The multiplication and division properties of inequality, split into the case of a positive multiplier, which preserves the direction, and a negative one, which reverses it. Includes solving one-step inequalities of both kinds, graphing the results, and recognising that only the sign of the multiplier decides whether the symbol turns.
Subject: Algebra 1 · 65 slides · symbolic lesson
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Title
Algebra 1 · Chapter 6 — Solving and Graphing Linear Inequalities
Solving Inequalities Using Multiplication or Division
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 6.2 Solving Inequalities Using Multiplication or Division §6.2, pp. 330-335 — the lesson these objectives are drawn from
Warm-up
Lesson 6.1 found that adding never changes the direction of an inequality. This lesson asks the same question about multiplying.
Discussion prompt
Start from the true statement 1 is less than 3. Multiply both sides by 2 and check. Then multiply both sides by negative 2 and check again.
Hint: Compute both products and compare them on a number line.
Answer:
\[ 1 < 3 \;\xrightarrow{\times 2}\; 2 < 6 \;\checkmark \qquad 1 < 3 \;\xrightarrow{\times (-2)}\; -2 \text{ and } -6 \]
The first stays true. The second gives negative two and negative six, and negative two is the larger — so the statement is only true written as negative two is greater than negative six. The direction had to be reversed, which is the one genuinely new rule in this chapter.
Concept
Multiplying or dividing each side of an inequality by the same positive number produces an equivalent inequality with the same direction. Doing it by a negative number requires reversing the direction.
multiplication property of inequality — Multiplying each side of an inequality by a positive number preserves its direction; multiplying by a negative number reverses it.
The Developing Concepts investigation on page 329 arrives at both halves by testing numbers.
Figure (svg): The multiplication and division properties for positive and negative multipliers
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.2 Solving Inequalities Using Multiplication or Division §6.2, pp. 329-331
Section
Section 1
Concept
For a positive multiplier, the properties look exactly like the ones from Lesson 6.1: whatever the direction was, it stays. Nothing about the inequality symbol changes.
\[ a > b, \; c > 0 \;\Longrightarrow\; ac > bc \]
Figure (svg): Two numbers multiplied by a positive number, keeping their order
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.2 Solving Inequalities Using Multiplication or Division §6.2, pp. 330-330 — the Properties of Inequality box for c greater than zero
Picture it
Both points move outward together.
Figure (svg): Two numbers multiplied by a positive number, keeping their order
Multiplying by two doubles every distance from zero without moving anything across zero, so the left-to-right order of the whole line is untouched.
Worked example
This is Example 1 from the textbook.
\[ \text{Solve } \; \tfrac{1}{4}a \le 10 \; \text{ and graph the solution.} \]
Write the original inequality
Why: The variable is multiplied by one quarter.
\[ (\frac{1}{4}) a \le 10 \]
Multiply each side by 4
Why: Four is positive, so the symbol does not change.
\[ 4(\frac{1}{4}) a \le 4(10) \]
Simplify
Why: The quarter is undone.
\[ a \le 40 \]
Graph
Why: Solid dot at forty, shading left.
\[ \text{solid at } 40 \]
Figure (svg): The solution to Worked example multiply by a positive number shown as a ladder of expressions, one row per algebraic move
\[ a \le 40 \]
Verify: test a number inside and one outside
Why: Twenty gives five, which is at most ten, so twenty is a solution. Fifty gives twelve and a half, which is not, so fifty is correctly excluded. The boundary of forty gives exactly ten, confirming the solid dot.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.2 Solving Inequalities Using Multiplication or Division §6.2, pp. 330-330
Sorting
Look at the sign of what you multiply or divide by.
Sort into buckets
Sort each operation by its effect on the inequality symbol.
Adding a negative number is still addition, so it belongs in the first bucket. Only multiplying or dividing by a negative causes a reversal.
Worked example
This is Example 2 from the textbook.
\[ \text{Solve } \; 4x > 20 \; \text{ and graph the solution.} \]
Write the original inequality
Why: The variable is multiplied by four.
\[ 4 x > 20 \]
Divide each side by 4
Why: Four is positive, so the symbol stays.
\[ 4 x / 4 > \frac{20}{4} \]
Simplify
Why: x is greater than five.
\[ x > 5 \]
Graph
Why: Open dot at five, shading right.
\[ \text{open at } 5 \]
Figure (svg): A one-step inequality solved by dividing by a positive number
\[ x > 5 \]
Verify: test the boundary
Why: Five gives twenty, and twenty is not greater than twenty, so five is not a solution and the dot is open. Testing the boundary is what settles which dot to draw, exactly as in Lesson 6.1.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.2 Solving Inequalities Using Multiplication or Division §6.2, pp. 331-331
Trap
\[ 4x > 20 \quad \text{with a negative on the right later in the exercise} \]
Divide by 4 and reverse, because the chapter is about reversing
Why: Once the reversal rule is learned it starts being applied everywhere.
\[ x < 5 \quad \text{(wrong)} \]
Testing six gives twenty-four, which is greater than twenty — so six is a solution and the answer wrongly excludes it.
\[ 4x > 20 \;\Longrightarrow\; x > 5 \]
Reverse only when the number you multiply or divide by is negative
Why: Four is positive, so nothing about the symbol changes.
Asking what am I dividing by, and is it negative, before touching the symbol is the whole of the decision.
Faded example
The symbol does not move.
Fill in the blanks
4x > 20 \;\Longrightarrow\; \dfrac4> > \dfrac______} \;\Longrightarrow\; x ___ 5
Why: Dividing both sides by four gives x greater than five, with the symbol unchanged because four is positive. Testing six gives twenty-four, which is greater than twenty, confirming that six is a solution.
Elimination
Solve 21 is less than 3y.
Eliminate the wrong options
Which is the solution?
Survives elimination: A
Why: Dividing both sides by three gives seven less than y, or y greater than seven. Testing eight gives twenty-four, which is greater than twenty-one, and testing six gives eighteen, which is not — both agree with the answer.
Socratic
The number line makes it visible.
Discussion prompt
Explain why multiplying two numbers by the same positive number cannot change which one is larger. Then say why the argument would fail for a multiplier of zero.
Hint: Think about what multiplying does to the number line.
Answer:
Multiplying by a positive number stretches or squashes the whole line towards or away from zero without folding it. Every point stays on the side of zero it started on and the order along the line is preserved, so whichever number was further right is still further right.
A multiplier of zero collapses the entire line onto the single point zero, so both sides become zero and the strict inequality becomes false. That is why the properties are stated for c greater than zero and c less than zero, with zero excluded from both — it is the one multiplier that destroys the information rather than transforming it.
Section
Section 2
Concept
Multiplying or dividing each side by the same negative number reverses the inequality. Less than becomes greater than, and less than or equal to becomes greater than or equal to.
\[ a > b, \; c < 0 \;\Longrightarrow\; ac < bc \]
Figure (svg): Two numbers multiplied by a negative number, reversing their order
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.2 Solving Inequalities Using Multiplication or Division §6.2, pp. 331-331 — the Properties of Inequality box for c less than zero and its Writing Algebra note
Picture it
Both points cross zero and swap order.
Figure (svg): Two numbers multiplied by a negative number, reversing their order
The larger number lands further from zero on the negative side, which puts it further left. Every order on the line is reversed at once, which is why the rule has no exceptions.
Worked example
This is Example 3 from the textbook.
\[ \text{Solve } \; -\tfrac{1}{2}y \le 5 \; \text{ and graph the solution.} \]
Write the original inequality
Why: The variable is multiplied by negative one half.
\[ -(\frac{1}{2}) y \le 5 \]
Multiply each side by -2
Why: Negative two is negative, so the symbol reverses.
\[ -2 \times\text{ both sides} \]
Write the reversed symbol
Why: Less than or equal to becomes greater than or equal to.
\[ y \ge - 10 \]
Graph
Why: Solid dot at negative ten, shading right.
\[ \text{solid at } -10 \]
Figure (svg): Two numbers multiplied by a negative number, reversing their order
\[ y \ge -10 \]
Verify: test a number on each side of the boundary
Why: Zero gives zero, which is at most five, so zero is a solution — and the answer includes it. Negative twenty gives ten, which is not at most five, so it is correctly excluded. Had the symbol not been reversed, the answer would have excluded zero and included negative twenty, exactly backwards.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.2 Solving Inequalities Using Multiplication or Division §6.2, pp. 331-331
Translation
The direction turns; the or equal to stays.
Match the pairs
Why: Reversing swaps which side the symbol points at and leaves the or equal to part alone. A strict inequality stays strict and an inclusive one stays inclusive, so the endpoint's status is never affected by a reversal.
Worked example
This is Example 4 from the textbook, both parts.
\[ \text{Solve } \; -12m > 18 \; \text{ and } \; -8x \le 20. \]
Divide the first by -12
Why: Negative, so the symbol reverses.
\[ m < \frac{18}{-12} \]
Simplify
Why: Eighteen over negative twelve is negative one and a half.
\[ m < -1.5 \]
Divide the second by -8
Why: Negative again, so the symbol reverses.
\[ x \ge \frac{20}{-8} \]
Simplify
Why: Twenty over negative eight is negative two and a half.
\[ x \ge - 2.5 \]
Figure (svg): A one-step inequality solved by dividing by a negative number, with the reversal
\[ m < -1.5 \qquad x \ge -2.5 \]
Verify: test the first answer at a convenient number
Why: Taking m equal to negative two gives twenty-four, which is greater than eighteen, so negative two is a solution — and the answer includes it since negative two is less than negative one and a half. Testing zero gives zero, which is not greater than eighteen, so zero is correctly excluded.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.2 Solving Inequalities Using Multiplication or Division §6.2, pp. 332-332
Error analysis
The student solved an inequality by dividing by a negative number.
Annotate
On: \( \begin{aligned} -12m &> 18 \\ \frac{-12m}{-12} &> \frac{18}{-12} \\ m &> -1.5 \end{aligned} \)
Writing the reversed symbol on the same line as the division, rather than afterwards, is what prevents this. A reversal remembered later is a reversal forgotten.
Faded example
Both changes on the same line.
Fill in the blanks
-12m > 18 \;\Longrightarrow\; m < \dfrac<-1.5 \;\Longrightarrow\; m ___ ___
Why: Dividing by negative twelve reverses greater-than to less-than, giving m less than negative one and a half. Testing negative two gives twenty-four, which is greater than eighteen, so negative two is a solution and the answer includes it.
Elimination
Solve -8x less than or equal to 20.
Eliminate the wrong options
Which is the solution?
Survives elimination: A
Why: Dividing by negative eight reverses the symbol and gives a negative boundary, so x is at least negative two and a half. Testing zero settles it in one line: zero satisfies the original and only option A accepts it.
Socratic
The rule is worth understanding rather than memorising.
Discussion prompt
Explain on the number line why multiplying two numbers by a negative reverses which is larger. Then say why the same thing happens for division.
Hint: Think about what multiplying by negative one does to the line.
Answer:
Multiplying by negative one reflects the whole line through zero: everything on the right lands on the left and vice versa. A number that was further right was further from zero in the positive direction, so after reflection it is further from zero in the negative direction — which puts it further left. Every order is reversed at once.
Dividing by a negative number is multiplying by its reciprocal, and the reciprocal of a negative is negative. So division by a negative is the same reflection followed by a stretch, and the reflection is what causes the reversal. That is why the two properties are stated identically and why one explanation covers both.
Section
Section 3
Concept
The presence of negative numbers in an inequality is not a reason to reverse it. The symbol turns only when you multiply or divide both sides by a negative number, or when you swap the two sides.
Swapping the two sides also reverses, as in Lesson 6.1.
Figure (svg): Two columns saying which operations reverse an inequality and which do not
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.2 Solving Inequalities Using Multiplication or Division §6.2, pp. 332-332 — the Properties of Inequality summary at the end of the lesson
Picture it
Four safe operations and two that turn the symbol.
Figure (svg): Two columns saying which operations reverse an inequality and which do not
The list is short and worth knowing exactly, because over-applying the reversal is as damaging as forgetting it — and both produce an answer that is precisely the wrong half of the line.
Worked example
Both contain negative numbers; only one needs the rule.
\[ \text{Solve } \; x - 5 \ge -3 \; \text{ and } \; -5x \ge -3. \]
Look at the first
Why: The five is being subtracted, so the move is an addition.
\[ \text{add } 5 \]
Solve it
Why: x is at least two; no reversal.
\[ x \ge 2 \]
Look at the second
Why: The variable is multiplied by negative five, so the move is a division by a negative.
\[ \text{divide by } -5 \]
Solve it
Why: The symbol reverses and the boundary is three fifths.
\[ x \le 0.6 \]
Figure (svg): An inequality containing negative numbers that does not require a reversal
\[ x \ge 2 \qquad x \le 0.6 \]
Verify: test zero in each original
Why: Zero minus five is negative five, which is not at least negative three, so zero is not a solution of the first — matching x at least two. Negative five times zero is zero, which is at least negative three, so zero is a solution of the second — matching x at most three fifths. Both answers are confirmed by one test number.
Discrimination
Look only at what you multiply or divide by.
Sort into buckets
Sort each inequality by whether solving it requires reversing the symbol.
Worked example
The right-hand side's sign is irrelevant to the reversal.
\[ \text{Solve } \; 3x > -12. \]
Identify the operation
Why: The variable is multiplied by three.
\[ \text{divide by } 3 \]
Check the divisor's sign
Why: Three is positive.
Divide
Why: Negative twelve over three is negative four.
\[ x > -4 \]
Note the negative that did not matter
Why: The negative twelve never affected the symbol.
Figure (svg): The solution to Worked example a negative on the right only shown as a ladder of expressions, one row per algebraic move
\[ x > -4 \]
Verify: test zero and negative five
Why: Zero gives zero, which is greater than negative twelve, so zero is a solution and the answer includes it. Negative five gives negative fifteen, which is not, so it is correctly excluded. The presence of a negative number in the problem changed nothing about the method.
Trap
\[ 3x > -12 \]
Divide by 3 and reverse, since there is a negative in the inequality
Why: The negative twelve is visible and the reversal rule is fresh, so the two get associated.
\[ x < -4 \quad \text{(wrong)} \]
Testing zero gives zero, which is greater than negative twelve, so zero is a solution — and the reported answer excludes it.
\[ 3x > -12 \;\Longrightarrow\; x > -4 \]
Ask only one question: is the number I am multiplying or dividing by negative?
Why: Here it is three, which is positive, so nothing changes.
The rule is about the multiplier alone, never about what else appears in the inequality.
Faded example
Name the operation, then its sign.
Fill in the blanks
To solve -14z >= 70 you divide by -14, which is negative, so the symbol reverses.
Why: Dividing by negative fourteen reverses greater-than-or-equal-to into less-than-or-equal-to, giving z at most negative five. Naming the divisor and its sign before doing any arithmetic is what makes the decision automatic.
Elimination
Only one operation among the four.
Eliminate the wrong options
Which step reverses an inequality?
Survives elimination: A
Why: Only multiplying or dividing by a negative reverses. Options B and C both involve a negative six and neither reverses anything, which is exactly the confusion the rule invites — the sign that matters belongs to the multiplier, not to any number that happens to be nearby.
Socratic
Both errors produce a definite-looking answer.
Discussion prompt
Explain what happens to the solution set when a reversal is applied that should not have been. Then say what single test catches both errors at once.
Hint: Compare the two halves of the number line.
Answer:
The answer becomes the exact complement of the truth: every number that should be a solution is excluded and every number that should not be is accepted. The boundary is in the right place, so the answer looks entirely plausible — which is what makes the error dangerous.
One test number catches both, provided it is chosen on a definite side of the boundary. Substituting it into the original tells you whether it is a solution, and comparing that with what your answer claims settles the direction immediately. Zero is usually the best choice, since the arithmetic is trivial and it is rarely the boundary.
Section
Section 4
Concept
After solving, graph the answer as in Lesson 6.1. The kind of dot depends only on whether the symbol includes or equal to, and a reversal never changes that.
Zero is usually the best test number, since it makes the arithmetic trivial.
Figure (svg): The four inequality symbols with their reversed counterparts
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.2 Solving Inequalities Using Multiplication or Division §6.2, pp. 331-332 — the Writing Algebra note on how to reverse each symbol
Picture it
Direction turns; inclusiveness does not.
Figure (svg): The four inequality symbols with their reversed counterparts
A strict inequality reverses into a strict one and an inclusive into an inclusive. So the dot on the graph is decided before the reversal and is unaffected by it.
Worked example
The whole routine on one problem.
\[ \text{Solve } \; -14z \ge 70, \text{ graph it, and check.} \]
Divide by -14 and reverse
Why: Greater than or equal to becomes less than or equal to.
\[ z \le - 5 \]
Choose the dot
Why: The symbol still includes or equal to, so the dot is solid.
\[ \text{solid at } -5 \]
Graph
Why: Shading left from negative five.
Check inside and outside
Why: Negative six gives eighty-four; zero gives zero.
Figure (svg): The solution to Worked example solve, graph and check shown as a ladder of expressions, one row per algebraic move
\[ z \le -5 \]
Verify: confirm both test results
Why: Negative six gives eighty-four, which is at least seventy, so it is a solution and the graph includes it. Zero gives zero, which is not at least seventy, so it is correctly excluded. And negative five gives exactly seventy, confirming the solid dot.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.2 Solving Inequalities Using Multiplication or Division §6.2, pp. 332-332
Faded example
Direction only.
Fill in the blanks
Dividing -14z >= 70 by -14 gives z <= -5, drawn with a solid dot.
Why: The direction turns and the or equal to survives, so the dot stays solid. Testing negative five gives exactly seventy, which satisfies the original, confirming that the endpoint is a solution.
Worked example
Two reversals can occur in one problem, for different reasons.
\[ \text{Solve } \; -24 \le -6t. \]
Divide both sides by -6
Why: Negative divisor, so the symbol reverses.
\[ 4 \ge t \]
Note what the reversal did
Why: Less than or equal to became greater than or equal to.
Turn the statement round
Why: Putting t first swaps the sides, reversing again.
\[ t \le 4 \]
Note why this is not double-counting
Why: One reversal came from the division and one from the swap.
Figure (svg): The solution to Worked example an answer with the variable on the right shown as a ladder of expressions, one row per algebraic move
\[ -24 \le -6t \;\Longrightarrow\; 4 \ge t \;\Longleftrightarrow\; t \le 4 \]
Verify: test zero and five
Why: Zero gives zero, and negative twenty-four is at most zero, so zero is a solution — and t at most four includes it. Five gives negative thirty, and negative twenty-four is not at most negative thirty, so five is correctly excluded. Both reversals were needed to land here.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.2 Solving Inequalities Using Multiplication or Division §6.2, pp. 332-332
Trap
\[ -14z \ge 70 \;\Longrightarrow\; z < -5 \]
Reverse to less than, dropping the or equal to as part of the change
Why: The whole symbol is being replaced, so all of it feels like it should change.
Reversing turns the direction only. Negative five gives exactly seventy, which satisfies greater than or equal to, so negative five is a solution and the dot must stay solid.
\[ -14z \ge 70 \;\Longrightarrow\; z \le -5 \]
Turn the direction and keep the underline
Why: Greater than or equal to becomes less than or equal to, not less than.
Testing the boundary settles it independently of any rule about symbols.
Prediction
You solved -3x > 9 and got x < -3.
Predict first
Which number best tests that answer?
Correct: 0, which the answer excludes and the original may accept.
\[ -3(0) = 0, \; \text{and } 0 > 9 \text{ is false, so } 0 \text{ is not a solution} \]
Why: Zero has trivial arithmetic and sits on the opposite side of the boundary from the answer's set, so it tests the direction directly. Negative three times zero is zero, which is not greater than nine, so zero is correctly excluded and the direction is confirmed. The boundary tests only the dot, and a number deep inside tests nothing about the direction.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| Changed by a reversal | Unchanged by a reversal | |
|---|---|---|
| The direction of the symbol | yes | no |
| The or equal to part | no | yes |
| The boundary number | no | yes |
Only one of the three features moves. The boundary comes from the arithmetic and the dot from the symbol's inclusiveness, and a reversal touches neither.
Socratic
Any number could be substituted.
Discussion prompt
Explain why zero is usually the best number to test an inequality's solution with, and say when it would be a poor choice.
Hint: Think about what makes a test both easy and informative.
Answer:
Zero makes every product zero and every multiple vanish, so the arithmetic takes a second. It is also on a definite side of most boundaries, which means substituting it distinguishes a correct answer from one with the direction reversed — the error this lesson invites.
It is a poor choice when zero is the boundary itself, since then it tests only the dot and says nothing about direction, and when the inequality involves division by the variable, where zero may not be permitted at all. In those cases pick the nearest convenient number on a definite side instead.
Section
Section 5
Concept
The summary on page 332 collects every property of inequality met so far: addition and subtraction preserve the direction, multiplication and division preserve it for a positive number and reverse it for a negative one.
Together they are enough to solve every one-step inequality.
Figure (svg): The multiplication and division properties for positive and negative multipliers
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.2 Solving Inequalities Using Multiplication or Division §6.2, pp. 332-332 — the Properties of Inequality summary
Picture it
Split by operation and by the sign of c.
Figure (svg): The multiplication and division properties for positive and negative multipliers
Multiplication and division always behave the same way as each other, which halves what has to be remembered — dividing by a number is multiplying by its reciprocal, and reciprocals keep their sign.
Worked example
Naming the property before solving makes the decision explicit.
\[ \text{Which property solves } \; x + 7 < 3, \quad 5x < 30, \quad -5x < 30? \]
Take the first
Why: Undo an addition by subtracting; the subtraction property.
Take the second
Why: Undo a multiplication by dividing by five, which is positive.
Take the third
Why: Undo a multiplication by dividing by negative five.
Solve all three
Why: x less than negative four, x less than six, x greater than negative six.
Figure (svg): The solution to Worked example choose the property for each problem shown as a ladder of expressions, one row per algebraic move
\[ x < -4, \quad x < 6, \quad x > -6 \]
Verify: compare the last two answers
Why: The second and third differ only by a minus sign in the problem and their answers are on opposite sides of a different boundary. Testing zero in each: zero satisfies the second and the third, and both answers include it — so both are consistent.
Matching
Name the operation and the sign.
Match the pairs
Why: The second and third differ only in the coefficient's sign, and that single difference decides whether the symbol turns. Naming the property before solving turns the reversal from something to remember into something you have already decided.
Worked example
Division is multiplication by a reciprocal.
\[ \text{Show that dividing by } -4 \text{ is the same as multiplying by } -\tfrac{1}{4}. \]
Write a division
Why: Dividing by negative four.
\[ \frac{a}{-4} \]
Rewrite it as a multiplication
Why: The reciprocal of negative four is negative one quarter.
\[ a \times(-\frac{1}{4}) \]
Check the sign
Why: The reciprocal of a negative is negative.
Conclude
Why: So the division property follows from the multiplication one.
Figure (svg): The solution to Worked example why division needs no separate rule shown as a ladder of expressions, one row per algebraic move
\[ \dfrac{a}{-4} = a \cdot \left(-\tfrac{1}{4}\right) \]
Verify: check on a true statement
Why: Starting from 1 less than 3, dividing by negative four gives negative a quarter and negative three quarters, and negative a quarter is the larger — the reversal. Multiplying by negative one quarter gives the same two numbers, so the two routes agree exactly as the argument predicts.
Trap
A student learns that dividing by a negative reverses the inequality and is unsure about multiplying by one.
Treat the two as different rules and hope the right one comes to mind
Why: The textbook states them as two properties, so they look like two facts.
They are the same fact, since dividing by a number is multiplying by its reciprocal and the reciprocal has the same sign. Remembering them separately doubles the load and invites the wrong one being recalled.
\[ \text{divide by } c \;\Longleftrightarrow\; \text{multiply by } \tfrac{1}{c}, \quad \text{same sign} \]
Remember one rule about the sign of the multiplier
Why: Whatever you do to both sides, ask only whether the number involved is negative.
One question, asked every time, covers all four of the properties in the summary.
Faded example
The sign of c decides.
Fill in the blanks
If a > b and c is positive, then ac > bc. If a > b and c is negative, then ac < bc.
Why: These are the two multiplication properties from the summary on page 332, and the division properties read identically. Only the sign of c distinguishes them, which is why one question settles every case.
Two truths and a lie
Three statements about solving inequalities. Two are true and one is not.
Eliminate the wrong options
Which statement is false?
Survives elimination: A
Why: The first is false, and it is the error the whole lesson guards against. In 3x greater than negative twelve there is a negative number and no reversal is needed, because the divisor is three. Only the sign of the number you multiply or divide by matters.
Socratic
Four properties, and one decision.
Discussion prompt
Reduce all four properties to a single question you can ask before every step. Then say what that question does not cover, so you know when to think harder.
Hint: Every property is about the same thing.
Answer:
The question is: is the number I am multiplying or dividing both sides by negative? If yes, reverse; if no, or if the step is an addition or subtraction, do not. That covers all four properties in the summary with one check.
It does not cover the swap: turning two greater than n into n less than two also reverses the symbol, for a completely different reason. So the full rule is the one question plus one exception, and it is worth writing both at the top of any page of inequality practice.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| Multiply or divide by a positive | Multiply or divide by a negative | |
|---|---|---|
| The direction | unchanged | reversed |
| The or equal to part | unchanged | unchanged |
| On the number line | a stretch | a reflection through zero |
Only the first row differs, and the third row explains why. A stretch preserves order and a reflection reverses it.
Pattern
Whether the coefficient is a whole number, a fraction, positive or negative, the same five moves cover it.
Step two before step three is deliberate. A reversal decided in advance is a reversal made; one left until after the arithmetic is the one that gets forgotten.
OpenStax Elementary Algebra 2e, §2.7 Solve Linear Inequalities §2.7
Check
Check the sign of the divisor.
Check your understanding
Solve -6x < 24.
Answer: A
Why: Dividing by negative six reverses the symbol and gives a boundary of negative four. Testing zero gives zero, which is less than twenty-four, so zero is a solution — and only this answer accepts it.
Check
A negative in the problem is not a reason to reverse.
Check your understanding
Solve 5x >= -20.
Answer: A
Why: The divisor is five, which is positive, so the symbol does not change. Testing zero gives zero, which is at least negative twenty, so zero is a solution and this answer includes it.
Check
Reversing turns the direction only.
Check your understanding
Dividing -2x >= 8 by -2 gives which symbol?
Answer: A
Why: Greater than or equal to reverses into less than or equal to, keeping the or equal to part. Testing negative four gives eight, which satisfies the original, so negative four is a solution and the symbol must include it.
Real world
This is Exercise 56's situation. Renting figure skates costs 5 dollars a session; buying a pair costs 120 dollars.
Discussion prompt
Write an inequality for the number of sessions after which buying costs less than renting, solve it, and say what your answer means. Then say what the boundary case represents.
Hint: Compare the two total costs.
Answer:
\[ 120 < 5n \;\Longrightarrow\; 24 < n \;\Longleftrightarrow\; n > 24 \]
Buying is cheaper once she skates more than twenty-four times. The divisor was five, which is positive, so no reversal was needed — the only reversal came from turning the statement round to put n first.
At exactly twenty-four sessions the two cost the same, one hundred and twenty dollars, which is why the boundary is excluded by a strict inequality. If the question had asked when buying costs no more than renting, the symbol would include twenty-four and the dot would be solid — a change of one word producing a change of one session.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
Does solving 3x > -12 require reversing the inequality symbol?
Correct: No, because the number you divide by is 3, which is positive.
\[ 3x > -12 \;\Longrightarrow\; x > -4 \]
\[ \text{compare } -3x > 12 \;\Longrightarrow\; x < -4 \quad \text{(reversed)} \]
Why: The rule depends only on the sign of the number you multiply or divide both sides by, which is three here. The negative twelve on the right and the negative boundary in the answer are both irrelevant to the direction. Testing zero settles it: zero is greater than negative twelve, so zero is a solution, and x greater than negative four includes it while x less than negative four would not.
Explain it
They have just learned that you sometimes flip the sign and are flipping it everywhere.
Discussion prompt
In no more than four sentences, tell them exactly when to flip and when not to, and give them a way to check without remembering the rule at all.
Hint: One question, asked every time.
Answer:
A usable answer: you flip only when you multiply or divide both sides by a negative number — never for adding or subtracting, and never just because a minus sign appears somewhere in the problem. So ask one question before each step: is the number I am multiplying or dividing by negative?
To check without the rule, substitute zero into the original inequality. If it comes out true, zero is a solution, so your answer had better include it — and if your answer excludes it, you flipped when you should not have, or failed to flip when you should.
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: Remembering the reversal is fixed by deciding it before doing the arithmetic and writing the new symbol on the same line. Not over-applying it is fixed by asking only about the divisor's sign. The or equal to is fixed by remembering that a reversal turns the direction and nothing else. The test number is fixed by choosing zero, or anything on the opposite side of the boundary from your solution set. 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 the true statement 1 is less than 3, and underneath it show what happens when you multiply both sides by 2 and by negative 2, drawing both results on small number lines and marking which one needed its symbol reversed. In the middle, solve four inequalities: one by multiplying by a positive, one by dividing by a positive, one by multiplying by a negative and one by dividing by a negative, writing the reversed symbol on the same line as the division wherever it is needed. Graph all four answers, marking each dot as open or solid and writing beside it which symbol produced it. In the lower half, write two inequalities that both contain negative numbers but only one of which needs a reversal, and beside each write the one question that decides it. Finally, in the margin, substitute zero into all four of your originals and write whether it is a solution, checking each answer against that.
Every one of your four answers should agree with what the zero test says. If one disagrees, the direction of that answer is wrong, and the reversal was either applied when it should not have been or omitted when it should have been.
Recap
Five things, and the second is the only genuinely new rule in the chapter.
| If the question says | Your first move is |
|---|---|
| Solve 4x > 20 | Divide by 4; no reversal |
| Solve -12m > 18 | Divide by -12 and reverse |
| A negative appears on the right | Ignore it; check only the divisor |
| Reverse the symbol | Turn the direction, keep the underline |
| Check your answer | Substitute zero into the original |
Lesson 6.3 combines the two lessons so far. A multi-step inequality needs additions and divisions in the same problem, so the reversal question has to be asked at exactly one of the steps rather than at all of them.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.2 Solving Inequalities Using Multiplication or Division §6.2, pp. 330-335 — everything on these slides traces back here
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