Simple and compound inequalities, their graphs with open and solid endpoints, the transformations that preserve them, the rule that reverses the symbol when you multiply or divide by a negative, and modelling a range of acceptable values.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 1 — Equations and Inequalities
Solve Linear Inequalities
Objectives
Five outcomes. The third is the only genuinely new rule in the whole lesson.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.6 Solve Linear Inequalities §1.6, pp. 41-47 — the lesson these objectives are drawn from
Warm-up
Everything from Lesson 1.3 carries over, with one exception you are about to meet.
Discussion prompt
Solve 4x plus 9 equals 25. Now solve 4x plus 9 is less than 25. What was identical about the two, and what is the only thing that changed?
Hint: Compare the moves you made, not the answers you got.
Answer:
\[ 4x + 9 = 25 \;\Longrightarrow\; 4x = 16 \;\Longrightarrow\; x = 4 \]
\[ 4x + 9 < 25 \;\Longrightarrow\; 4x < 16 \;\Longrightarrow\; x < 4 \]
The moves are identical: subtract nine, divide by four. What changed is the answer — a single number becomes an entire ray of numbers. That is the whole difference, until you divide by a negative.
Concept
An equation asks which value makes the two sides equal. An inequality asks which values make one side smaller, and the answer is usually infinitely many of them. That is why the answer is drawn on a number line rather than written as a single value.
linear inequality in one variable — An inequality that can be written as a linear expression compared to zero, using one of the four symbols for less than, greater than, at most, or at least.
The graph is not an illustration of the answer. For an inequality, the graph IS a complete statement of the answer, and it says one thing the symbols do not say as loudly: whether the endpoint is included.
Figure (svg): Two number lines: one graphing x less than 2 with an open dot, one graphing x at least negative one with a solid dot
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.6 Solve Linear Inequalities §1.6, pp. 41-41
Section
Section 1
Concept
The endpoint of the graph is the boundary value. Draw it open when the inequality is strict, and solid when the inequality allows equality. That single choice is the difference between two different answer sets.
solution of an inequality — A value that, when substituted for the variable, makes the inequality a true statement. The graph is the set of all of them.
The four symbols come in two pairs: strict ones, which get open dots, and inclusive ones, which get solid dots. Nothing else about the drawing changes.
Figure (svg): Two number lines: one graphing x less than 2 with an open dot, one graphing x at least negative one with a solid dot
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.6 Solve Linear Inequalities §1.6, pp. 41-41 — Graph simple inequalities
Picture it
Example 1: x less than 2, and x at least negative one.
Figure (svg): Two number lines: one graphing x less than 2 with an open dot, one graphing x at least negative one with a solid dot
Test the endpoint itself. Is 2 less than 2? No, so the dot is open. Is negative one greater than or equal to negative one? Yes, so the dot is solid. That test never fails.
Worked example
Example 1, both parts. The reasoning for the dot is written out each time.
\[ \text{Graph } x < 2 \text{ and } x \geq -1. \]
Locate the boundary value for each
Why: Two for the first, negative one for the second. The boundary is whatever number the variable is compared with.
Test whether the boundary itself is a solution
Why: Is 2 less than 2? No. Is -1 greater than or equal to -1? Yes.
Draw the dots accordingly
Why: Open where the boundary fails the test, solid where it passes.
\[ \text{open at } 2,\text{ solid at } -1 \]
Shade the side that contains the solutions
Why: Less than means everything to the left; greater than means everything to the right.
\[ \text{left of } 2;\text{ right of } -1 \]
Figure (svg): The solution to Worked example graph two simple inequalities shown as a ladder of expressions, one row per algebraic move
\[ x < 2 \qquad \text{and} \qquad x \geq -1 \]
Verify: test one point from each shaded region and one from outside
Why: For the first: 0 is shaded and 0 is less than 2, correct; 5 is unshaded and 5 is not less than 2, correct. For the second: 3 is shaded and 3 is at least -1, correct; -4 is unshaded and -4 is not, correct. Testing one point on each side of the boundary confirms the shading direction.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.6 Solve Linear Inequalities §1.6, pp. 41-41
Sorting
Decide the endpoint style for each without drawing anything.
Sort into buckets
Sort each inequality by the dot its graph needs.
Two of these are written with the variable on the right. Reversing how a sentence is read never changes whether the endpoint is included.
Worked example
Guided Practice 1 through 4, run in reverse — because tests ask it both ways.
\[ \text{A graph shows a solid dot at } 3 \text{ with everything to the left shaded. Write the inequality.} \]
Read the boundary value off the line
Why: The dot sits at 3, so 3 is the number the variable is compared with.
\[ \text{boundary is } 3 \]
Read the direction from the shading
Why: Everything to the left is shaded, and left means smaller, so the symbol is less than.
\[ x < 3\text{ so far} \]
Read inclusion from the dot style
Why: The dot is solid, so 3 itself is a solution and the symbol must allow equality.
\[ x\text{ is at most } 3 \]
Write it
Why: Less than or equal to three.
\[ x \le 3 \]
Figure (svg): The solution to Worked example read a graph back into symbols shown as a ladder of expressions, one row per algebraic move
\[ x \leq 3 \]
Verify: substitute the endpoint and one point on each side
Why: Three itself satisfies at most three, matching the solid dot. Two is shaded and satisfies it; four is unshaded and does not. All three tests agree with the drawing, so the translation is right.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.6 Solve Linear Inequalities §1.6, pp. 41-41
Trap
\[ \text{Graph } x > 4. \]
Draw a solid dot at 4 and shade right
Why: The dot style is chosen from where the boundary is rather than from whether it is included.
The graph now claims that 4 is a solution. Substituting gives four is greater than four, which is false.
\[ \text{Graph } x > 4. \]
Test the boundary before choosing the dot
Why: Four is not greater than four, so the boundary is excluded and the dot must be open.
Open dot at four, shading to the right. The answer set starts immediately past four but never includes it.
\[ 4.0001 > 4 \quad \checkmark \qquad 4 > 4 \quad \times \]
Matching
Four symbols, four ways they are usually spoken.
Match the pairs
Why: The everyday phrasings are worth learning because word problems use them rather than symbols. At most and at least are the two inclusive ones, and both appear constantly in tolerance and budget problems. More than and less than are the strict pair, and in ordinary speech they genuinely do exclude the boundary, which is why the translation is reliable.
Prediction
Commit before you reason.
Predict first
How many numbers are in the solution set of x is less than 2 but not in the solution set of x is at most 2?
Correct: None — the second set contains the first, plus the single value 2.
This is why the dot matters: the two answer sets differ by a single point out of infinitely many, and yet in a tolerance problem that point is often the whole question — whether a part measuring exactly the limit passes inspection.
Why: Every number less than 2 is also at most 2, so nothing in the first set is missing from the second. The difference runs the other way: the second set contains 2 itself, which the first does not. So the two sets differ by exactly one number, and that one number is the entire meaning of the dot style.
Explain it to yourself
An equation's answer fits on one line of a page.
\[ 4x + 9 = 25 \qquad \text{versus} \qquad 4x + 9 < 25 \]
Discussion prompt
Explain in your own words why the second one needs a picture and the first does not. Then say what information the picture carries that the symbols x is less than 4 do not carry as obviously.
Hint: Think about what a reader has to reconstruct from each form.
Answer:
The equation has one solution, so writing it down is a complete answer. The inequality has infinitely many, so the answer is a region, and a region is faster to read as a picture than as a sentence.
What the picture makes obvious is the endpoint and the direction — the two things that get mixed up. A reader glancing at an open dot with left shading knows instantly both which side and whether the edge counts, without parsing a symbol.
Section
Section 2
Concept
A compound inequality is two simple inequalities connected by and or by or. The word decides everything: and means both must hold, which gives one band; or means either is enough, which gives two pieces.
compound inequality — Two simple inequalities joined by the word and or the word or. An and-inequality may be written as a single three-part statement with the variable in the middle.
The three-part form, with the variable between two bounds, is another way of writing an and-inequality. It exists precisely because and-solutions are always a single connected band.
Figure (svg): Two number lines showing an and compound inequality as a single band and an or compound inequality as two separate rays
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.6 Solve Linear Inequalities §1.6, pp. 41-41 — Graph compound inequalities
Picture it
Example 2: negative one less than x less than two, and x at most negative two or x greater than one.
Figure (svg): Two number lines showing an and compound inequality as a single band and an or compound inequality as two separate rays
If a graph is one connected piece it came from an and; if it is two pieces going opposite ways it came from an or. That is a reliable way to check you copied the word correctly.
Worked example
Example 2a. The three-part form is read as two statements at once.
\[ \text{Graph } -1 < x < 2. \]
Split the three-part form into its two statements
Why: The middle term is compared with each end separately, and both must hold.
\[ x > -1 AND x < 2 \]
Mark both boundaries with the right dot style
Why: Both comparisons are strict, so both dots are open.
\[ \text{open at } -1,\text{ open at } 2 \]
Shade only where BOTH conditions hold
Why: To the right of negative one and to the left of two, so the overlap between the two.
\[ \text{shade between } -1\text{ and } 2 \]
Read the result
Why: All real numbers greater than negative one and less than two.
Figure (svg): The solution to Worked example graph an and-inequality shown as a ladder of expressions, one row per algebraic move
\[ -1 < x < 2 \]
Verify: test one point inside and one on each side
Why: Zero is inside and satisfies both conditions. Negative five is outside and fails the first. Three is outside and fails the second. And the endpoints themselves both fail, matching the open dots.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.6 Solve Linear Inequalities §1.6, pp. 41-41
Discrimination
Do not solve. Just say what shape the answer will have.
Sort into buckets
Sort each compound inequality by the shape of its solution set.
Worked example
Example 2b. Two pieces, and the dots may differ between them.
\[ \text{Graph } x \leq -2 \text{ or } x > 1. \]
Graph the first piece on its own
Why: At most negative two: solid dot at negative two, shading left.
\[ \text{solid at } -2,\text{ shade left} \]
Graph the second piece on the same line
Why: Greater than one: open dot at one, shading right.
\[ \text{open at } 1,\text{ shade right} \]
Keep everything that either piece shaded
Why: Or means either condition is enough, so the two shadings are combined rather than overlapped.
Note that the two dots differ
Why: One is solid and one is open, because the two pieces came from different symbols.
\[ \text{solid at } -2,\text{ open at } 1 \]
Figure (svg): The solution to Worked example graph an or-inequality shown as a ladder of expressions, one row per algebraic move
\[ x \leq -2 \; \text{or} \; x > 1 \]
Verify: test a point in each ray and one in the gap
Why: Negative five is in the left ray and satisfies the first condition. Three is in the right ray and satisfies the second. Zero is in the gap and satisfies neither, so it is correctly unshaded. The gap is exactly the numbers that fail both.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.6 Solve Linear Inequalities §1.6, pp. 41-41
Trap
\[ x \leq -2 \; \text{or} \; x > 1 \]
Compress it into a three-part statement
Why: The three-part form looks like a general shorthand for any compound inequality.
\[ 1 < x \leq -2 \quad \text{(meaningless)} \]
This claims x is greater than one and at most negative two at the same time, which no number is.
\[ x \leq -2 \; \text{or} \; x > 1 \]
Leave an or-inequality as two separate statements
Why: The three-part form is shorthand for AND only, because it asserts both comparisons at once.
Or-solutions are two disconnected pieces, and no single three-part statement can describe two disconnected pieces.
\[ -1 < x < 2 \;\Longleftrightarrow\; x > -1 \;\text{and}\; x < 2 \]
Translation
Everyday phrasings, translated.
Match the pairs
Why: The word inclusive makes both endpoints solid; between on its own is usually strict, which is why the word is added when it matters. Under and over both exclude, giving an or with open dots. No more than and no less than are both inclusive despite the word no, which is the one phrasing worth reading twice.
Elimination
The inequality is x at most negative two, or x greater than one.
Eliminate the wrong options
Which description of the graph is correct?
Survives elimination: A
Why: At most negative two gives a solid endpoint shading left, and greater than one gives an open endpoint shading right. Or combines the two shadings rather than intersecting them, so the gap between negative two and one is left unshaded — and testing zero confirms it satisfies neither condition.
Counterexample
A classmate offers a shortcut.
\[ \text{every or-inequality has two separate pieces} \]
Discussion prompt
Find a pair of simple inequalities joined by or whose graph is NOT two separate pieces, and say what has to be true about the two conditions for the pieces to merge.
Hint: Try two conditions whose shaded regions overlap.
Answer:
\[ x < 4 \;\text{or}\; x > 1 \;\Longrightarrow\; \text{every real number} \]
When the two shaded regions overlap, combining them produces one piece, and if they cover the whole line the answer is all real numbers. The two-pieces picture only applies when the regions are disjoint — which is common but not guaranteed.
The same caution applies to and: if the two regions do not overlap at all, the answer is empty rather than a band.
Section
Section 3
Concept
Adding and subtracting never disturb an inequality. Multiplying or dividing by a positive number never disturbs it either. Multiplying or dividing by a negative number reflects the whole number line, and a reflection swaps left and right — so the symbol has to reverse to stay true.
\[ 2 < 6 \;\xrightarrow{\;\times(-1)\;}\; -2 > -6 \]
The rule is not arbitrary. It is the same fact from Lesson 1.1 that negating a pair of numbers reverses their order, applied to both sides of an inequality at once.
Figure (svg): A table of the transformations that produce equivalent inequalities, with the two that reverse the symbol highlighted
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.6 Solve Linear Inequalities §1.6, pp. 42-42 — Transformations That Produce Equivalent Inequalities
Picture it
Two points, and what negating does to them.
Figure (svg): A number line showing that multiplying both sides of an inequality by a negative reflects the numbers across zero and reverses their order
Two was to the left of six. After negation, negative two is to the right of negative six. Nothing about the numbers changed except which side of zero they are on, and that alone reverses the comparison.
Worked example
Example 4. Everything is ordinary until the last line.
\[ \text{Solve } 5x + 2 > 7x - 4 \text{ and graph the solution.} \]
Subtract 7x from each side
Why: Subtraction never affects the symbol, so it stays as it is.
\[ -2 x + 2 > -4 \]
Subtract 2 from each side
Why: Again a subtraction, so again the symbol is untouched.
\[ -2 x > -6 \]
Divide each side by -2 and REVERSE the symbol
Why: This is the only step where the symbol changes, and it changes because the divisor is negative.
\[ x < 3 \]
Graph it
Why: Strict inequality, so an open dot at three, shading left.
\[ \text{open dot at } 3,\text{ shade left} \]
Figure (svg): The solution to Worked example a variable on both sides, ending in a reversal shown as a ladder of expressions, one row per algebraic move
\[ x < 3 \]
Verify: test a value inside and a value outside
Why: At x equal to 0 the original reads 2 greater than -4, which is true, and 0 is in the solution set. At x equal to 5 it reads 27 greater than 31, which is false, and 5 is outside. Testing on both sides of the boundary is what confirms the direction after a reversal.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.6 Solve Linear Inequalities §1.6, pp. 43-43
Sorting
For each move, say whether the inequality symbol reverses.
Sort into buckets
Sort each transformation.
The first item is the one that catches people: adding a negative number is addition, and addition never flips.
Worked example
Guided Practice 6. The reversal happens once, and nothing after it flips again.
\[ \text{Solve } 1 - 3x \geq -14 \text{ and graph the solution.} \]
Subtract 1 from each side
Why: The symbol is unaffected by subtraction.
\[ -3 x \ge - 15 \]
Divide each side by -3 and REVERSE
Why: Negative fifteen divided by negative three is positive five, and the at-least becomes an at-most.
\[ x \le 5 \]
Graph it
Why: The symbol allows equality, so the dot at five is solid, shading left.
\[ \text{solid dot at } 5,\text{ shade left} \]
Figure (svg): The solution to Worked example a reversal in the middle, not at the end shown as a ladder of expressions, one row per algebraic move
\[ x \leq 5 \]
Verify: test the endpoint and a point on each side
Why: At x equal to 5 the original reads 1 minus 15, which is -14, and -14 is greater than or equal to -14 — true, so the solid dot is right. At x equal to 0 it reads 1 at least -14, true. At x equal to 10 it reads -29 at least -14, false. All three agree with the graph.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.6 Solve Linear Inequalities §1.6, pp. 43-43
Trap
\[ x - 7 < 4 \]
Add 7 to both sides and reverse the symbol, because there is a negative in the problem
Why: The reversal is triggered by the sight of a minus sign rather than by the operation performed.
\[ x > 11 \quad \text{(wrong)} \]
Testing zero: zero minus seven is negative seven, which is less than four, so zero IS a solution — but the wrong answer excludes it.
\[ x - 7 < 4 \]
Add 7 to both sides; the symbol is untouched
Why: Addition and subtraction never reverse an inequality, no matter what signs are lying around.
\[ x < 11 \]
The reversal is triggered by multiplying or dividing by a negative, never by a negative merely being present. Adding a negative is still addition.
Error analysis
A student solves an inequality and gets an answer that excludes zero. One line is wrong.
Annotate
On: \( \begin{aligned} 3 - x &> x - 9 \\ 3 - 2x &> -9 \\ -2x &> -12 \\ x &> 6 \end{aligned} \)
After any solve, substitute one easy value - zero is usually easiest - and check it lands on the side your answer claims.
Commit first
Answer, then rate your confidence honestly.
Predict first
You solve an inequality and the last step is to divide both sides by x. Is that allowed?
Correct: No — you do not know the sign of x, so you cannot know whether to flip.
\[ x^2 > x \;\xrightarrow{\;\div x\;}\; x > 1 \quad \text{fails at } x = -2 \]
At x equal to negative two the original reads 4 greater than -2, which is true, but the divided version reads -2 greater than 1, which is false. One division destroyed a whole family of solutions.
Why: The reversal rule depends entirely on whether the multiplier is positive or negative. A letter could be either, so dividing by it forks the problem into two separate cases rather than producing one answer. This is why every technique in this lesson gets the variable out of the multiplier position first, by collecting variable terms, rather than dividing by a variable. Chapter 8 handles the general case properly.
Fill the middle
Guided Practice 7. One step has been erased.
Fill in the blanks
5x - 7 \leq 6x \;\longrightarrow\; -7 \leq x \;\longrightarrow\; x \geq -7
Why: Subtracting 5x from both sides gives negative seven at most x. No multiplication or division by a negative happened, so nothing reversed — the final line simply reads the same statement from the other end, which is a rewrite rather than a flip. Reading an inequality backwards is always allowed and is not the same operation as reversing it.
Section
Section 4
Concept
A three-part and-inequality is solved by applying the same move to all three parts at once, keeping the variable in the middle. An or-inequality is two separate problems whose answers are then reported together.
\[ -4 < 6x - 10 \leq 14 \]
If a step in a three-part inequality multiplies or divides by a negative, both symbols reverse together — the whole statement turns round.
Figure (svg): Two number lines showing an and compound inequality as a single band and an or compound inequality as two separate rays
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.6 Solve Linear Inequalities §1.6, pp. 43-43
Picture it
Whatever you started with, the answer keeps the same shape.
Figure (svg): Two number lines showing an and compound inequality as a single band and an or compound inequality as two separate rays
An and-problem cannot produce two pieces, and an or-problem cannot produce a band unless the pieces overlapped. That is a useful check on your answer.
Worked example
Example 5. Every move is applied to all three parts.
\[ \text{Solve } -4 < 6x - 10 \leq 14 \text{ and graph the solution.} \]
Add 10 to each of the three parts
Why: Not just the middle: all three, so the statement stays balanced.
\[ 6 < 6 x \le 24 \]
Divide each of the three parts by 6
Why: Six is positive, so neither symbol reverses.
\[ 1 < x \le 4 \]
Read the two dot styles from the two symbols
Why: The left comparison is strict, the right one is inclusive, so the dots differ.
\[ \text{open at } 1,\text{ solid at } 4 \]
Graph the band
Why: All real numbers greater than one and at most four.
\[ \text{band from } 1\text{ to } 4 \]
Figure (svg): The solution to Worked example an and-compound inequality shown as a ladder of expressions, one row per algebraic move
\[ 1 < x \leq 4 \]
Verify: test both endpoints and one interior point
Why: At x equal to 4 the middle is 14, and 14 is at most 14 — true, so the solid dot is right. At x equal to 1 the middle is -4, and -4 is not greater than -4 — false, so the open dot is right. At x equal to 2 the middle is 2, comfortably inside both bounds.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.6 Solve Linear Inequalities §1.6, pp. 43-43
Ranking
Solving negative eight at most negative x minus five at most six, from Guided Practice 10.
Put in order
Why: Adding five clears the constant from the middle without touching the symbols. Multiplying by negative one is the only step that reverses, and it reverses BOTH symbols at once because the whole statement is being reflected. Rewriting from smallest to largest is cosmetic but makes the band easy to read, and the check confirms the direction survived the reversal.
Worked example
Example 6. Two separate solves, reported together.
\[ \text{Solve } 3x + 5 \leq 11 \text{ or } 5x - 7 \geq 23 \text{ and graph the solution.} \]
Solve the first inequality on its own
Why: Subtract five, then divide by three; both are ordinary moves with no reversal.
\[ 3 x \le 6,\text{ so } x \le 2 \]
Solve the second inequality on its own
Why: Add seven, then divide by five; again no reversal.
\[ 5 x \ge 30,\text{ so } x \ge 6 \]
Report both, joined by or
Why: A solution of the compound statement is a solution of either part.
\[ x \le 2\text{ or } x \ge 6 \]
Graph both pieces on one line
Why: Solid dots at both endpoints, shading outward in opposite directions.
Figure (svg): The solution to Worked example an or-compound inequality shown as a ladder of expressions, one row per algebraic move
\[ x \leq 2 \;\text{ or }\; x \geq 6 \]
Verify: test a point in each ray and one in the gap
Why: At x equal to 0 the first part reads 5 at most 11 — true, so 0 is a solution. At x equal to 10 the second reads 43 at least 23 — true. At x equal to 4 the first gives 17 at most 11, false, and the second gives 13 at least 23, also false, so 4 correctly sits in the unshaded gap.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.6 Solve Linear Inequalities §1.6, pp. 43-43
Error analysis
A student solves a three-part inequality and the answer is far too wide.
Annotate
On: \( -4 < 6x - 10 \leq 14 \;\Longrightarrow\; -4 < 6x \leq 14 \;\Longrightarrow\; -\tfrac{2}{3} < x \leq \tfrac{7}{3} \)
The check is one substitution: pick a value your answer allows and confirm the middle expression really does land between the two outer numbers.
Comparison
Fill the blanks. The structural difference is the point.
Comparison matrix
| Feature | AND compound | OR compound |
|---|---|---|
| Can be written three-part | yes | no |
| Shape of the graph | one connected band | two pieces (usually) |
| How you solve it | same move on all three parts | solve each part separately |
| A value in the gap or outside | fails at least one part | fails both parts |
| Degenerate case | can be empty | can be all real numbers |
The last row is the pair of edge cases worth having met once, so that an empty answer or an everything answer does not look like a mistake.
Prediction
Commit before you solve.
Predict first
Solving 3x - 1 < -1 or 2x + 5 is at least 11 will produce what shape?
Correct: Two rays with a gap between them: x less than 0, or x at least 3.
\[ 3x - 1 < -1 \;\Longrightarrow\; x < 0 \qquad 2x + 5 \geq 11 \;\Longrightarrow\; x \geq 3 \]
Why: The first part gives 3x less than 0, so x is less than 0. The second gives 2x at least 6, so x is at least 3. The two regions do not overlap, so the answer is two rays with a gap from 0 to 3. Predicting the shape before solving is a real check: an or-problem that came out as a single narrow band would mean a sign error somewhere.
Explain it
A classmate keeps applying moves to only the middle part.
Discussion prompt
In three sentences, explain why every move in a three-part inequality has to be applied to all three parts, using the two-sided rule they already accept for equations. Give them a one-substitution check they can run afterwards.
Hint: A three-part statement is really two statements sharing a term.
Answer:
A three-part inequality is two inequalities that happen to share a middle expression, so a move applied to the middle is a move applied to one side of each of them. To keep both true you must apply it to the other side of each as well, which is the two outer parts.
The check: take any value their answer allows, substitute it into the middle expression, and confirm the result really does lie between the two outer numbers of the ORIGINAL statement.
Section
Section 5
Concept
Real constraints are rarely equations. Budgets, tolerances, safe operating ranges and eligibility rules all say at most, at least or between, and each of those translates into an inequality with a specific dot style.
\[ 18 \leq C \leq 34 \]
Once the range is written, any conversion applied to the quantity is applied to all three parts, which is how a Celsius range becomes a Fahrenheit range.
Figure (svg): A thermometer style band showing the monitor lizard temperature range in Celsius converted to the equivalent Fahrenheit range
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.6 Solve Linear Inequalities §1.6, pp. 44-44 — Write and use a compound inequality
Picture it
Example 7: a monitor lizard's temperature ranges from 18 to 34 degrees Celsius.
Figure (svg): A thermometer style band showing the monitor lizard temperature range in Celsius converted to the equivalent Fahrenheit range
The conversion formula is substituted for the middle term, and then the same moves that solve any three-part inequality convert the bounds.
Worked example
Example 7. The Celsius range is given; the Fahrenheit range is wanted.
\[ \text{A lizard's temperature ranges from } 18\degree\text{C to } 34\degree\text{C. Give the range in } \degree\text{F.} \]
Write the Celsius range as a compound inequality
Why: From and to are inclusive here, so both endpoints belong to the range.
\[ 18 \le C \le 34 \]
Substitute the conversion formula for C
Why: Celsius equals five ninths of the quantity F minus 32, which is the formula rearranged for C.
\[ 18 \le(\frac{5}{9}) (F - 32) \le 34 \]
Multiply all three parts by 9/5
Why: Nine fifths is positive, so neither symbol reverses. Nine fifths of 18 is 32.4 and of 34 is 61.2.
\[ 32.4 \le F - 32 \le 61.2 \]
Add 32 to all three parts
Why: The last move, and again the symbols are untouched.
\[ 64.4 \le F \le 93.2 \]
Figure (svg): The solution to Worked example convert a temperature range shown as a ladder of expressions, one row per algebraic move
\[ 64.4 \leq F \leq 93.2 \]
Verify: convert one endpoint back independently
Why: Take 64.4 Fahrenheit: five ninths of 64.4 minus 32 is five ninths of 32.4, which is exactly 18 degrees Celsius — the lower bound we started from. The upper bound checks the same way, and the fact that the Fahrenheit range is wider than the Celsius one is expected, since a Fahrenheit degree is smaller.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.6 Solve Linear Inequalities §1.6, pp. 44-44
Real world
A lift is rated for a maximum load of 630 kilograms. The car itself weighs 180 kilograms and the average passenger is 70 kilograms.
Discussion prompt
Write an inequality for the number of passengers the lift may carry, solve it, and say what the answer means in practice. What does the dot style tell you about a lift carrying exactly the number you found?
Hint: Maximum means the limit itself is allowed.
Answer:
\[ 180 + 70p \leq 630 \;\Longrightarrow\; 70p \leq 450 \;\Longrightarrow\; p \leq 6.43 \]
At most about 6.43 passengers — so at most six, since passengers come whole. Six is allowed because the symbol is inclusive and six is comfortably under the limit; the fractional bound simply tells you seven would exceed it.
This is a case where the algebra gives a range and the situation narrows it. The inequality's inclusive endpoint matters at 6.43 only in principle; what matters in practice is that the whole numbers up to 6 all satisfy it.
Worked example
Guided Practice 13. Same method, different bounds.
\[ \text{A lizard's temperature ranges from } 15\degree\text{C to } 30\degree\text{C. Give the range in } \degree\text{F.} \]
Write the range and substitute the formula
Why: The same conversion applies; only the bounds change.
\[ 15 \le(\frac{5}{9}) (F - 32) \le 30 \]
Multiply all three parts by 9/5
Why: Nine fifths of 15 is 27 and of 30 is 54.
\[ 27 \le F - 32 \le 54 \]
Add 32 to all three parts
Why: Twenty-seven plus 32 is 59, and 54 plus 32 is 86.
\[ 59 \le F \le 86 \]
Figure (svg): The solution to Worked example a narrower range shown as a ladder of expressions, one row per algebraic move
\[ 59 \leq F \leq 86 \]
Verify: check both endpoints through the forward formula
Why: Nine fifths of 15 plus 32 is 27 plus 32, which is 59. Nine fifths of 30 plus 32 is 54 plus 32, which is 86. Both bounds convert forward to exactly the values found, and the range is again wider in Fahrenheit than in Celsius, as it must be.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.6 Solve Linear Inequalities §1.6, pp. 44-44
Error analysis
A club's rule is that members must be at least 13 years old. A student writes the eligibility condition.
Annotate
On: \( a > 13 \quad \text{where } a \text{ is the age in years} \)
In modelling problems the dot style is not a detail - it decides what happens at exactly the limit, which is precisely the case rules are written to settle.
Matching
Each rule is a constraint someone actually wrote down.
Match the pairs
Why: Three of these are inclusive and one is strict, and the wording is what decides it every time. At least, inclusive and no more than all admit the boundary; under excludes it. A package weighing exactly two kilograms fails the second rule, which is the kind of edge case shipping companies write rules to settle.
Estimation
A Celsius range of 18 to 34 degrees was converted to Fahrenheit.
Predict first
Before computing: will the Fahrenheit range be wider or narrower, and roughly by what factor?
Correct: Wider, by about 1.8 times.
\[ 34 - 18 = 16 \qquad 93.2 - 64.4 = 28.8 = \tfrac{9}{5} \cdot 16 \]
Why: A Celsius degree is nine fifths of a Fahrenheit degree, so any temperature interval measured in Fahrenheit contains 1.8 times as many degrees. The Celsius range spans 16 degrees, so the Fahrenheit range should span about 28.8 — and 93.2 minus 64.4 is exactly 28.8. Note that the plus-32 shifts where the range sits but does not change its width at all, which is why the fourth option is wrong.
Definition probe
A shipping rule: packages must weigh at least 0.5 kg and at most 20 kg.
Sort into buckets
Sort each package by how it fares against the rule.
Comparison
Fill the blanks. Every row is decided by two questions: which direction, and is the boundary included?
Comparison matrix
| Symbol | Spoken as | Endpoint |
|---|---|---|
| x < a | less than a | open |
| x <= a | at most a | solid |
| x > a | more than a | open |
| x >= a | at least a | solid |
| a < x < b | strictly between a and b | both open |
The everyday phrasings in the middle column are what word problems actually use, so they are the ones worth over-learning.
Pattern
One routine solves every inequality in this chapter.
Step five is not optional. After a reversal, a single substitution is the only cheap way to confirm the answer points the way you think it does.
OpenStax Algebra and Trigonometry 2e, §2.7 Linear Inequalities and Absolute Value Inequalities §2.7
Check
A reversal is hiding in here. Solve it on paper first.
Check your understanding
Solve 5x + 2 > 7x - 4.
Answer: A
Why: Subtracting 7x gives -2x + 2 > -4, subtracting 2 gives -2x > -6, and dividing by -2 reverses the symbol to give x < 3. Testing x = 0: 2 > -4 is true, and 0 is indeed less than 3.
Check
A three-part inequality. Every move goes on all three parts.
Check your understanding
Solve -4 < 6x - 10 and 6x - 10 is at most 14.
Answer: A
Why: Adding 10 to all three parts gives 6 < 6x at most 24, and dividing all three by 6 gives 1 < x at most 4. The left endpoint is excluded and the right is included, matching the two original symbols.
Check
A modelling question. Read the wording for inclusion.
Check your understanding
A lizard's body temperature ranges from 15 degrees Celsius to 30 degrees Celsius. Using C = (5/9)(F - 32), what is the range in degrees Fahrenheit?
Answer: A
Why: Multiplying all three parts by 9/5 gives 27 at most F minus 32 at most 54, and adding 32 gives 59 at most F at most 86. Checking: nine fifths of 15 plus 32 is 59.
Real world
A bolt is specified as 12.00 mm in diameter with a tolerance of 0.05 mm. A batch is measured and any bolt outside specification is rejected.
Discussion prompt
Write the acceptable diameters as a compound inequality, say whether the endpoints are included and why that matters on a factory floor, and then write the condition for a bolt to be REJECTED as a separate compound inequality.
Hint: The rejection condition is the opposite of the acceptance condition, and opposites turn and into or.
Answer:
\[ 11.95 \leq d \leq 12.05 \quad \text{(accepted)} \]
\[ d < 11.95 \;\text{ or }\; d > 12.05 \quad \text{(rejected)} \]
The endpoints are included because a tolerance of plus or minus 0.05 states the largest deviation that is still acceptable. A bolt measuring exactly 12.05 passes; one at 12.051 does not, which is precisely the kind of edge a measuring instrument has to be able to resolve.
Notice the structural point: negating an and-condition produces an or-condition, and the inclusive endpoints become strict. That swap is worth remembering — it is the same logic that will govern absolute value inequalities in the next lesson.
Commit first
Answer, then rate your confidence honestly.
Predict first
You multiply both sides of an inequality by a negative number twice in a row. What happens to the symbol?
Correct: It reverses twice, so it ends up pointing the way it started.
\[ 2 < 6 \;\xrightarrow{\times(-1)}\; -2 > -6 \;\xrightarrow{\times(-1)}\; 2 < 6 \]
Why: Each multiplication by a negative reflects the line and reverses the symbol, so two of them reflect twice and restore the original order. This is the same reason the product of two negatives is positive, seen from the inequality side. It also explains why multiplying by a positive never reverses: a positive multiplier is equivalent to an even number of reflections, namely zero of them.
Explain it
They can solve equations but keep flipping the inequality symbol at random.
Discussion prompt
In four sentences or fewer, tell them exactly when the symbol reverses and when it does not, explain WHY using the number line, and give them a one-substitution check they can run on any answer.
Hint: The why is one sentence about reflection.
Answer:
The symbol reverses only when you multiply or divide both sides by a negative number. Adding or subtracting anything, and multiplying or dividing by anything positive, leaves it alone — even when the number involved is itself negative.
The reason is that multiplying by a negative reflects the whole number line across zero, and a reflection swaps left and right, which is what the symbol records. The check: substitute zero into the original inequality and see whether it is true, then confirm your answer agrees about zero.
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 dots, substitute the boundary value and see whether it makes the statement true. For reversals, remember that only multiplying or dividing by a negative does it — never adding. For three-part problems, write all three parts on every line so that skipping one is visible. For and against or, sketch the two regions and ask whether you want the overlap or everything covered. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
Draw four short number lines down the left of a page and graph x less than 2, x at most 2, x greater than 2 and x at least 2, labelling each dot open or solid and writing the everyday phrase beside it. To the right, draw two more lines: one showing an and-inequality as a band and one showing an or-inequality as two rays, and write beneath each what a value in the gap does. In the middle of the page write the six transformations, and box the two that reverse the symbol. At the bottom, take a real rule you actually live with — an age limit, a weight limit, a budget — write it as an inequality, graph it, and write one sentence about what happens to someone sitting exactly on the boundary.
That last sentence is the whole reason dot style is worth caring about. If it was easy to write, you have the lesson.
Recap
Five things, and the third is the only rule in this lesson that equations did not already give you.
| If you see | Then |
|---|---|
| A strict symbol | Open dot |
| At least, at most, inclusive | Solid dot |
| The word and | One band; three-part form allowed |
| The word or | Two separate solves |
| Division by a negative | Reverse the symbol |
Lesson 1.7 combines this with distance on the number line: absolute value equations and inequalities, where an and or an or falls out of the symbol itself.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.6 Solve Linear Inequalities §1.6, pp. 41-47 — everything on these slides traces back here
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