Graphing one-variable inequalities on a number line with open and solid endpoints, the addition and subtraction properties of inequality, solving one-step inequalities by adding or subtracting, checking a solution set with numbers inside and outside it, and writing an inequality from a described situation.
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 Inequalities Using Addition or Subtraction
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 6.1 Solving Inequalities Using Addition or Subtraction §6.1, pp. 323-328 — the lesson these objectives are drawn from
Warm-up
Lesson 3.1 solved equations by adding or subtracting the same amount from both sides. This lesson does the same thing to a statement that is not an equation.
Discussion prompt
You know that 3 is less than 7. Add 2 to each side and check whether the statement is still true. Then subtract 10 from each side and check again.
Hint: Draw both pairs on a number line if it helps.
Answer:
\[ 3 < 7 \;\Longrightarrow\; 5 < 9 \qquad 3 < 7 \;\Longrightarrow\; -7 < -3 \]
Both stay true. Adding or subtracting the same amount slides both numbers the same distance along the line, so whichever was on the left is still on the left — which is the entire reason the properties in this lesson work.
Concept
The graph of an inequality in one variable is the set of points on a number line representing all its solutions. A solid dot marks an endpoint that is a solution, an open dot one that is not, and an arrowhead shows the graph continues indefinitely.
graph of an inequality — The set of points on a number line that represent all solutions of an inequality in one variable.
An equation usually had one answer; an inequality has infinitely many, so it is drawn rather than listed.
Figure (svg): Four one-variable inequalities graphed on number lines
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.1 Solving Inequalities Using Addition or Subtraction §6.1, pp. 323-323
Section
Section 1
Concept
Graphing an inequality means marking the endpoint, choosing the right kind of dot, and shading in the direction of the solutions with an arrowhead to show they continue.
Every one-variable inequality has infinitely many solutions, which is why a graph is the natural way to present them.
Figure (svg): Four one-variable inequalities graphed on number lines
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.1 Solving Inequalities Using Addition or Subtraction §6.1, pp. 323-323 — Example 1, Graph an Inequality in One Variable, and its Reading Algebra note
Picture it
Two open endpoints, two solid.
Figure (svg): Four one-variable inequalities graphed on number lines
Reading each graph back into words is worth doing once: all real numbers less than two, all real numbers greater than negative two, and so on. The verbal phrase and the picture carry the same information.
Worked example
This is Example 1 from the textbook.
\[ \text{Describe and graph } \; x < 2, \quad a > -2, \quad z \le 1, \quad d \ge 0. \]
Take x less than 2
Why: All real numbers less than two; an open dot at two, shading left.
\[ \text{open at } 2 \]
Take a greater than -2
Why: All real numbers greater than negative two; an open dot, shading right.
\[ \text{open at } -2 \]
Take z less than or equal to 1
Why: All real numbers less than or equal to one; a solid dot at one.
\[ \text{solid at } 1 \]
Take d greater than or equal to 0
Why: All real numbers greater than or equal to zero; a solid dot at zero.
\[ \text{solid at } 0 \]
Figure (svg): Four one-variable inequalities graphed on number lines
\[ x < 2, \; a > -2, \; z \le 1, \; d \ge 0 \]
Verify: test each endpoint in its own inequality
Why: Substituting two into x less than two gives two less than two, which is false, so the dot is open. Substituting one into z less than or equal to one gives a true statement, so that dot is solid. Testing the endpoint is the reliable way to choose the dot.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.1 Solving Inequalities Using Addition or Subtraction §6.1, pp. 323-323
Sorting
Test whether the endpoint itself is a solution.
Sort into buckets
Sort each inequality by the kind of endpoint its graph has.
The direction of shading plays no part in this sorting, and three of these shade left while three shade right. The two decisions are independent.
Worked example
Guided Practice 1 to 4. Two of each kind.
\[ \text{Describe and graph } \; t > 1, \quad x \le -1, \quad n < 0, \quad y \ge 4. \]
Take t greater than 1
Why: Open dot at one, shading right.
Take x less than or equal to -1
Why: Solid dot at negative one, shading left.
Take n less than 0
Why: Open dot at zero, shading left.
Take y greater than or equal to 4
Why: Solid dot at four, shading right.
Figure (svg): The solution to Worked example four from guided practice shown as a ladder of expressions, one row per algebraic move
\[ t > 1, \; x \le -1, \; n < 0, \; y \ge 4 \]
Verify: check the direction against the symbol
Why: The symbol points towards the smaller side, so greater-than graphs shade right and less-than shades left. Reading the symbol as an arrowhead pointing at the smaller quantity is a memory aid that survives having the variable on either side.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.1 Solving Inequalities Using Addition or Subtraction §6.1, pp. 323-323
Trap
\[ z \le 1 \]
Draw an open dot, since the graph goes off to the left
Why: The arrow at the far end is open in the sense of unbounded, which gets confused with the endpoint's dot.
The dot records whether the endpoint itself is a solution, and one is a solution here since one is less than or equal to one. The direction of shading is a separate question entirely.
\[ z \le 1: \; \text{solid dot at } 1, \text{ shading left} \]
Substitute the endpoint into the inequality and see whether it is true
Why: A true statement means the endpoint belongs, so the dot is solid.
Two independent decisions are being made here — which dot and which direction — and testing settles the first while the symbol settles the second.
Faded example
Describe the solution set in words.
Fill in the blanks
The graph of z <= 1 is all real numbers less than or equal to 1, drawn with a solid dot at 1.
Why: The or equal to part is what makes one itself a solution, and a solution at the endpoint is drawn as a solid dot. Every strict inequality gets an open dot for the opposite reason.
Elimination
Two decisions: the dot and the direction.
Eliminate the wrong options
Which description is right?
Survives elimination: A
Why: Greater than gives an open dot and shades towards the larger numbers. Each distractor changes exactly one of the three decisions, which is why all three parts of the graph — the position, the dot and the direction — have to be checked separately.
Socratic
An equation's answer was written on one line.
Discussion prompt
Explain why the solution of a one-variable inequality is presented as a graph rather than as a list. Then say what the arrowhead adds that the shading alone does not.
Hint: Count the solutions.
Answer:
An equation like x plus five equals seven has one solution, which fits on a line. An inequality like x is greater than two has infinitely many — every number past two, including fractions and irrationals — so no list can give them all. A picture shows the whole set at once, which is the only practical way to present it.
The arrowhead records that the solutions continue past the edge of the drawing rather than stopping there. Without it the picture would suggest the set ends wherever the paper does, which is the same reason Lesson 4.2 insisted on extending a line past its plotted points.
Section
Section 2
Concept
Adding the same number to, or subtracting the same number from, each side of an inequality produces an equivalent inequality — one with the same solutions and the same direction.
Equivalent inequalities have the same solutions, exactly as equivalent equations did in Chapter 3.
Figure (svg): The addition and subtraction properties of inequality stated together
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.1 Solving Inequalities Using Addition or Subtraction §6.1, pp. 324-324 — the Properties of Inequality box and its Study Tip
Picture it
Add or subtract; the sign does not move.
Figure (svg): The addition and subtraction properties of inequality stated together
The Study Tip notes that although the properties are written for strict inequalities, they hold for the or equal to versions too. That is worth remembering, since most exercises use all four symbols.
Worked example
The number line makes the reason visible.
\[ \text{Show that } 3 < 7 \text{ stays true when } 2 \text{ is added to each side.} \]
Mark both numbers
Why: Three is to the left of seven.
\[ 3\text{ left of } 7 \]
Add two to each
Why: Both slide two units right, to five and nine.
\[ 5\text{ and } 9 \]
Compare the new positions
Why: Five is still to the left of nine.
\[ 5 < 9 \]
State the reason
Why: Both moved the same distance, so their order cannot change.
Figure (svg): An inequality on a number line with the same amount added to both sides
\[ 3 < 7 \;\Longrightarrow\; 5 < 9 \]
Verify: try a negative shift
Why: Subtracting ten gives negative seven and negative three, and negative seven is still to the left. The argument never depended on the direction of the slide, only on both numbers moving together — which is why the property covers subtraction as well.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.1 Solving Inequalities Using Addition or Subtraction §6.1, pp. 324-324
Sorting
Start from 3 is less than 7 each time.
Sort into buckets
Sort each operation by whether the inequality stays true as written.
Adding a negative number is still addition, so it is safe. It is multiplying by a negative that causes the reversal, and Lesson 6.2 is entirely about that case.
Worked example
Seeing one failure shows what the properties are actually claiming.
\[ \text{Take } 3 < 7 \text{ and multiply each side by } -1. \text{ Is the result still true?} \]
Multiply each side
Why: Negative three and negative seven.
\[ -3\text{ and } -7 \]
Compare
Why: Negative three is to the right of negative seven.
\[ -3 > -7 \]
Note the reversal
Why: The direction of the inequality has flipped.
Draw the lesson
Why: Not every operation preserves the direction; addition and subtraction do.
Figure (svg): The solution to Worked example an operation that does not preserve the direction shown as a ladder of expressions, one row per algebraic move
\[ 3 < 7 \quad \text{but} \quad -3 > -7 \]
Verify: say what makes the two cases different
Why: Adding slides both numbers the same way, so their order survives. Multiplying by a negative reflects them both through zero, which reverses every order at once. Lesson 6.2 makes this precise; here it is enough to see that the properties in this lesson are a real claim rather than a formality.
Error analysis
The student solved an inequality using the subtraction property.
Annotate
On: \( \begin{aligned} x + 4 &< 9 \\ x + 4 - 4 &< 9 \\ x &< 9 \end{aligned} \)
The property is about doing the same thing to both sides, exactly as in Chapter 3. Checking one number from just outside the reported answer catches a one-sided operation immediately.
Faded example
The same number from each side.
Fill in the blanks
x + 4 < 9 \;\Longrightarrow\; x + 4 - 4 < 9 - 4 \;\Longrightarrow\; x < 5
Why: Subtracting four from both sides leaves x less than five. Doing it to only one side gives x less than nine, which wrongly accepts numbers such as seven — and testing seven in the original catches that immediately.
Elimination
Equivalent inequalities have the same solutions.
Eliminate the wrong options
Which of these is equivalent to x minus 3 greater than 5?
Survives elimination: A
Why: Adding three to both sides gives x greater than eight. Testing nine confirms it: nine minus three is six, which is greater than five. Each distractor fails that test, and one number is enough to reject all three.
Socratic
The word was used for equations in Chapter 3.
Discussion prompt
Say what it means for two inequalities to be equivalent, and why that is exactly what solving needs. Then say what would go wrong if a step produced an inequality that was not equivalent.
Hint: Think about what solving is trying to preserve.
Answer:
Two inequalities are equivalent when they have exactly the same solutions. Solving means replacing a complicated inequality with a simpler one that has the same solution set, so every step has to preserve that set — otherwise the simple answer at the end would describe a different set of numbers.
A non-equivalent step would either add solutions that do not satisfy the original or lose ones that do. Subtracting from only one side does the first, and it is why the answer x less than nine wrongly accepts seven. Checking a number from just outside the reported answer is a direct test of whether that has happened.
Section
Section 3
Concept
To solve a one-step inequality, undo the addition or subtraction attached to the variable by doing the opposite to both sides. Then graph the resulting solution set.
The inequality symbol is carried through every line unchanged.
Figure (svg): An inequality solved in two lines and graphed
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.1 Solving Inequalities Using Addition or Subtraction §6.1, pp. 324-325 — Examples 2 and 3, using subtraction and addition
Picture it
One line of algebra, one picture.
Figure (svg): An inequality solved in two lines and graphed
The answer x greater than or equal to two is a description of infinitely many numbers, so the graph is part of the answer rather than an illustration of it.
Worked example
This is Example 2 from the textbook.
\[ \text{Solve } \; x - 5 \ge -3 \; \text{ and graph the solution.} \]
Write the original inequality
Why: The five is being subtracted from x.
\[ x - 5 \ge - 3 \]
Add 5 to each side
Why: This is the addition property of inequality.
\[ x - 5 + 5 \ge - 3 + 5 \]
Simplify
Why: Negative three plus five is two.
\[ x \ge 2 \]
Graph the answer
Why: Solid dot at two, shading right.
\[ \text{solid at } 2 \]
Figure (svg): An inequality solved in two lines and graphed
\[ x \ge 2 \]
Verify: test a number inside and one outside
Why: Four gives negative one, which is greater than or equal to negative three, so four is a solution. Zero gives negative five, which is not, so zero is correctly excluded. Both tests agree with the graph.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.1 Solving Inequalities Using Addition or Subtraction §6.1, pp. 324-324
Translation
Undo the addition or subtraction.
Match the pairs
Why: Each is one addition or subtraction applied to both sides. The last has the variable on the right, so after solving it reads negative five is less than y, which turned round is y greater than negative five — the symbol reverses when the sides swap.
Worked example
This is Example 3 from the textbook.
\[ \text{Solve } \; -2 > n - 4 \; \text{ and graph the solution.} \]
Write the original inequality
Why: The variable is on the right this time.
\[ -2 > n - 4 \]
Add 4 to each side
Why: The addition property applies as usual.
\[ -2 + 4 > n - 4 + 4 \]
Simplify
Why: Two is greater than n.
\[ 2 > n \]
Rewrite with the variable first
Why: Turning it round reverses the symbol: n is less than two.
\[ n < 2 \]
Figure (svg): An inequality with the variable on the right, read in both directions
\[ 2 > n \;\Longleftrightarrow\; n < 2 \]
Verify: test a number on each side of 2
Why: Zero gives negative four, and negative two is greater than negative four, so zero is a solution. Five gives one, and negative two is not greater than one, so five is not. The graph shades left of two, matching both tests.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.1 Solving Inequalities Using Addition or Subtraction §6.1, pp. 325-325
Trap
\[ 2 > n \]
Rewrite it as n > 2, keeping the symbol as it was
Why: The symbol looks like part of the statement rather than something that depends on the order of the two sides.
That says n is greater than two, which is the opposite set of numbers. The two sides have swapped places, so the symbol has to swap direction with them.
\[ 2 > n \;\Longleftrightarrow\; n < 2 \]
Read the statement aloud after turning it round
Why: Two is greater than n says the same thing as n is less than two.
The textbook's Writing Algebra note makes this point explicitly, and testing one number settles it: zero satisfies the original and satisfies n less than two.
Faded example
Add to both sides.
Fill in the blanks
-2 > n - 4 \;\Longrightarrow\; -2 + 4 > n \;\Longrightarrow\; 2 > n \;\Longleftrightarrow\; n < 2
Why: Adding four gives two greater than n, and turning the statement round to put the variable first reverses the symbol to less than. Both forms describe the same set of numbers, and writing the variable first is the usual convention.
Elimination
Solving n minus 6 greater than or equal to -2.
Eliminate the wrong options
What is the right first move?
Survives elimination: A
Why: Adding six undoes the subtraction and isolates n, giving n greater than or equal to four. Option D is worth naming because the habit of reversing symbols comes from Lesson 6.2's rule, which does not apply to addition or subtraction at all.
Socratic
In Lesson 6.2 it will, sometimes.
Discussion prompt
Explain why adding or subtracting the same number never changes the direction of an inequality, using the number line. Then say what the two circumstances are in which the symbol does change.
Hint: Think about both numbers sliding together.
Answer:
Adding a fixed amount moves every point on the line the same distance in the same direction, so the relative positions of the two sides are untouched. Whichever was to the left stays to the left, which is exactly what the inequality symbol records.
The symbol changes in two circumstances. One is turning the whole statement round, as in rewriting two greater than n as n less than two, where the sides have swapped so the symbol must too. The other is multiplying or dividing by a negative number, which reflects both sides through zero and reverses every order — the subject of Lesson 6.2.
Section
Section 4
Concept
You cannot check every solution of an inequality. Instead choose several numbers from the solution set and confirm they work, and several from outside it and confirm they do not.
Only the second kind of test can catch a boundary in the wrong place.
Figure (svg): An inequality's solution checked with a solution and with a non-solution
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.1 Solving Inequalities Using Addition or Subtraction §6.1, pp. 324-324 — the paragraph on checking, and the Study Tip on choosing easy numbers
Picture it
One number in, one number out.
Figure (svg): An inequality's solution checked with a solution and with a non-solution
The failing test is the informative one. An answer whose boundary is too generous still passes every test taken from inside it, so the outside test is where the real check lives.
Worked example
This is the checking recommended alongside Example 2.
\[ \text{Check the solution } x \ge 2 \text{ of } \; x - 5 \ge -3. \]
Choose a number inside
Why: Four is a solution according to the answer.
\[ x = 4 \]
Test it
Why: Four minus five is negative one, which is at least negative three.
Choose a number outside
Why: Zero is not a solution according to the answer.
\[ x = 0 \]
Test it
Why: Zero minus five is negative five, which is not at least negative three.
Figure (svg): An inequality's solution checked with a solution and with a non-solution
\[ 4: \;\checkmark \qquad 0: \;\times \]
Verify: test the endpoint itself
Why: Two gives negative three, which is greater than or equal to negative three, so two is a solution and the dot is solid. Testing the endpoint is what confirms which kind of dot the graph needs, and it is worth doing every time.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.1 Solving Inequalities Using Addition or Subtraction §6.1, pp. 324-324
Elimination
The claimed solution of x plus 4 less than 9 is x less than 9.
Eliminate the wrong options
Which number best tests that claim?
Survives elimination: A
Why: A useful test is a number the two candidate answers disagree about, which means one near the claimed boundary. Seven is accepted by x less than nine and rejected by the original, so it settles the question in one substitution while the other three cannot.
Worked example
A wrong boundary passes half the tests and fails the other half.
\[ \text{A student solves } x + 4 < 9 \text{ and reports } x < 9. \text{ Check it.} \]
Test a number the answer accepts
Why: Two gives six, which is less than nine.
Test another
Why: Four gives eight, which is less than nine.
Test a number near the claimed boundary
Why: Seven gives eleven, which is not less than nine.
Diagnose
Why: The answer accepts seven and the original does not, so the boundary is too high.
Figure (svg): The solution to Worked example a check that catches an error shown as a ladder of expressions, one row per algebraic move
\[ x < 5, \text{ not } x < 9 \]
Verify: notice which tests caught it
Why: The first two tests passed even though the answer was wrong, because small numbers satisfy both the correct and the incorrect version. Only a number between the true boundary and the claimed one exposes it, which is why tests should be chosen near the boundary rather than at random.
Trap
\[ x + 4 < 9 \text{, reported as } x < 9 \]
Test 0 and 1, both of which work, and accept the answer
Why: Small numbers are easy to substitute and both do satisfy the original inequality.
Both also satisfy the correct answer, so neither test can distinguish the two. The check has confirmed nothing about where the boundary is.
Test a number just inside the claimed boundary, and one just outside
Why: Seven satisfies x less than nine and fails the original, which exposes the error at once.
Choosing test numbers near the boundary rather than deep inside the set is what makes a check informative.
Faded example
One inside, one outside.
Fill in the blanks
For x - 5 >= -3 with answer x >= 2: at x = 4, -1 >= -3 is true; at x = 0, -5 >= -3 is false.
Why: The inside number gives a true statement and the outside one a false statement, which is exactly what the answer predicts. Both results agreeing with the graph is what makes the check complete.
Prediction
A number the answer excludes turns out to satisfy the original.
Predict first
What has gone wrong?
Correct: The solution set is too small; the boundary is in the wrong place.
Losing solutions usually comes from an operation applied to one side only, or from reversing the symbol when nothing required it.
Why: A number that satisfies the original inequality is a genuine solution, so an answer excluding it has lost solutions and is too small. The fourth option is the dangerous one: a correct solution set contains every solution and nothing else, so an excluded number satisfying the original always means an error.
Socratic
In Chapter 3 checking the single answer was enough.
Discussion prompt
Explain why an inequality cannot be checked completely by substitution, and say what the substitution check can and cannot establish. Then say what would establish it completely.
Hint: Count the solutions again.
Answer:
There are infinitely many solutions, so no finite number of substitutions can test them all. A check can establish that a particular number is or is not a solution, and by testing near the boundary it can strongly suggest the boundary is right — but it cannot prove the whole set is correct.
What establishes it completely is the algebra itself: each step used a property that produces an equivalent inequality, so the final simple inequality has exactly the same solutions as the original. The substitution check is looking for a slip in applying those properties, not supplying the proof — which is why one well-chosen number is worth more than five badly chosen ones.
Section
Section 5
Concept
To model a situation with an inequality, name the variable, write the relationship in words, and translate. The phrase used decides which of the four symbols to write.
The word farther excludes the boundary itself, which is why Example 4 uses an open dot.
Figure (svg): A number line of light years with a boundary at Sirius
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.1 Solving Inequalities Using Addition or Subtraction §6.1, pp. 325-325 — Example 4, Write and Graph an Inequality in One Variable
Picture it
One boundary, excluded.
Figure (svg): A number line of light years with a boundary at Sirius
Sirius is about 8.8 light years away, and a light year is about six trillion miles. The graph's open dot says that Sirius itself is not among the points described as farther than Sirius.
Worked example
This is Example 4 from the textbook.
\[ \text{Sirius is about } 8.8 \text{ light years away. Describe distances to points farther from Earth than Sirius.} \]
Name the variable
Why: Let d be the distance in light years.
Write the relationship in words
Why: The distance is greater than 8.8.
Translate into symbols
Why: Greater than gives a strict symbol.
\[ d > 8.8 \]
Graph it
Why: Open dot at 8.8, shading right.
\[ \text{open at } 8.8 \]
Figure (svg): A number line of light years with a boundary at Sirius
\[ d > 8.8 \]
Verify: ask whether Sirius itself is described
Why: Sirius is 8.8 light years away and is not farther from Earth than itself, so it should be excluded — and the strict symbol with its open dot excludes exactly that one value. The word farther is doing the work, and changing it to at least would change the dot.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.1 Solving Inequalities Using Addition or Subtraction §6.1, pp. 325-325
Matching
Each phrase corresponds to one symbol.
Match the pairs
Why: At least and at most include the boundary and give solid dots; farther than and fewer than exclude it and give open ones. Testing whether the boundary value itself satisfies the description is the reliable way to choose, since the wording alone can mislead.
Worked example
Guided Practice 11 and 12. One changes the symbol; one changes the number.
\[ \text{What if the phrase were } \textit{at least as far as Sirius}? \text{ And what about Deneb, at about } 1600 \text{ light years?} \]
Take at least as far
Why: At least means greater than or equal to.
\[ d \ge 8.8 \]
Say what changes on the graph
Why: The dot becomes solid; the shading is unchanged.
Take Deneb
Why: Farther than Deneb, at 1600 light years.
\[ d > 1600 \]
Say what changes on the graph
Why: The boundary moves and the scale of the number line must change with it.
Figure (svg): The solution to Worked example change the wording shown as a ladder of expressions, one row per algebraic move
\[ d \ge 8.8 \qquad d > 1600 \]
Verify: check which feature each change affected
Why: Changing the wording changed only the dot, and changing the star changed only the position. The two features are independent, which is why a graph carries both pieces of information rather than one.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.1 Solving Inequalities Using Addition or Subtraction §6.1, pp. 325-325
Trap
A ride requires you to be at least 48 inches tall.
Write h < 48, since at least sounds like a small amount
Why: The word least suggests smallness, so the less-than symbol feels right.
At least 48 means 48 or more, so the inequality is h greater than or equal to 48. The written version would let in anyone shorter than the limit, which is the opposite of the rule.
\[ h \ge 48 \]
Ask whether the boundary value itself is allowed, and which side is allowed
Why: Someone exactly 48 inches tall may ride, and taller people may too.
Testing the boundary and one number on each side of it turns a wording question into three quick checks.
Faded example
Name the variable, then choose the symbol.
Fill in the blanks
Points farther from Earth than Sirius, which is 8.8 light years away: d > 8.8, drawn with an open dot.
Why: Farther than excludes Sirius's own distance, so the symbol is strict and the dot is open. Changing the phrase to at least as far would make both of those change together, since the dot always follows the symbol.
Hypothesis
Predict before you decide.
Predict first
A sign reads: children under 12 travel free. Which inequality describes who travels free?
Correct: a < 12.
Testing the boundary is the reliable method: ask whether a child of exactly twelve travels free, and the answer picks the symbol.
Why: Under twelve means below twelve and not twelve itself, so a child who has turned twelve pays. The symbol is strict and the dot open. This wording is worth dwelling on because under and up to differ by exactly one value: up to and including twelve would be the or equal to version, and signs in the real world are often ambiguous precisely here.
Socratic
Farther than and at least as far differ by a single value.
Discussion prompt
Explain why changing farther than to at least as far changes the graph, even though it adds only one number to an infinite set. Then say when that one number actually matters.
Hint: Think about what the boundary case represents.
Answer:
The set gains exactly one point, which is invisible as a proportion of an infinite set and entirely visible on the graph, since it is the endpoint. The dot is the only part of the drawing that records it, which is why the two kinds of dot exist at all.
It matters whenever the boundary is the case being decided — a height limit at exactly forty-eight inches, a speed limit at exactly the posted number, a budget spent exactly to the last dollar. Those are precisely the situations that end up in dispute, so the single point the dot records is very often the one that counts.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| Equation (Chapter 3) | Inequality (Chapter 6) | |
|---|---|---|
| Number of solutions | usually one | infinitely many |
| How the answer is shown | a single value | a graph on a number line |
| How to check | substitute the one answer | test numbers inside and outside |
The solving moves are identical; what changes is that the answer is a set, so it has to be drawn and checked differently.
Pattern
Whether the inequality comes from symbols or from words, the same five moves cover it.
Step four is the only place in this lesson where the symbol changes, and it changes because the two sides swapped rather than because of anything arithmetic.
OpenStax Elementary Algebra 2e, §2.7 Solve Linear Inequalities §2.7
Check
Test the endpoint.
Check your understanding
Which graph shows x greater than or equal to -1?
Answer: A
Why: The or equal to part makes negative one itself a solution, so the dot is solid, and greater than shades towards the larger numbers. Substituting negative one gives a true statement, which confirms the dot.
Check
Do the same thing to both sides.
Check your understanding
Solve x + 7 < 3.
Answer: A
Why: Subtracting seven from both sides gives x less than negative four. Testing negative five gives two, which is less than three, and testing zero gives seven, which is not — both agree with the answer.
Check
The sides have swapped.
Check your understanding
Rewrite 5 > y with the variable first.
Answer: A
Why: Five is greater than y says the same thing as y is less than five. Turning the statement round swaps the two sides, so the symbol reverses to keep the meaning. Testing zero confirms it: five is greater than zero, and zero is less than five.
Real world
A lift is rated to carry at most 1200 pounds. Four people weighing a total of 640 pounds are already inside, and one more person wants to get in.
Discussion prompt
Write an inequality for the weight the next person may have, solve it, graph the solution, and say what the graph's dot means. Then say what part of the graph is not physically meaningful.
Hint: At most means the boundary is allowed.
Answer:
\[ 640 + w \le 1200 \;\Longrightarrow\; w \le 560 \]
The next person may weigh up to and including 560 pounds. The dot is solid, because a person weighing exactly 560 pounds would bring the total to exactly the rating, which at most permits.
The graph shades everything below 560, including negative weights, which no person has. So the physically meaningful part is between zero and 560 — a restriction that comes from the situation rather than the algebra, exactly as with the taxi fare in Lesson 4.2 and the budget line in Lesson 4.4.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
Solving -2 > n - 4 gives 2 > n. What is the answer with the variable first?
Correct: n < 2, reversing the symbol with the swap.
\[ 2 > n \;\Longleftrightarrow\; n < 2 \]
\[ \text{check } n = 0: \; -2 > 0 - 4 = -4 \;\checkmark \]
Why: Two is greater than n and n is less than two say the same thing: the symbol points at the smaller quantity, so moving the sides moves the symbol with them. Testing zero settles it, since zero satisfies the original and satisfies n less than two. The first option is the standard error and it produces exactly the opposite solution set — worth noticing that nothing arithmetic happened here, only a change in reading order.
Explain it
They can solve x plus 4 equals 9 and have never seen an inequality.
Discussion prompt
In no more than four sentences, explain what changes when the equals sign becomes a less-than sign, and what does not. Then tell them the one thing to check about the endpoint.
Hint: The solving is the same; the answer is different.
Answer:
A usable answer: you solve it exactly the same way, by doing the same thing to both sides until the letter is alone. What changes is the answer — instead of one number you get every number on one side of a boundary, so you draw it on a number line with an arrow rather than writing it on a line.
The thing to check is whether the boundary itself is a solution. Substitute it: if the statement comes out true, fill the dot in, and if it comes out false, leave it open. That is the only decision the picture asks you to make that the algebra does not.
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 dot is fixed by substituting the endpoint and seeing whether the statement is true. The swap is fixed by reading the statement aloud after turning it round. Checking is fixed by choosing a number just outside the claimed boundary rather than deep inside the set. Wording is fixed by asking whether the boundary value itself satisfies the description. Pick yours and do five of that kind tonight rather than twenty mixed ones.
Connect it up
Do this on paper. It is worth more than rereading the slides.
Draw it
Draw four number lines down one side of a page and graph one inequality of each kind on them — less than, greater than, less than or equal to, greater than or equal to — labelling each with its inequality and writing beside each dot whether the endpoint is a solution. In the middle of the page, solve one inequality with a subtraction and one with an addition, showing the same operation written on both sides of each line, and graph both answers. Beside each answer write two test numbers, one inside the set and one just outside it, with the true or false verdict for each. In the lower half, write a real situation of your own involving the words at least or at most, translate it into an inequality, graph it, and mark on the graph the part that is not physically meaningful. Finally, in the margin, write one sentence saying why adding the same number to both sides never changes the direction of the symbol.
Your two test numbers should give opposite verdicts every time. If both come out true, the outside number was not far enough outside — move it closer to the boundary and try again.
Recap
Five things, and the first is the one the graph asks you to decide.
| If the question says | Your first move is |
|---|---|
| Graph the inequality | Substitute the endpoint to choose the dot |
| Solve x - 5 >= -3 | Add 5 to each side |
| The answer is 2 > n | Turn it round to n < 2 |
| Check your solution | Test a number just outside the boundary |
| At least, or at most | Ask whether the boundary itself is allowed |
Lesson 6.2 solves inequalities by multiplying and dividing instead. That turns out to need one genuinely new rule, because multiplying by a negative number reverses the direction of the inequality.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.1 Solving Inequalities Using Addition or Subtraction §6.1, pp. 323-328 — everything on these slides traces back here
Want this taught 1-on-1? Alexander tutors Algebra 1 — $55/session, free consultation.