Compound inequalities joined by and, whose solutions must satisfy both parts. Includes writing them as a single statement with the variable between two bounds, graphing the overlap, solving by separating the parts or by operating on all three expressions at once, and recognising when the overlap is empty.
Subject: Algebra 1 · 65 slides · symbolic lesson
Open the interactive version of this deck
Title
Algebra 1 · Chapter 6 — Solving and Graphing Linear Inequalities
Solving Compound Inequalities Involving “And”
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 6.4 Solving Compound Inequalities Involving “And” §6.4, pp. 342-347 — the lesson these objectives are drawn from
Warm-up
Every inequality so far had one boundary and a solution set stretching to infinity. This lesson bounds it at both ends.
Discussion prompt
Graph x greater than or equal to 0 on one number line and x less than 4 on another. Which numbers appear on both graphs, and what shape is that collection?
Hint: Look for the overlap rather than the union.
Answer:
\[ 0 \le x \;\text{ and }\; x < 4 \;\Longrightarrow\; 0 \le x < 4 \]
The numbers from zero up to but not including four appear on both. That is a bounded segment rather than a ray, which is the shape every compound and inequality produces when it has any solutions at all.
Concept
A compound inequality consists of two inequalities connected by the word and or the word or. A number solves an and compound inequality only if it solves both of the parts.
compound inequality — Two inequalities connected by and or or. With and, a number is a solution only when it satisfies both parts.
The or version is the subject of Lesson 6.5.
Figure (svg): Two one-sided inequalities and the overlap that satisfies both
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.4 Solving Compound Inequalities Involving “And” §6.4, pp. 342-342
Section
Section 1
Concept
When both parts of an and compound inequality point the same way, they can be written as a single statement with the variable in the middle and a bound on each side.
\[ 0 \le x \;\text{ and }\; x < 4 \;\Longleftrightarrow\; 0 \le x < 4 \]
It is read as x is greater than or equal to zero and less than four.
Figure (svg): Two separate inequalities written as one compound statement
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.4 Solving Compound Inequalities Involving “And” §6.4, pp. 342-342 — Example 1, Write Compound Inequalities with And, and its Study Tip
Picture it
One line, two bounds.
Figure (svg): Two separate inequalities written as one compound statement
The combined form makes the shape of the solution obvious: the variable is trapped between two numbers, so the graph must be a segment.
Worked example
This is Example 1 from the textbook.
\[ \text{Write a compound inequality for all real numbers at least } 0 \text{ and less than } 4, \text{ then graph it.} \]
Write each condition separately
Why: Zero is at most x, and x is less than four.
\[ 0 \le x\text{ and } x < 4 \]
Check the two symbols point the same way
Why: Both put x above the first bound and below the second.
Combine into one statement
Why: Put x between the two bounds.
\[ 0 \le x < 4 \]
Graph the overlap
Why: Solid dot at zero, open dot at four, shaded between.
Figure (svg): Two one-sided inequalities and the overlap that satisfies both
\[ 0 \le x < 4 \]
Verify: test the two endpoints
Why: Zero satisfies both parts, since zero is at least zero and less than four, so its dot is solid. Four fails the second part, so its dot is open. The two ends are decided independently, which is why they can differ.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.4 Solving Compound Inequalities Involving “And” §6.4, pp. 342-342
Translation
A lower bound and an upper bound each time.
Match the pairs
Why: At least and at most include the bound; greater than and less than exclude it. Each end is decided by its own phrase, which is why three of these four have one open end and one closed end.
Worked example
Guided Practice 1 to 3. Describing each in words.
\[ \text{Describe } \; -2 < y < 0, \quad -7 \le t < 8, \quad -4 \le n \le 11. \]
Take the first
Why: y is greater than negative two and less than zero.
Take the second
Why: t is at least negative seven and less than eight.
Take the third
Why: n is at least negative four and at most eleven.
Note the pattern
Why: Each end is described by its own symbol.
Figure (svg): The solution to Worked example read three compound inequalities aloud shown as a ladder of expressions, one row per algebraic move
\[ -2 < y < 0, \quad -7 \le t < 8, \quad -4 \le n \le 11 \]
Verify: check how many integers each contains
Why: The first contains only the integer negative one, since negative two and zero are both excluded. The third contains sixteen integers, from negative four to eleven inclusive. Counting the integers is a quick way to confirm that the symbols were read correctly.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.4 Solving Compound Inequalities Involving “And” §6.4, pp. 342-342
Trap
\[ x > 5 \;\text{ and }\; x < 2 \]
Write it as 5 < x < 2
Why: Both parts have been written into one line in the order they were given.
That statement claims five is less than two, which is false. The two conditions do not overlap, so no combined form exists and the answer is that there is no solution.
Leave them separate, and report that no number satisfies both.
Check that the left bound is genuinely below the right one before combining
Why: The combined form asserts an ordering between the two bounds as well as trapping the variable.
A combined form with the larger number on the left is always a warning sign, and it is worth reading the whole statement to see whether it makes sense.
Faded example
Put the variable between the bounds.
Fill in the blanks
0 \le x \;\text0\; x < 4 \;\Longleftrightarrow\; < \le x ___ 4
Why: The lower bound goes on the left with its own symbol and the upper bound on the right with its own. Each half of the combined statement is exactly one of the original two inequalities, which is why the symbols are not required to match.
Elimination
The combined form needs the two parts to point the same way.
Eliminate the wrong options
Which pair cannot be combined?
Survives elimination: A
Why: Writing five less than x less than two would assert that five is below two, which is false, and the two conditions have no overlap at all. When the lower bound is not actually lower, there is no combined form and no solution.
Socratic
The notation looks like it should accept anything.
Discussion prompt
Explain what the statement a less than x less than b claims, beyond the two separate inequalities. Then say why a compound inequality with and does not always have such a form.
Hint: Read the outer two expressions against each other.
Answer:
It claims three things at once: a is less than x, x is less than b, and — implicitly, by the chain — a is less than b. The notation reads as a chain, so the outer two numbers are being compared as well as each being compared with x.
If the two conditions point opposite ways, as in x greater than five and x less than two, the chain would assert five less than two, which is false. There is no combined form because the chain cannot be written truthfully, and that failure is itself the information that no number satisfies both.
Section
Section 2
Concept
To graph a compound and inequality, graph each part separately and keep the numbers covered by both. The result is the overlap of the two rays.
Each endpoint keeps the dot from its own condition.
Figure (svg): Two one-sided inequalities and the overlap that satisfies both
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.4 Solving Compound Inequalities Involving “And” §6.4, pp. 342-342 — the graph accompanying Example 1
Picture it
The bottom line keeps only the shared part.
Figure (svg): Two one-sided inequalities and the overlap that satisfies both
The two rays point at each other, so their overlap is bounded. Rays pointing away from each other would overlap in nothing, which is the empty case.
Worked example
Building the picture in three stages.
\[ \text{Graph } \; -4 < x \le 2. \]
Graph the first part
Why: x greater than negative four: open dot, shading right.
\[ \text{open at } -4 \]
Graph the second part
Why: x at most two: solid dot, shading left.
\[ \text{solid at } 2 \]
Find the overlap
Why: The numbers between negative four and two.
Draw the result
Why: Open at negative four, solid at two, shaded between.
Figure (svg): The solution of a compound and inequality graphed as a bounded segment
\[ -4 < x \le 2 \]
Verify: test the two endpoints and a number outside
Why: Negative four fails the first part and two satisfies the second, which is why one dot is open and the other solid. Testing three shows it fails the upper bound, so it is correctly outside — and testing negative five shows it fails the lower one.
Sorting
It must satisfy both parts of -4 < x at most 2.
Sort into buckets
Sort each number by whether it solves the compound inequality.
The two endpoints land in different buckets, which is the whole point of having two kinds of dot. A number failing either part fails the whole compound inequality.
Worked example
This is Example 2 from the textbook, on Mount Rainier.
\[ \text{Trees grow from } 2000 \text{ to below } 6000 \text{ ft; alpine flowers from } 6000 \text{ to below } 7500; \text{ neither above that to } 14\,410. \]
Write the tree band
Why: At least 2000 and below 6000.
\[ 2000 \le y < 6000 \]
Write the flower band
Why: At least 6000 and below 7500.
\[ 6000 \le y < 7500 \]
Write the bare band
Why: At least 7500 and at most the summit.
\[ 7500 \le y \le 14410 \]
Check the bands fit together
Why: Each band's upper bound is the next one's lower bound.
Figure (svg): A mountain divided into elevation bands by plant life
\[ 2000 \le y < 6000, \; 6000 \le y < 7500, \; 7500 \le y \le 14\,410 \]
Verify: ask which band contains exactly 6000 feet
Why: The first band excludes it and the second includes it, so 6000 feet belongs to the alpine flower band alone. Having one bound open and the next closed is what makes the bands cover every elevation without any of them overlapping.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.4 Solving Compound Inequalities Involving “And” §6.4, pp. 343-343
Error analysis
The student graphed the compound inequality -4 < x at most 2.
Annotate
On: \( \begin{aligned} &\text{open dot at } -4, \text{ shading left} \\ &\text{solid dot at } 2, \text{ shading right} \\ &\text{shade everything covered by either} \end{aligned} \)
The two errors compound each other: rays pointing outward and a union instead of an overlap turn a small bounded segment into everything. Testing one number, such as ten, catches it immediately.
Faded example
Two endpoints, each with its own dot.
Fill in the blanks
A graph open at -4 and solid at 2, shaded between, is the inequality -4 < x <= 2.
Why: An open dot gives a strict symbol and a solid dot an inclusive one, and the two ends are read independently. That is why a single compound inequality can have one of each.
Elimination
Two rays are drawn on the same line.
Eliminate the wrong options
Which part is shaded?
Survives elimination: A
Why: And requires both conditions, so a number must appear on both graphs to survive. Option B is the natural confusion and it is exactly what Lesson 6.5 will do — the connecting word is the only thing that distinguishes the two rules.
Socratic
Each part on its own is unbounded.
Discussion prompt
Explain why the overlap of two rays is a bounded segment when it is not empty. Then say what the overlap looks like if the two rays point the same way.
Hint: Think about which directions the rays point.
Answer:
For the overlap to be non-empty the two rays must point towards each other: one covers everything above a lower bound and the other everything below an upper bound. Their common part is squeezed between those two numbers, so it is bounded at each end — which is why every compound and inequality with solutions gives a segment.
If the two rays point the same way, one contains the other entirely. Then the overlap is the smaller of the two rays, which is still unbounded — for example x greater than one and x greater than three is simply x greater than three. That is a legitimate compound inequality whose combined form does not exist, since the variable is not between two bounds.
Section
Section 3
Concept
One way to solve a compound and inequality is to split it into its two parts, solve each with the methods of Lessons 6.1 to 6.3, and combine the two answers.
This method makes every step an ordinary one-variable inequality.
Figure (svg): A compound inequality solved by separating it and by working on all three parts
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.4 Solving Compound Inequalities Involving “And” §6.4, pp. 343-343 — Example 3, Method 1
Picture it
The left-hand column of the method comparison.
Figure (svg): A compound inequality solved by separating it and by working on all three parts
Nothing in the left column is new. Each half is a one-step inequality from Lesson 6.1, which is why this method is the safer one when the problem is complicated.
Worked example
This is Example 3, Method 1, from the textbook.
\[ \text{Solve } \; -2 < x + 2 \le 4 \; \text{ by separating the parts.} \]
Write the two parts
Why: x plus two is greater than negative two, and at most four.
Solve the first
Why: Subtract two from each side.
\[ x > -4 \]
Solve the second
Why: Subtract two from each side.
\[ x \le 2 \]
Combine the answers
Why: Both must hold.
\[ -4 < x \le 2 \]
Figure (svg): A compound inequality solved by separating it and by working on all three parts
\[ -4 < x \le 2 \]
Verify: test a number inside and one outside
Why: Zero gives two, which is greater than negative two and at most four, so zero is a solution. Three gives five, which exceeds four, so three fails the upper part and is correctly excluded.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.4 Solving Compound Inequalities Involving “And” §6.4, pp. 343-343
Faded example
Two separate one-step solves.
Fill in the blanks
x + 2 > -2 \;\Longrightarrow\; x > -4 \qquad x + 2 \le 4 \;\Longrightarrow\; x \le 2
Why: Each part is solved by subtracting two, giving a lower bound of negative four and an upper bound of two. Combining them gives negative four less than x at most two, with each end keeping its own symbol.
Worked example
The middle expression need not be the same in both parts.
\[ \text{Solve } \; 2x > 6 \;\text{ and }\; x + 1 < 9. \]
Note the two parts differ
Why: One involves 2x and the other x plus one.
Solve the first
Why: Divide by two.
\[ x > 3 \]
Solve the second
Why: Subtract one.
\[ x < 8 \]
Combine
Why: Both must hold.
\[ 3 < x < 8 \]
Figure (svg): The solution to Worked example when separating is clearly better shown as a ladder of expressions, one row per algebraic move
\[ 3 < x < 8 \]
Verify: ask whether the other method was available
Why: The combined form could not have been written at the start, because the two parts contain different expressions. Separating is the only route here, which is why it is worth being fluent in it rather than relying on the shortcut.
Trap
\[ -2 < x + 2 \le 4 \]
Solve the more interesting-looking half and report x <= 2
Why: The upper bound looks like the binding one, so the lower half gets forgotten.
That answer accepts negative ten, and negative ten plus two is negative eight, which is not greater than negative two. Half the condition has been lost.
\[ -4 < x \le 2 \]
Solve both halves and combine, keeping both bounds
Why: And requires both, so the answer needs both ends.
A compound answer with only one bound is a warning sign in itself, since the overlap of two opposing rays is always bounded at both ends.
Elimination
One method is not always available.
Eliminate the wrong options
Which compound inequality cannot be written in combined form?
Survives elimination: A
Why: The combined form requires the same expression in both parts, and here one part has 2x while the other has x plus one. Separating is the only route, which is worth knowing before reaching for the shortcut automatically.
Translation
Both parts, then combine.
Match the pairs
Why: These are Guided Practice 4 to 6 plus Example 3. The first two have the same bounds and different symbols, which is a reminder that the dots are carried through the solving unchanged and are decided by the original statement.
Socratic
One of them often looks redundant.
Discussion prompt
Explain why a compound and inequality's answer must report both bounds, even when one seems obvious. Then say when one bound genuinely does make the other redundant.
Hint: Ask which numbers each bound excludes.
Answer:
Each bound excludes a different set of numbers, so dropping one lets in everything that bound was keeping out. In negative four less than x at most two, dropping the lower bound admits negative ten, which fails the original — the two halves are doing independent work.
One bound makes the other redundant when the two rays point the same way: x greater than one and x greater than three is just x greater than three, since every number above three is already above one. That is a legitimate compound inequality whose answer is a single ray, and it is worth checking for before assuming the answer must be a segment.
Section
Section 4
Concept
The second method keeps the compound inequality in one line and isolates the variable in the middle. To perform any operation, it must be performed on all three expressions.
The textbook's Study Tip states this requirement explicitly.
Figure (svg): An operation applied to all three expressions of a compound inequality
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.4 Solving Compound Inequalities Involving “And” §6.4, pp. 343-343 — Example 3, Method 2, and its Study Tip
Picture it
One on each expression.
Figure (svg): An operation applied to all three expressions of a compound inequality
The middle expression is the one most often forgotten, because it is the one being changed rather than being a bound. Writing all three arrows makes the omission visible.
Worked example
This is Example 3, Method 2, from the textbook.
\[ \text{Solve } \; -2 < x + 2 \le 4 \; \text{ without separating it.} \]
Write the original inequality
Why: Three expressions in one line.
\[ -2 < x + 2 \le 4 \]
Subtract 2 from each expression
Why: All three, including the middle.
Simplify
Why: Negative four, x, and two.
\[ -4 < x \le 2 \]
State the answer
Why: Greater than negative four and at most two.
\[ -4 < x \le 2 \]
Figure (svg): An operation applied to all three expressions of a compound inequality
\[ -4 < x \le 2 \]
Verify: compare with the separated method
Why: Method 1 gave exactly this answer, as it must, since both are valid sequences of steps on the same statement. Working a problem both ways is a genuine self-check, and it costs only a few extra lines.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.4 Solving Compound Inequalities Involving “And” §6.4, pp. 343-343
Faded example
Subtract from each part.
Fill in the blanks
-2 - 2 < x + 2 - 2 \le 4 - 2 \;\Longrightarrow\; -4 < x \le 2
Why: The same two is subtracted from all three expressions, which is what keeps the statement equivalent to the original. Subtracting from only the outer two would change what the middle says and break the equivalence.
Worked example
Guided Practice 5. A coefficient rather than a constant.
\[ \text{Solve } \; -6 \le 3x \le 12. \]
Identify the operation
Why: The middle is multiplied by three.
\[ \text{divide by } 3 \]
Check the divisor's sign
Why: Three is positive, so no reversal.
Divide all three expressions
Why: Negative six over three, 3x over three, twelve over three.
Simplify
Why: Negative two at most x at most four.
\[ -2 \le x \le 4 \]
Figure (svg): The solution to Worked example dividing all three parts shown as a ladder of expressions, one row per algebraic move
\[ -2 \le x \le 4 \]
Verify: test both endpoints
Why: At negative two the middle gives negative six, which the lower bound includes, and at four it gives twelve, which the upper bound includes. Both dots are solid, as the two inclusive symbols require.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.4 Solving Compound Inequalities Involving “And” §6.4, pp. 343-343
Trap
\[ -2 < x + 2 \le 4 \]
Subtract 2 from the outer numbers only: -4 < x + 2 <= 2
Why: The two bounds look like the sides of an inequality, and the middle looks like the answer already.
The middle expression still has its plus two, so the statement no longer says what the original did. Testing zero: it satisfies the original and fails this version, which claims x plus two is at most two.
\[ -4 < x \le 2 \]
Apply every operation to all three expressions
Why: The compound inequality has three parts, and an operation on two of them breaks the equivalence.
Drawing three arrows, one under each expression, before doing any arithmetic makes the omission impossible.
Prediction
Solving -6 < -2x <= 4.
Predict first
What happens to the symbols when you divide all three parts by -2?
Correct: Both symbols reverse.
\[ -6 < -2x \le 4 \;\Longrightarrow\; 3 > x \ge -2 \;\Longleftrightarrow\; -2 \le x < 3 \]
Why: Dividing by a negative reverses every comparison in the statement, and a compound inequality contains two of them. Dividing gives 3 greater than x at least negative two, which is more usually written negative two at most x less than three. Reversing only one symbol would produce a statement whose two bounds are in the wrong order.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| Separate the parts | All three at once | |
|---|---|---|
| Number of lines | more | fewer |
| Available when the parts differ | yes | no |
| Main risk | forgetting one part | forgetting the middle expression |
Each method has one characteristic error and they are different errors. Knowing which one your chosen method invites is worth more than a general resolution to be careful.
Socratic
The middle is the answer, after all.
Discussion prompt
Explain why an operation applied to only the two outer expressions breaks a compound inequality. Then say what the corresponding rule was for equations in Chapter 3.
Hint: Ask what the statement is comparing.
Answer:
The statement compares the middle expression with each bound. Changing a bound without changing the middle changes what is being compared, so the new statement is about a different question — it is no longer equivalent, and its solutions differ from the original's.
For an equation the rule was to do the same thing to both sides, because an equation compares two expressions. A compound inequality compares three, so the rule becomes do the same thing to all three. The principle is identical: every expression involved in the comparison must change together.
Section
Section 5
Concept
If the two conditions of an and compound inequality do not overlap, no number satisfies both and the inequality has no solution. That is a complete answer.
This is the same reading of an impossible statement as in Lesson 3.9.
Figure (svg): Two conditions whose overlap is empty
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.4 Solving Compound Inequalities Involving “And” §6.4, pp. 342-347 — the definition of an and compound inequality and its exercises
Picture it
Nothing is covered twice.
Figure (svg): Two conditions whose overlap is empty
The two shaded stripes never share a number. Drawing them is the fastest way to see that the answer is empty rather than merely small.
Worked example
Two conditions that cannot both hold.
\[ \text{Solve } \; x > 5 \;\text{ and }\; x < 2. \]
Graph the first
Why: Everything above five.
Graph the second
Why: Everything below two.
Look for the overlap
Why: The two stripes never meet.
State the answer
Why: There is no solution.
Figure (svg): Two conditions whose overlap is empty
\[ \text{no solution} \]
Verify: try to write the combined form
Why: It would read five less than x less than two, which asserts five is less than two — a false statement. The impossibility of writing the combined form is the algebra reporting the same thing the graph shows.
Sorting
Look for an overlap.
Sort into buckets
Sort each by whether any number satisfies both parts.
The third and fourth items differ only in their symbols and land in different buckets. When the two bounds coincide, whether the solution set is a single point or empty is decided entirely by the dots.
Worked example
The conflict is not always visible at the start.
\[ \text{Solve } \; 2x + 1 > 11 \;\text{ and }\; x + 3 < 7. \]
Solve the first
Why: Subtract one, divide by two.
\[ x > 5 \]
Solve the second
Why: Subtract three.
\[ x < 4 \]
Compare the two answers
Why: Above five and below four cannot both hold.
State the answer
Why: No solution.
Figure (svg): The solution to Worked example an emptiness that only appears after solving shown as a ladder of expressions, one row per algebraic move
\[ x > 5 \;\text{ and }\; x < 4: \;\text{ no solution} \]
Verify: check a number that satisfies one part
Why: Six satisfies the first part, since thirteen is greater than eleven, and fails the second, since nine is not less than seven. Every number satisfying one part fails the other, which is what an empty overlap means.
Trap
\[ x > 5 \;\text{ and }\; x < 2 \;\Longrightarrow\; 5 < x < 2 \]
Write the combined form and hand it in
Why: The notation accepts any two numbers, so the statement can be written down.
Read aloud it says x is greater than five and less than two, which no number is. Writing it does not make it a solution set.
There is no solution: no number is both above five and below two.
Check that the left bound is below the right one before combining
Why: If it is not, the overlap is empty and that is the answer.
Saying no solution is a complete answer, exactly as it was for the equations in Lesson 3.9.
Elimination
The parts are x greater than 5 and x less than 2.
Eliminate the wrong options
Which statement is correct?
Survives elimination: A
Why: An empty solution set is a legitimate answer to a legitimate question. Option C is the one worth naming: the number zero and the empty set are different objects, and confusing them is a habit worth breaking early.
Hypothesis
Predict before you check.
Predict first
Which compound inequality has exactly one solution?
Correct: x >= 3 and x <= 3.
\[ x \ge 3 \;\text{ and }\; x \le 3 \;\Longleftrightarrow\; x = 3 \]
Why: A number at least three and at most three must be exactly three, so the solution set is the single point three. The other three all exclude three from at least one side, so nothing survives and each has no solution. The two bounds coincide in all four, and only the dots decide whether anything is left.
Socratic
Adding a condition never makes things easier.
Discussion prompt
Explain what joining two conditions with and does to the collection of solutions, compared with either condition alone. Then predict what or will do, before you meet it in the next lesson.
Hint: Think about whether a condition can add solutions.
Answer:
Requiring both conditions can only remove solutions. Every solution of the compound inequality is a solution of each part, so the set is contained in both — it is at most as large as the smaller of the two, and often much smaller.
Or should do the opposite: a number needs only one of the conditions, so every solution of either part survives and the set is at least as large as the larger of the two. That is why and gives bounded segments and or gives spread-out pairs of rays, which is the shape difference Lesson 6.5 is built around.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| Simple inequality | Compound with and | |
|---|---|---|
| Number of conditions | one | two, both required |
| Shape of the graph | a ray | a bounded segment, or empty |
| Expressions in a solving step | two | three |
Each row is a consequence of the first. Two conditions bound the set at both ends, and three expressions appear because the variable now sits between two bounds.
Pattern
Whether the compound inequality arrives combined or as two statements, the same five moves cover it.
Step five's check should use a number that fails only one of the two parts, since that is what tests whether both bounds survived the solving.
OpenStax Intermediate Algebra 2e, §2.6 Solve Compound Inequalities §2.6
Check
Both parts must hold.
Check your understanding
Which number solves -4 < x <= 2?
Answer: A
Why: Two is greater than negative four and at most two, so it satisfies both parts. The upper bound is inclusive, which is why the endpoint itself counts.
Check
Every operation, three times.
Check your understanding
Solve 1 < x + 3 < 7.
Answer: A
Why: Subtracting three from all three expressions gives negative two less than x less than four. Testing zero gives three, which is between one and seven, and zero is between negative two and four — consistent.
Check
Check whether the rays meet.
Check your understanding
What is the solution of x > 6 and x < 1?
Answer: A
Why: No number is both above six and below one, so the two conditions have no overlap and the solution set is empty. Writing it as a combined form would assert that six is less than one, which is false.
Real world
A recipe says the oven should be between 180 and 200 degrees Celsius. Your oven's dial is marked in Fahrenheit, and the conversion is F equals nine fifths C plus 32.
Discussion prompt
Write the temperature requirement as a compound inequality in Celsius, convert it to Fahrenheit by operating on all three expressions, and state the range on the dial.
Hint: Apply each step of the conversion to all three parts.
Answer:
\[ 180 \le C \le 200 \;\xrightarrow{\times \frac{9}{5}}\; 324 \le \tfrac{9}{5}C \le 360 \]
\[ \;\xrightarrow{+32}\; 356 \le F \le 392 \]
So the dial should read between 356 and 392 degrees Fahrenheit. Both operations were applied to all three expressions, and both multipliers were positive, so neither symbol reversed at any stage.
Notice that the width of the range changed: twenty Celsius degrees became thirty-six Fahrenheit degrees, because the multiplier of nine fifths stretched the interval. A compound inequality's bounds move independently, and the distance between them is scaled by whatever the middle expression is multiplied by.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
Solving -2 < x + 2 <= 4, what must be subtracted from what?
Correct: 2 from all three expressions.
\[ -2 - 2 < x + 2 - 2 \le 4 - 2 \;\Longrightarrow\; -4 < x \le 2 \]
Why: A compound inequality compares the middle expression with each bound, so every expression involved in those comparisons has to change together. Subtracting from the outer two only would leave x plus two in the middle, making the statement say something different — and testing zero would then reject a genuine solution. This is the same principle as doing the same thing to both sides of an equation, extended from two expressions to three.
Explain it
They can solve a single inequality and have just met one with two symbols in it.
Discussion prompt
In no more than four sentences, explain what a compound inequality with and means and how to solve one. Then tell them the one thing that goes wrong most often.
Hint: Two conditions, both required.
Answer:
A usable answer: it is two conditions at once, and a number counts only if it passes both — so the graph is the part where the two shaded regions overlap, which is a segment with a bound at each end. To solve it you can split it into two ordinary inequalities and solve each, or keep it as one line and do every step to all three expressions.
The thing that goes wrong is forgetting the middle. If you subtract from the two outer numbers and leave the middle alone, the statement no longer means what it did — so write the operation under all three parts before you start.
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 three expressions are fixed by writing the operation under all three before doing any arithmetic. The dots are fixed by testing each endpoint against its own half of the statement. Recognising emptiness is fixed by checking whether the lower bound is genuinely below the upper one. The combined form is fixed by asking whether the same expression appears in both parts. 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
At the top of a page write two one-sided inequalities that overlap, graph each on its own number line, and beneath them draw a third line showing only the overlap, labelling it with the combined compound inequality. In the middle, solve one compound inequality twice: once by separating it into two parts and solving each, and once by operating on all three expressions, boxing both answers to show they match. Beside the second method, draw three small arrows under the three expressions to record that every step was applied to all of them. In the lower half, write two conditions that do not overlap, graph both, and write beside the picture why the answer is no solution and what the impossible combined form would have claimed. Finally, in the margin, write one compound inequality whose solution is a single point and say what makes it so.
Your two boxed answers must match, and your single-point example should have both bounds equal and both dots solid. Changing either dot to open turns it into a no-solution case, which is worth trying once to see.
Recap
Five things, and the third is where the middle expression gets lost.
| If the question says | Your first move is |
|---|---|
| Between two values | Write a compound inequality with and |
| Solve -2 < x + 2 <= 4 | Subtract 2 from all three expressions |
| Graph the compound inequality | Graph both parts and keep the overlap |
| The bounds are in the wrong order | Report no solution |
| The two parts have different expressions | Separate them and solve each |
Lesson 6.5 changes the connecting word. With or, a number needs only one of the two conditions, which turns the overlap into a union and the bounded segment into a pair of rays spreading outwards.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.4 Solving Compound Inequalities Involving “And” §6.4, pp. 342-347 — everything on these slides traces back here
Want this taught 1-on-1? Alexander tutors Algebra 1 — $55/session, free consultation.