
A solution of a system of linear inequalities is an ordered pair that solves each inequality in the system. The graph of the system is where all the half-planes overlap.
This is the same and requirement as a system of equations, applied to regions.
Figure (svg): Two columns relating a system of inequalities to compound and inequalities
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.6 Systems of Linear Inequalities §7.6, pp. 424-424 — the definitions of a system of linear inequalities and its solutions
Picture it
The same rule, a different picture.
Figure (svg): Two columns relating a system of inequalities to compound and inequalities
Both columns keep the overlap. Moving from a number line to a plane changes the answer from a segment to a region and leaves the logic untouched.
Worked example
Every inequality has to hold.
\[ \text{Is } (0, 0) \text{ a solution of } \; y < 2, \; x \ge -1, \; y > x - 2? \]
Test the first
Why: Zero is less than two.
Test the second
Why: Zero is at least negative one.
Test the third
Why: Zero is greater than negative two.
Conclude
Why: All three hold, so the origin is a solution.
Figure (svg): One point tested against every inequality of a system
\[ (0, 0) \text{ is a solution} \]
Verify: find a point that fails one condition
Why: The pair (0, 5) satisfies the second and third and fails the first, since five is not less than two. One failure is enough, so it is not a solution — which is what makes the overlap smaller than any single half-plane.
Sorting
The system is y < 2, x at least -1, and y > x - 2.
Sort into buckets
Sort each point by whether it solves the whole system.
Three of the six qualify, and each rejection is caused by a different one of the three conditions. Every inequality is doing real work in shaping the region.
Worked example
Satisfying one inequality is not enough.
\[ \text{Is } (5, 0) \text{ a solution of } \; x + y < 3 \; \text{ and } \; x - 4y \le 0? \]
Test the first
Why: Five plus zero is five, which is not less than three.
Stop there
Why: One failure rejects the pair.
Test the second anyway
Why: Five minus zero is five, which is not at most zero.
State the verdict
Why: It fails both, and failing one would have been enough.
Figure (svg): The solution to Worked example a point in only one region shown as a ladder of expressions, one row per algebraic move
\[ (5, 0): \; \text{Ineq 1} \;\times \]
Verify: find a point failing only one
Why: The pair (0, -1) satisfies x plus y less than three and fails x minus 4y at most zero, since zero plus four is four. It lies in one shaded region and not the other, which is exactly the case the overlap excludes.
Trap
\[ \text{Is } (0, -1) \text{ a solution of the system?} \]
Check the first inequality, find it holds, and answer yes
Why: One true statement looks like confirmation.
The second inequality gives four, which is not at most zero, so the pair fails. Every inequality is a condition, and a point in one half-plane but not the other is not in the overlap.
Check every inequality, stopping only at a failure
Why: The system joins its inequalities with and, so all of them are required.
This is the same discipline as checking both equations of a system in Lesson 7.1, and the same reason applies.
Faded example
One failure rejects the point.
Fill in the blanks
At (0, 0): y < 2 is true, x >= -1 is true, and y > x - 2 gives 0 > -2, which is true.
Why: All three conditions hold, so the origin lies in the overlap and is a solution. Had any one of them failed, the point would have been rejected however well the others held.
Elimination
Two or more conditions on the same variables.
Eliminate the wrong options
Which statement is correct?
Survives elimination: A
Why: The system joins its inequalities with and, so every one is a condition. Option C is worth noticing because the crossing points do matter — they are the corners of the region — but they are a feature of the answer rather than the answer itself.
Socratic
A system of equations gave a single point.
Discussion prompt
Explain why replacing the equations of a system with inequalities turns a point into a region. Then say what would happen if you used one equation and one inequality.
Hint: Ask what each condition's own solution set looks like.
Answer:
An equation's solutions form a line, so two lines meet in a point. An inequality's solutions form a half-plane — a two-dimensional set — so two half-planes meet in a two-dimensional region. Each condition is weaker, so what survives is bigger.
One equation and one inequality would give the part of the line lying inside the half-plane, which is a ray or a segment. That is the intermediate case: one condition cuts to a line and the other trims it, so the answer is one-dimensional rather than a point or a region.
Section
Section 2
Concept
Graph each inequality exactly as in Lesson 6.8 — boundary line dashed or solid, then the correct half-plane shaded — and take the region shaded by both.
Each inequality is handled on its own before anything is combined.
Figure (svg): The three steps of graphing a system of inequalities
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.6 Systems of Linear Inequalities §7.6, pp. 425-425 — the Graphing a System of Linear Inequalities summary
Picture it
Two familiar, one new.
Figure (svg): The three steps of graphing a system of inequalities
The first two steps are Lesson 6.8 repeated once per inequality. Only the third is new, and it is a matter of reading rather than of drawing.
Worked example
This is Example 1 from the textbook.
\[ \text{Graph } \; x + y < 3 \; \text{ and } \; x - 4y \le 0. \]
Graph the first boundary
Why: x plus y equals three, dashed because the symbol is strict.
Shade for the first
Why: Testing the origin gives zero, which is less than three, so shade the origin's side.
Graph the second boundary
Why: x minus 4y equals zero, solid because the symbol includes or equal to.
Shade and combine
Why: The solution is the overlap of the two shadings.
Figure (svg): Two half-planes overlapping, with the overlap shaded darker
\[ \text{the intersection of the two half-planes} \]
Verify: test a point in the overlap
Why: The origin satisfies the first, and zero minus zero is zero which is at most zero, so it satisfies the second too — it lies on that boundary. A point on a solid boundary is a solution, which is why the line style matters.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.6 Systems of Linear Inequalities §7.6, pp. 424-424
Sorting
The symbol decides each boundary.
Sort into buckets
Sort each inequality by the kind of boundary line it needs.
The decision is made once per inequality and is entirely independent of which side gets shaded. Both have to be got right for the region to be correct.
Worked example
The origin cannot be the test point here.
\[ \text{Shade } \; x - 4y \le 0 \; \text{ when its boundary passes through the origin.} \]
Check the origin
Why: Zero minus zero is zero, which is on the boundary.
Choose another point
Why: The point (0, 1) is off the line.
\[ \text{test } (0, 1) \]
Substitute it
Why: Zero minus four is negative four, which is at most zero.
Shade its side
Why: The half-plane containing (0, 1) is the answer.
Figure (svg): The solution to Worked example a boundary through the origin shown as a ladder of expressions, one row per algebraic move
\[ (0, 1): \; -4 \le 0 \;\checkmark \]
Verify: test a point on the other side
Why: The pair (0, -1) gives four, which is not at most zero, so it is correctly outside. Testing on both sides once confirms the shading as well as the boundary, which is worth doing when the origin was unavailable.
Error analysis
The student graphed a system of two inequalities.
Annotate
On: \( \begin{aligned} &\text{graph both boundary lines} \\ &\text{shade the correct half-plane for each} \\ &\text{report everything shaded by either as the solution} \end{aligned} \)
The union is the rule for an or condition, which Lesson 6.5 used. A system joins its inequalities with and, so the overlap is what survives.
Faded example
Two of them are Lesson 6.8.
Fill in the blanks
Graph each boundary, shade the correct half-plane for each, and take the intersection of the half-planes as the solution of the system.
Why: The intersection is the region satisfying every inequality, which is what a system requires. Taking the union instead would be the rule for an or condition and would produce a much larger region.
Elimination
Two half-planes are shaded on one plane.
Eliminate the wrong options
Which part is the solution of the system?
Survives elimination: A
Why: Both conditions must hold, so only the doubly shaded region qualifies. Option B is the natural confusion with Lesson 6.5's or, and choosing it would report a region several times too large.
Socratic
They could all be drawn on one plane at once.
Discussion prompt
Explain what is gained by shading each half-plane on its own before combining them. Then say what makes a combined drawing hard to read.
Hint: Think about what happens with three or four shadings.
Answer:
Each inequality gets its own boundary decision and its own test point, and doing them one at a time keeps those decisions separate. On a single plane the shadings overlap in several ways, and it becomes difficult to tell a region shaded twice from one shaded three times.
With three or more inequalities a combined drawing has many overlapping areas, and the eye cannot reliably count layers of shading. Drawing them separately and then identifying the common region is slower and far more reliable, which is what the textbook's Example 2 does explicitly.
Section
Section 3
Concept
A system may have three or more inequalities. Each is graphed separately and the solution is the region satisfying all of them, which gets smaller as conditions are added.
Three conditions often produce a bounded region such as a triangle.
Figure (svg): Three half-planes shaded separately before being combined
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.6 Systems of Linear Inequalities §7.6, pp. 425-425 — Example 2, Graph a System of Three Linear Inequalities
Picture it
Then read what all three cover.
Figure (svg): Three half-planes shaded separately before being combined
Drawing them apart makes the combining step a matter of comparison rather than of squinting at overlapping colours, which is what makes three conditions manageable.
Worked example
This is Example 2 from the textbook.
\[ \text{Graph } \; y < 2, \quad x \ge -1, \quad y > x - 2. \]
Graph the first
Why: The half-plane below the dashed line y equals two.
\[ \text{below } y = 2 \]
Graph the second
Why: The half-plane on and to the right of the solid line x equals negative one.
\[ \text{right of } x = -1 \]
Graph the third
Why: The half-plane above the dashed line y equals x minus two.
Combine
Why: The solution is the region satisfying all three.
Figure (svg): Three half-planes overlapping in a triangular region
\[ \text{the intersection of the three half-planes} \]
Verify: test a point inside and one just outside
Why: The origin satisfies all three, so it is inside. The point (0, 3) fails the first, since three is not less than two, so it is outside — and it satisfies the other two, which is exactly how a single condition trims the region.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.6 Systems of Linear Inequalities §7.6, pp. 425-425
Prediction
A region satisfies two inequalities, and a third is added.
Predict first
What happens to the solution region?
Correct: It shrinks, or stays the same.
Adding enough conditions can shrink the region to nothing at all, which is the empty case.
Why: Every point must now pass an extra test, so points can only be lost and never gained. If the new condition happens to be satisfied everywhere in the existing region, nothing is lost and it stays the same — which happens when the new condition is redundant. This is the same shrinking as adding a condition to a compound and inequality in Lesson 6.4.
Worked example
Guided Practice 3. Two axis conditions and one slanted.
\[ \text{Graph } \; x \ge 0, \quad y \ge 0, \quad 2x + 3y \le 12. \]
Graph x at least zero
Why: The half-plane on and right of the vertical axis.
Graph y at least zero
Why: The half-plane on and above the horizontal axis.
Graph the third
Why: On and below the line through (6, 0) and (0, 4).
Combine
Why: A triangle with corners at the origin, (6, 0) and (0, 4).
Figure (svg): A bounded solution region with its corner points marked
\[ \text{corners } (0, 0), (6, 0), (0, 4) \]
Verify: find the corners from the boundaries
Why: The intercepts of 2x plus 3y equals twelve are six and four, and the two axes meet at the origin. Each corner is where two of the three boundaries cross, which is how corners are found in general.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.6 Systems of Linear Inequalities §7.6, pp. 425-425
Trap
\[ y < 2, \quad x \ge -1, \quad y > x - 2 \]
Graph the first two, see a sensible region and stop
Why: Two conditions already produce a recognisable shape, so the third looks like a refinement.
The third condition cuts the region off below the slanted line. Without it the region extends downwards forever, which is a different answer entirely.
Graph every inequality and take the region common to all of them
Why: Each condition is doing work, and omitting one enlarges the answer.
Testing a point that the omitted condition would exclude is the way to catch this — here, any point well below the slanted line.
Faded example
Each corner is where two boundaries meet.
Fill in the blanks
For x >= 0, y >= 0 and 2x + 3y <= 12, the corners are (0, 0), (6, 0) and (0, 4).
Why: The two intercepts of 2x plus 3y equals twelve give two corners, and the two axes cross at the origin for the third. Finding corners is finding where pairs of boundaries meet, which is a system of two equations each time.
Elimination
A region is bounded by three lines.
Eliminate the wrong options
How do you find the coordinates of a corner exactly?
Survives elimination: A
Why: A corner is where two boundary lines cross, so it is the solution of the system formed by their two equations — which the earlier lessons of this chapter solve exactly. That connects the two halves of the chapter directly.
Socratic
The shape is not an accident.
Discussion prompt
Explain why three half-planes typically bound a triangular region. Then say what would have to be true for three conditions to leave an unbounded region instead.
Hint: Count the sides of the region.
Answer:
Each inequality contributes one straight boundary, and a region bounded by exactly three straight edges is a triangle. So if all three conditions are doing work and the region is closed, three edges is what you get.
The region is unbounded if the three half-planes do not close it off — for instance if all three boundaries are parallel, or if the shadings all point outwards in a way that leaves an escape in some direction. The three inequalities of Example 2 do close it, while y less than two on its own leaves everything below open forever.
Section
Section 4
Concept
To check a region, pick a point inside it and confirm every inequality holds, and pick one outside and confirm at least one fails. Some systems have no solution at all.
An empty region is a complete answer, exactly as in Lesson 6.4.
Figure (svg): Two half-planes that do not overlap
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.6 Systems of Linear Inequalities §7.6, pp. 424-430 — the definition of a solution and the exercises on systems of inequalities
Picture it
Two conditions with no common ground.
Figure (svg): Two half-planes that do not overlap
The two half-planes are parallel and point away from each other, so nothing is shaded twice. The system has no solution, and saying so is the answer.
Worked example
One point inside and one outside.
\[ \text{Check the region for } \; y < 2, \; x \ge -1, \; y > x - 2. \]
Choose a point inside
Why: The origin looks well inside.
\[ (0, 0) \]
Test every condition
Why: Zero is below two, at least negative one, and above negative two.
Choose a point outside
Why: The point (-3, 0) is left of the vertical boundary.
\[ (-3, 0) \]
Test it
Why: Negative three is not at least negative one, so it fails.
Figure (svg): One point tested against every inequality of a system
\[ (0, 0) \;\checkmark \qquad (-3, 0) \;\times \]
Verify: say which condition rejected the outside point
Why: The second one, x at least negative one, was the only one it failed — it satisfies the other two. Naming which condition rejects a point tells you which boundary you are outside, which is useful when a region has been drawn wrongly.
Discrimination
The system is x at least 0, y at least 0, and 2x + 3y at most 12.
Sort into buckets
Sort each point by whether it lies in the solution region.
Worked example
Two conditions that cannot both hold.
\[ \text{Graph } \; x + y < 1 \; \text{ and } \; x + y > 10. \]
Graph the first
Why: The half-plane below the dashed line x plus y equals one.
Graph the second
Why: The half-plane above the dashed line x plus y equals ten.
Look for the overlap
Why: The two boundaries are parallel and the shadings point apart.
Conclude
Why: No point satisfies both, so the system has no solution.
Figure (svg): Two half-planes that do not overlap
\[ \text{no solution} \]
Verify: read the two conditions together
Why: The expression x plus y would have to be both less than one and greater than ten, which no number is. That is the two-variable version of the empty compound and inequality from Lesson 6.4, and it is spotted the same way.
Trap
\[ \text{Is } (0, 5) \text{ in the region for } y < 2, \; x \ge -1, \; y > x - 2? \]
Check that x is at least -1, find it is, and answer yes
Why: One condition held, and checking three feels repetitive.
The first condition fails, since five is not less than two. Every inequality is a test the point has to pass, and stopping at a success proves nothing.
Test every inequality, stopping only when one fails
Why: A single failure settles it; a single success settles nothing.
This is the same search strategy as for a compound and condition: look for a failure, not for a success.
Faded example
It must fail at least one condition.
Fill in the blanks
At (5, 3): 2(5) + 3(3) = 19, which is not at most 12, so the point is outside the region.
Why: The point satisfies both axis conditions and fails the slanted one, which is enough to put it outside. Naming the condition that rejected it tells you which boundary the point is on the wrong side of.
Hypothesis
Predict before you decide.
Predict first
Which system has no solution?
Correct: x + y < 1 and x + y > 10.
The two boundaries are parallel and the shadings point away from each other, so nothing is shaded twice.
Why: The expression x plus y would have to be both below one and above ten, which is impossible. The second option is the same pair with the bounds the other way round, and it gives a band between the two parallel lines — a perfectly ordinary non-empty region. The last two give quarter-planes and unbounded regions respectively.
Socratic
Each condition on its own has plenty of solutions.
Discussion prompt
Explain how two conditions each with infinitely many solutions can have none in common. Then say what the boundaries look like when that happens.
Hint: Ask what each half-plane leaves out.
Answer:
Each half-plane contains infinitely many points and excludes infinitely many others. If what one excludes contains everything the other includes, the overlap is empty — the sets are large and disjoint, which is entirely possible for two halves of a plane.
It happens when the two boundaries are parallel and the shadings point away from each other, so the two half-planes lie on opposite sides of a gap. If the shadings pointed towards each other, they would overlap in a band, so the directions decide it — exactly as the ray directions did in Lesson 6.5.
Section
Section 5
Concept
Real problems often impose several conditions at once. The solution region is the set of combinations satisfying all of them, and its corners are usually the interesting cases.
A budget or capacity limit produces a bounded region.
Figure (svg): A bounded solution region with its corner points marked
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.6 Systems of Linear Inequalities §7.6, pp. 424-430 — Exercises 34 to 36 on ordering spotlights for a theatre
Picture it
Two axes and one budget line.
Figure (svg): A bounded solution region with its corner points marked
Every point in the triangle is a combination you could actually choose. Its corners are the extreme options: none of one, none of the other, or the whole budget spent.
Worked example
A theatre orders two kinds of spotlight, as in Exercises 34 to 36.
\[ \text{Spotlights cost } 2 \text{ and } 3 \text{ hundred dollars. The budget is } 12 \text{ hundred. Model the possible orders.} \]
Name the variables
Why: Numbers of each kind of spotlight.
Write the budget constraint
Why: Two x plus three y is at most twelve.
\[ 2 x + 3 y \le 12 \]
Write the non-negativity conditions
Why: Neither count can be negative.
\[ x \ge 0, y \ge 0 \]
Describe the region
Why: A triangle with corners at the origin, (6, 0) and (0, 4).
Figure (svg): A bounded solution region with its corner points marked
\[ 2x + 3y \le 12, \; x \ge 0, \; y \ge 0 \]
Verify: read three points out of the region
Why: The orders (6, 0), (0, 4) and (3, 2) all satisfy every condition, and the last spends exactly twelve hundred as well. Reading concrete combinations out of a region is what makes it an answer rather than a picture.
Matching
Each restriction becomes one condition.
Match the pairs
Why: Budget limits give at-most conditions and minimum requirements give at-least ones. The non-negativity conditions come from the situation rather than from any sentence, which is why they are the ones most often left out.
Worked example
Each corner is an extreme case.
\[ \text{Interpret the corners } (0, 0), \; (6, 0) \text{ and } (0, 4). \]
Take the origin
Why: Buying nothing at all.
Take (6, 0)
Why: Six of the cheaper kind and none of the dearer.
Take (0, 4)
Why: Four of the dearer kind and none of the cheaper.
Note what they share
Why: Each is where two constraints are met exactly.
Figure (svg): A bounded solution region with its corner points marked
\[ (0, 0), \; (6, 0), \; (0, 4) \]
Verify: check which constraints are tight at each corner
Why: At (6, 0) the budget is exactly spent and y is exactly zero, so two constraints hold with equality. That is what makes a corner a corner, and it is why corners are found by solving pairs of boundary equations.
Trap
\[ 2x + 3y \le 12 \]
Report the whole half-plane below the line as the answer
Why: The budget is the interesting constraint, so it looks like the whole problem.
The region extends into negative numbers of spotlights, which cannot be ordered. The conditions x at least zero and y at least zero are part of the model even though the problem does not state them.
\[ 2x + 3y \le 12, \quad x \ge 0, \quad y \ge 0 \]
Add a non-negativity condition for every quantity that cannot be negative
Why: The situation imposes them even when the wording does not.
This is the same restriction as in Lessons 4.4 and 6.8, now written as explicit inequalities rather than added as a remark.
Faded example
One inequality per restriction.
Fill in the blanks
\text2 2 \text3 3 \text___ 12: \quad ___x + ___y \le 12, \; x \ge 0, \; y \ge 0
Why: Each coefficient is a cost per item and each variable a count, so the left side is in hundreds of dollars and matches the budget on the right. The two non-negativity conditions come from the situation and complete the model.
Elimination
The constraints are 2x + 3y at most 12, with x and y at least 0.
Eliminate the wrong options
Which order can the theatre place?
Survives elimination: A
Why: Six plus six is twelve, exactly the budget, and both counts are non-negative — so the order sits on the budget boundary and is permitted by the at-most symbol. The other three each fail one of the three conditions.
Socratic
The region contains infinitely many valid combinations.
Discussion prompt
Say why the corners of a feasible region tend to matter more than the points inside it. Then say what kind of question would be answered by a corner rather than by the region as a whole.
Hint: Ask what is true at a corner that is not true inside.
Answer:
Inside the region no constraint is tight — you could always buy a little more of something. At a corner two constraints are met exactly, so no further improvement is possible in either direction. Corners are the extreme options, and extremes are usually what a decision comes down to.
A question asking for the most of something, or the cheapest way to meet a requirement, is answered at a corner rather than in the interior. That observation is the foundation of linear programming, which later courses develop — and the region drawn here is exactly the object it works with.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| System of equations (7.1) | System of inequalities (7.6) | |
|---|---|---|
| Each condition's graph | a line | a half-plane |
| The solution set | usually one point | usually a region |
| How to find it | solve algebraically or read the crossing | shade each and read the overlap |
Weakening each condition from an equation to an inequality raises the answer's dimension from a point to a region, and the and rule is unchanged throughout.
Pattern
Whether there are two conditions or five, the same five moves cover it.
Step one is where real problems go wrong most often, because non-negativity conditions are imposed by the situation rather than written in the wording.
OpenStax Elementary Algebra 2e, §5.6 Graphing Systems of Linear Inequalities §5.6
Check
Every inequality must hold.
Check your understanding
Is (2, 1) a solution of the system y < 2, x >= -1 and y > x - 2?
Answer: A
Why: One is less than two, two is at least negative one, and one is greater than zero, which is two minus two. All three conditions hold, so the point lies in the region.
Check
And, not or.
Check your understanding
What is the graph of a system of two linear inequalities?
Answer: A
Why: A solution has to satisfy both inequalities, so it must lie in both half-planes. The region shaded twice is the answer.
Check
The situation imposes conditions too.
Check your understanding
Modelling a purchase of two items with a budget, which condition is easiest to forget?
Answer: A
Why: The budget is stated explicitly in the problem and the non-negativity conditions are not, so they are the ones that get left out. Without them the region extends into negative counts, which cannot be ordered.
Real world
This is the situation in Exercises 34 to 36. A theatre is ordering two kinds of spotlight, costing 200 and 300 dollars each, with a budget of 1200 dollars, and it needs at least two of the cheaper kind.
Discussion prompt
Write the full system, describe the region, and name three orders it permits. Then say which corner represents the largest total number of spotlights.
Hint: Four conditions: a budget, a minimum, and two non-negativity conditions.
Answer:
\[ 200x + 300y \le 1200, \quad x \ge 2, \quad y \ge 0 \]
Dividing the budget condition by a hundred gives 2x plus 3y at most twelve. The region is bounded on the left by x equals two, below by the horizontal axis, and above by the budget line, giving a triangle with corners at (2, 0), (6, 0) and (2, 8/3).
Permitted orders include six of the cheaper and none of the dearer, three of each, and two cheaper with two dearer. The corner (6, 0) gives six spotlights in total, the most of any corner — which is what you would expect, since the cheaper kind buys more units per dollar. That kind of question is always answered at a corner rather than inside the region.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
Two half-planes are shaded on one plane. Which part is the solution of the system?
Correct: The part shaded by both inequalities.
A point in one shading and not the other satisfies one inequality and fails the other, so it is not a solution.
Why: A system joins its inequalities with and, so a solution must satisfy every one of them and therefore lie in every half-plane. The first option is the union, which is the rule for the or conditions of Lesson 6.5, and choosing it would report a region several times too large — including points that fail one of the two conditions outright. Testing such a point in both inequalities settles it in one line.
Explain it
They can shade one inequality and are unsure what to do with two.
Discussion prompt
In no more than four sentences, explain how to graph a system of inequalities and which part is the answer. Then tell them the check that catches the commonest mistake.
Hint: Shade each, then keep what is shared.
Answer:
A usable answer: graph each inequality exactly as you already do — draw the boundary dashed or solid, test a point, shade the right side. Do that for every inequality on the same plane, and then the answer is the part that ended up shaded by all of them, not by any of them.
To check, pick a point in the region you have shaded and put it into every inequality: all of them should come out true. If one comes out false, you have kept a part that only some of the conditions cover — which is the mistake almost everyone makes first.
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 overlap is fixed by testing a point from your shaded region in every inequality. Boundary styles are fixed by checking each symbol separately before drawing. Three conditions are fixed by shading each on its own plane first and then comparing. Implied constraints are fixed by asking which quantities cannot be negative and writing those conditions down. 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 three small coordinate planes across the top of a page and on each graph one inequality of a three-inequality system, labelling every boundary as dashed or solid and writing the test point and verdict beside each. Underneath, draw one larger plane and shade the region satisfying all three, marking its corners. Beside it, write two test points — one inside the region and one outside — with the verdict of every inequality for each. In the lower half, invent a real purchase or capacity problem with two quantities, write every constraint as an inequality including the ones the situation implies rather than states, graph the region, and mark and interpret each corner in a sentence. Finally, in the margin, write down two inequalities whose region is empty and say what their boundaries look like.
Your inside point should make every inequality true and your outside point should fail at least one. If the outside point passes all of them, the region you shaded is too small; if the inside point fails one, it is too large or in the wrong place.
Recap
Five things, and the first is the one that separates and from or.
| If the question says | Your first move is |
|---|---|
| Graph the system of inequalities | Graph each one separately first |
| Which region is the solution | The part shaded by all of them |
| Is this point a solution | Test it in every inequality |
| Find the corners | Solve pairs of boundary equations |
| A real purchase or capacity problem | Add the non-negativity conditions |
That completes Chapter 7. Chapter 8 leaves lines behind and returns to exponents, building the rules for multiplying and dividing powers that Chapter 9's quadratic work will need.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.6 Systems of Linear Inequalities §7.6, pp. 424-430 — everything on these slides traces back here
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