Inequalities requiring more than one operation: undoing an addition and a multiplication in sequence, distributing to clear brackets, collecting variable terms from both sides, and choosing which side to collect on so that no reversal is needed. Includes a profit model whose answer must be rounded in the direction the situation requires.
Subject: Algebra 1 · 65 slides · symbolic lesson
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Title
Algebra 1 · Chapter 6 — Solving and Graphing Linear Inequalities
Solving Multi-Step Inequalities
Objectives
Five outcomes, each one you can test yourself on with a pencil and no answer key.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.3 Solving Multi-Step Inequalities §6.3, pp. 336-341 — the lesson these objectives are drawn from
Warm-up
Lessons 6.1 and 6.2 each used one operation. This lesson puts them in the same problem.
Discussion prompt
To solve 2y minus 5 is less than 7, two things have to be undone. Which do you undo first, and does either step change the inequality symbol?
Hint: Recall the order Lesson 3.3 used for equations.
Answer:
\[ 2y - 5 < 7 \;\xrightarrow{+5}\; 2y < 12 \;\xrightarrow{\div 2}\; y < 6 \]
Undo the subtraction first and the multiplication second, which is the reverse of the order of operations. Neither step divides by a negative, so the symbol stays exactly as it was through both lines.
Concept
A multi-step inequality requires more than one operation to solve. The operations are the same ones as in Chapter 3, applied in the same order, with the reversal question asked at each step.
multi-step inequality — An inequality requiring more than one operation to solve, such as one with both a coefficient and a constant attached to the variable.
The reversal question is asked step by step, and for most steps the answer is no.
Figure (svg): A two-step inequality solved with the addition first and the division second
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.3 Solving Multi-Step Inequalities §6.3, pp. 336-336
Section
Section 1
Concept
To solve a two-step inequality, first undo whatever is added to or subtracted from the variable term, then undo the coefficient. This is the reverse of the order of operations, as in Lesson 3.3.
Every line carries the inequality symbol forward, changed only when a step requires it.
Figure (svg): A two-step inequality solved with the addition first and the division second
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.3 Solving Multi-Step Inequalities §6.3, pp. 336-336 — Example 1, Solve a Multi-Step Inequality
Picture it
Add, then divide.
Figure (svg): A two-step inequality solved with the addition first and the division second
Neither five nor two is negative, so the symbol is unchanged from the first line to the last. That is the ordinary case, and it is worth having seen before meeting the exception.
Worked example
This is Example 1 from the textbook.
\[ \text{Solve } \; 2y - 5 < 7. \]
Write the original inequality
Why: Two operations are attached to y.
\[ 2 y - 5 < 7 \]
Add 5 to each side
Why: Five is positive; the symbol does not change.
\[ 2 y < 12 \]
Divide each side by 2
Why: Two is positive; the symbol does not change.
\[ y < 6 \]
State the answer
Why: All real numbers less than six.
\[ y < 6 \]
Figure (svg): A two-step inequality solved with the addition first and the division second
\[ y < 6 \]
Verify: test a number inside and one outside
Why: Zero gives negative five, which is less than seven, so zero is a solution and the answer includes it. Ten gives fifteen, which is not, so ten is correctly excluded. Both tests agree with y less than six.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.3 Solving Multi-Step Inequalities §6.3, pp. 336-336
Faded example
Addition first, then division.
Fill in the blanks
2y - 5 < 7 \;\Longrightarrow\; 2y < 12 \;\Longrightarrow\; y < 6
Why: Adding five to both sides gives twelve on the right, and dividing by two gives six. Neither number is negative, so the symbol is the same on every line.
Worked example
Guided Practice 1 and 2. Same structure, different numbers.
\[ \text{Solve } \; 3x + 5 > 4 \; \text{ and } \; 10 - n \le 5. \]
Take the first
Why: Subtract five, then divide by three.
\[ 3 x > -1 \]
Finish it
Why: x is greater than negative one third.
\[ x > -\frac{1}{3} \]
Take the second
Why: Subtract ten from each side.
\[ -n \le - 5 \]
Finish it
Why: Multiply by negative one and reverse.
\[ n \ge 5 \]
Figure (svg): The solution to Worked example two from guided practice shown as a ladder of expressions, one row per algebraic move
\[ x > -\tfrac{1}{3} \qquad n \ge 5 \]
Verify: test zero in each
Why: Zero gives five in the first, which is greater than four, and zero is greater than negative one third — consistent. Zero gives ten in the second, which is not at most five, and zero is not at least five — also consistent. One test number confirms both directions.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.3 Solving Multi-Step Inequalities §6.3, pp. 336-336
Trap
\[ 2y - 5 < 7 \]
Divide everything by 2 first: y - 5 < 3.5
Why: Dividing looks like progress towards isolating y, so it gets done first.
The five was never divided, so the line is not equivalent to the original. Testing y equal to eight in the original gives eleven, which is not less than seven, while the corrupted version accepts it.
\[ 2y - 5 < 7 \;\Longrightarrow\; 2y < 12 \;\Longrightarrow\; y < 6 \]
Undo the addition or subtraction first, then the coefficient
Why: This is the reverse of the order of operations, exactly as in Lesson 3.3.
Dividing first is not wrong if every term is divided, and it usually produces fractions for no benefit.
Elimination
Solving 3x plus 5 greater than 4.
Eliminate the wrong options
What is the right first move?
Survives elimination: A
Why: Undoing the addition first leaves 3x greater than negative one, and then a single division finishes it. Option C is worth noticing because it confuses a coefficient with an added term, which is the distinction the whole two-step routine depends on.
Sorting
Each attachment has an opposite.
Sort into buckets
Sort each attachment by the operation that undoes it.
The two buckets are undone in a fixed order: everything in the first, then everything in the second. Only one item in the whole list can move the inequality symbol.
Socratic
The other order can be made to work.
Discussion prompt
Explain why undoing the addition before the coefficient is the usual order. Then say what happens if you divide first, and whether the answer changes.
Hint: Think about what each order does to the arithmetic.
Answer:
It is the reverse of the order of operations. To evaluate two y minus five you would multiply first and subtract second, so to undo it you add first and divide second — each step peeling off the outermost operation, exactly as in Lesson 3.3.
Dividing first works provided every term on both sides is divided, giving y minus two and a half less than three and a half, and then y less than six — the same answer. The arithmetic is worse, because the constants become fractions, and the risk is higher, because it is easy to divide the variable term and forget the constant. The answer never changes; only the effort does.
Section
Section 2
Concept
In a multi-step inequality, some steps reverse the symbol and others do not. The question has to be asked at each step separately, and for most steps the answer is no.
A typical three-step solve reverses exactly once, or not at all.
Figure (svg): A multi-step solve with the reversal question asked at each line
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.3 Solving Multi-Step Inequalities §6.3, pp. 336-336 — Example 2, where only the last step reverses
Picture it
The question, asked line by line.
Figure (svg): A multi-step solve with the reversal question asked at each line
Writing the reversal decision beside each line, even when the answer is no, turns a rule that must be remembered into one that has already been applied.
Worked example
This is Example 2 from the textbook.
\[ \text{Solve } \; 5 - x > 4. \]
Write the original inequality
Why: The variable has a coefficient of negative one.
\[ 5 - x > 4 \]
Subtract 5 from each side
Why: Subtraction never reverses.
\[ -x > -1 \]
Multiply each side by -1
Why: Negative multiplier, so reverse.
\[ x < 1 \]
State the answer
Why: All real numbers less than one.
\[ x < 1 \]
Figure (svg): A two-step inequality in which only the second step reverses the symbol
\[ x < 1 \]
Verify: test zero and two
Why: Zero gives five, which is greater than four, so zero is a solution — and x less than one includes it. Two gives three, which is not greater than four, so two is correctly excluded. Had the reversal been forgotten, the answer would have excluded zero and included two, exactly backwards.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.3 Solving Multi-Step Inequalities §6.3, pp. 336-336
Discrimination
Ask the question for each step separately.
Sort into buckets
Sort each step by whether it reverses the symbol.
Worked example
The same shape of problem with a positive coefficient.
\[ \text{Solve } \; 5 + x > 4 \; \text{ and compare with the previous example.} \]
Subtract 5 from each side
Why: No reversal.
\[ x > -1 \]
Note that the solve is finished
Why: The coefficient is already one.
Compare the two problems
Why: They differ by one sign and by one reversal.
State the answers
Why: x greater than negative one, against x less than one.
Figure (svg): The solution to Worked example two steps, no reversal at all shown as a ladder of expressions, one row per algebraic move
\[ 5 + x > 4 \;\Longrightarrow\; x > -1 \]
Verify: test zero in both originals
Why: Zero satisfies both, since five is greater than four either way, and both answers include zero. Testing a number that distinguishes them, such as two: it satisfies the plus version and not the minus version, which is what the two different answers say.
Error analysis
The student solved a two-step inequality with a negative coefficient.
Annotate
On: \( \begin{aligned} 5 - x &> 4 \\ -x &> -1 \\ x &> 1 \end{aligned} \)
The first step here was a subtraction and required nothing; only the second step needed the rule. Asking the question at each line, rather than once at the start, is what catches this.
Faded example
Write the decision beside each line.
Fill in the blanks
For 5 - x > 4: subtracting 5 gives -x > -1 with no reversal; multiplying by -1 gives x < 1 with a reversal.
Why: Only the second step turns the symbol, because only it multiplies by a negative. Writing the decision beside each line, even when it is no, makes the reversal something already decided rather than something to remember at the end.
Prediction
Solving -3x + 7 <= 1.
Predict first
How many of the steps reverse the symbol?
Correct: One: the division by -3.
\[ -3x + 7 \le 1 \;\Longrightarrow\; -3x \le -6 \;\Longrightarrow\; x \ge 2 \]
Why: Subtracting seven from both sides leaves negative 3x at most negative six, with no reversal, and dividing by negative three reverses it to x greater than or equal to two. The number of negative signs visible in the problem is irrelevant; only the sign of what you divide by counts.
Socratic
Some problems turn the symbol more than once.
Discussion prompt
Give a worked situation in which an inequality's symbol is reversed twice during a solve, and say what the net effect is. Then say why that does not mean the reversals cancel out in general.
Hint: Two negative divisions, or a negative division and a swap.
Answer:
Solving negative twenty-four at most negative six t divides by negative six, reversing to four at least t, and then turning the statement round to put t first reverses again to t at most four. Two reversals, and the final symbol points the same way as the original — so the net effect here is none.
They cancel only because there happened to be exactly two. A solve with one reversal ends pointing the other way, and one with three ends pointing the other way again. Counting them is not a shortcut; asking the question at each step is, because the count is only known once every step has been examined anyway.
Section
Section 3
Concept
When an inequality contains brackets, use the distributive property to clear them before solving. What remains is an ordinary multi-step inequality.
Distributing a negative factor changes the sign of every term inside, and does not reverse the inequality.
Figure (svg): An inequality with brackets expanded before the solving begins
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.3 Solving Multi-Step Inequalities §6.3, pp. 337-337 — Example 3, Use the Distributive Property
Picture it
Distribute, then two ordinary steps.
Figure (svg): An inequality with brackets expanded before the solving begins
Distributing is the Lesson 2.6 move applied inside an inequality. It never affects the symbol, whatever the sign of the factor outside the brackets.
Worked example
This is Example 3 from the textbook.
\[ \text{Solve } \; 3(x - 2) < -7. \]
Write the original inequality
Why: There are brackets to clear.
\[ 3(x - 2) < -7 \]
Distribute the 3
Why: Three times x is 3x, and three times negative two is negative six.
\[ 3 x - 6 < -7 \]
Add 6 to each side
Why: No reversal.
\[ 3 x < -1 \]
Divide each side by 3
Why: Three is positive; no reversal.
\[ x < -\frac{1}{3} \]
Figure (svg): An inequality with brackets expanded before the solving begins
\[ x < -\tfrac{1}{3} \]
Verify: test a number on each side
Why: Taking x equal to negative one gives three times negative three, which is negative nine, and negative nine is less than negative seven — so negative one is a solution. Taking zero gives negative six, which is not less than negative seven, so zero is correctly excluded.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.3 Solving Multi-Step Inequalities §6.3, pp. 337-337
Faded example
Every term inside the brackets.
Fill in the blanks
3(x - 2) < -7 \;\Longrightarrow\; 3x - 6 < -7 \;\Longrightarrow\; 3x < -1
Why: Distributing gives 3x minus six, and adding six to both sides leaves 3x less than negative one. Dividing by three then gives x less than negative one third, with no reversal anywhere since three is positive.
Worked example
Guided Practice 3 and 4. One has a negative factor outside.
\[ \text{Solve } \; 3(n + 4) \ge 6 \; \text{ and } \; -2(x - 1) < 2. \]
Distribute in the first
Why: Three n plus twelve.
\[ 3 n + 12 \ge 6 \]
Finish it
Why: Subtract twelve, then divide by three.
\[ n \ge - 2 \]
Distribute in the second
Why: Negative two times x is negative 2x; negative two times negative one is positive two.
\[ -2 x + 2 < 2 \]
Finish it
Why: Subtract two, then divide by negative two and reverse.
\[ x > 0 \]
Figure (svg): The solution to Worked example two from guided practice shown as a ladder of expressions, one row per algebraic move
\[ n \ge -2 \qquad x > 0 \]
Verify: notice where the reversal came from
Why: The second problem's negative sign outside the brackets did not reverse anything during the distribution — it only changed the signs inside. The reversal came later, when dividing by negative two. Distributing and dividing are different operations even when the same negative number is involved.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.3 Solving Multi-Step Inequalities §6.3, pp. 337-337
Trap
\[ -2(x - 1) < 2 \]
Distribute and reverse at the same time: -2x + 2 > 2
Why: A negative number is involved, so the reversal rule seems to apply.
Distributing is a multiplication inside one side, not a multiplication of both sides. The symbol is untouched by it, and reversing here gives x less than zero, the wrong half of the line.
\[ -2(x - 1) < 2 \;\Longrightarrow\; -2x + 2 < 2 \;\Longrightarrow\; -2x < 0 \;\Longrightarrow\; x > 0 \]
Reverse only when both sides are multiplied or divided by a negative
Why: Distributing affects one side only, so it cannot change the relationship between the two.
Testing one gives zero, which is less than two, so one is a solution — and only the corrected answer includes it.
Elimination
Expanding -2(x - 1) inside an inequality.
Eliminate the wrong options
Which statement is correct?
Survives elimination: A
Why: Negative two times negative one is positive two, and the symbol is untouched because only one side was changed. The distinction between operating on one side and operating on both is exactly what decides whether the symbol moves.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| Distributing a -2 | Dividing both sides by -2 | |
|---|---|---|
| How many sides change | one | both |
| Does the symbol reverse | no | yes |
| Why | the relationship between the sides is untouched | both sides reflect through zero |
The same negative number appears in both columns and does completely different things. What matters is whether the operation is applied to the relationship or inside one side of it.
Socratic
It multiplies by a negative, after all.
Discussion prompt
Explain why multiplying inside brackets on one side never changes the inequality symbol, while multiplying both sides does. Then say what distributing actually is, in terms of Lesson 2.6.
Hint: Ask what the inequality is comparing.
Answer:
An inequality compares the value of the left side with the value of the right. Distributing rewrites one side as an equivalent expression with the same value, so nothing being compared has changed and the comparison cannot change either. Multiplying both sides changes both values, which is why the comparison can flip.
Distributing is simply the distributive property from Lesson 2.6 — a rule about rewriting an expression, not about equations or inequalities at all. It could be applied to a bare expression with no relation symbol in sight, which is the clearest sign that it cannot affect one.
Section
Section 4
Concept
When the variable appears on both sides, collect it on one side. Choosing the side whose variable term has the greater coefficient leaves a positive number to divide by, so no reversal is needed.
Collecting on the other side is equally correct and requires a reversal at the end.
Figure (svg): Two routes through the same inequality, collecting the variable on each side in turn
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.3 Solving Multi-Step Inequalities §6.3, pp. 337-337 — Example 4, Collect Variable Terms, and its Study Tip
Picture it
One route reverses; one does not.
Figure (svg): Two routes through the same inequality, collecting the variable on each side in turn
Both columns end at the same answer. The right-hand route never produces a negative coefficient, so the reversal question never has to be answered yes.
Worked example
This is Example 4 from the textbook, shown both ways.
\[ \text{Solve } \; 2x - 3 \ge 4x - 1. \]
Method 1: collect on the left
Why: Add three, then subtract 4x.
\[ -2 x \ge 2 \]
Finish method 1
Why: Divide by negative two and reverse.
\[ x \le - 1 \]
Method 2: collect on the right
Why: Subtract 2x, then add one.
\[ -2 \ge 2 x \]
Finish method 2
Why: Divide by two; no reversal, then turn it round.
\[ x \le - 1 \]
Figure (svg): Two routes through the same inequality, collecting the variable on each side in turn
\[ x \le -1 \]
Verify: test the boundary and one number on each side
Why: At negative one both sides give negative five, so the boundary satisfies the or equal to and the answer includes it. At zero the left is negative three and the right is negative one, and negative three is not at least negative one, so zero is correctly excluded.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.3 Solving Multi-Step Inequalities §6.3, pp. 337-337
Sorting
Compare the two coefficients.
Sort into buckets
Sort each inequality by the side to collect the variable on.
Every one of these can be solved either way. Choosing by the coefficients removes the reversal question from the problem entirely, which is worth the two seconds the comparison takes.
Worked example
Guided Practice 5 to 8. Choosing the side each time.
\[ \text{Solve } \; 5n + 21 < 8n, \quad 3z - 15 > -2z, \quad x + 3 \le 2x - 4, \quad 4y - 3 < y + 12. \]
Take the first
Why: Collect on the right, where the coefficient is eight.
\[ 21 < 3 n,\text{ so } n > 7 \]
Take the second
Why: Collect on the left, where three exceeds negative two.
\[ 5 z > 15,\text{ so } z > 3 \]
Take the third
Why: Collect on the right, where the coefficient is two.
\[ 7 \le x,\text{ so } x \ge 7 \]
Take the fourth
Why: Collect on the left, where four exceeds one.
\[ 3 y < 15,\text{ so } y < 5 \]
Figure (svg): The solution to Worked example four from guided practice shown as a ladder of expressions, one row per algebraic move
\[ n > 7, \; z > 3, \; x \ge 7, \; y < 5 \]
Verify: notice that none of the four needed a reversal
Why: In every case the variable was collected on the side with the larger coefficient, so the number to divide by came out positive. The Study Tip's advice removed the reversal question from all four problems at once.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.3 Solving Multi-Step Inequalities §6.3, pp. 337-337
Trap
\[ 5n + 21 < 8n \]
Subtract 8n from both sides, since the variable is usually collected on the left
Why: Putting the variable on the left is a habit from Chapter 3.
\[ -3n + 21 < 0 \;\Longrightarrow\; -3n < -21 \;\Longrightarrow\; n > 7 \]
The answer is right, and the route passed through a negative coefficient and needed a reversal — one more chance to lose the symbol for no gain.
\[ 5n + 21 < 8n \;\Longrightarrow\; 21 < 3n \;\Longrightarrow\; 7 < n \]
Compare the two coefficients first and collect on the larger side
Why: Eight exceeds five, so subtracting 5n leaves a positive coefficient.
Both routes are correct; one of them simply has no step at which the symbol can go wrong.
Faded example
Subtract the smaller variable term.
Fill in the blanks
5n + 21 < 8n \;\Longrightarrow\; 21 < 3n \;\Longrightarrow\; 7 < n
Why: Subtracting 5n from both sides leaves 3n on the right, a positive coefficient, so dividing by three needs no reversal. Turning the result round gives n greater than seven.
Elimination
Both routes reach the same answer.
Eliminate the wrong options
What is the reason for the Study Tip's advice?
Survives elimination: A
Why: Collecting on the larger side leaves a positive coefficient, so the division never triggers a reversal and there is one fewer thing to get wrong. The two routes are equally valid and equally long; one simply removes the step where the symbol can be lost.
Socratic
The two routes look nothing alike in the middle.
Discussion prompt
Explain why collecting on either side must give the same solution set. Then say what it would mean if a student got two different answers from the two routes.
Hint: Think about what each step preserves.
Answer:
Every step in each route is an application of a property of inequality, which produces an equivalent inequality — one with exactly the same solutions. So both routes are chains of equivalences starting from the same inequality, and they must end at the same solution set even though the intermediate lines differ.
Two different answers would mean at least one step was not an equivalence, which in practice means a reversal was missed or applied wrongly. Working both routes is therefore a genuine self-check, and if they disagree, substituting one test number into the original identifies which route went wrong.
Section
Section 5
Concept
Real problems often ask for a minimum or a maximum, which is an inequality rather than an equation. Build the verbal model first, then translate, solve, and interpret the answer in the situation.
A boundary of 537.5 flies means at least 538, since half a fly cannot be sold.
Figure (svg): Income, expenses and profit assembled into an inequality
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.3 Solving Multi-Step Inequalities §6.3, pp. 338-338 — Example 5, Write and Use a Linear Model, on selling fishing flies
Picture it
Income, expenses and a target.
Figure (svg): Income, expenses and profit assembled into an inequality
The fifteen dollars covers the book at 13.95 plus 1.05 for shipping, and it is paid once rather than per fly — which is why it sits outside the term multiplied by x.
Worked example
This is Example 5 from the textbook.
\[ \text{Flies sell for } 0.60 \text{ and cost } 0.20 \text{ each, plus a } 15 \text{ dollar book. How many for a profit of at least } 200? \]
Write the verbal model
Why: Income minus total expenses is at least the desired profit.
Translate
Why: Sixty cents times x, minus twenty cents times x plus fifteen, is at least two hundred.
\[ 0.60 x - (0.20 x + 15) \ge 200 \]
Simplify
Why: Distribute the minus sign and combine like terms.
\[ 0.40 x - 15 \ge 200 \]
Solve
Why: Add fifteen, then divide by four tenths.
\[ x \ge 537.5 \]
Figure (svg): Income, expenses and profit assembled into an inequality
\[ x \ge 537.5 \;\Longrightarrow\; \text{at least } 538 \text{ flies} \]
Verify: check 537 and 538 against the target
Why: At 537 flies the profit is 0.40 times 537 minus 15, which is 199.80 — just short of 200. At 538 it is 200.20, which meets the target. So 538 is genuinely the first whole number that works, and the rounding was not a matter of convention.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.3 Solving Multi-Step Inequalities §6.3, pp. 338-338
Faded example
Income minus expenses, at least the target.
Fill in the blanks
0.60x - (0.20x + 15) \ge 200 \;\Longrightarrow\; 0.40x - 15 \ge 200 \;\Longrightarrow\; x \ge 537.5
Why: The two per-fly amounts combine into a profit of forty cents each, and the fifteen dollar book is subtracted once. Dividing 215 by 0.40 gives 537.5, which the situation rounds up to 538 whole flies.
Worked example
Guided Practice 9 and 10. The same structure, different numbers.
\[ \text{Candles sell for } 2 \text{ and cost } 0.50 \text{ each, plus } 12 \text{ for instructions. How many for a profit of at least } 300? \]
Write the income
Why: Two dollars times the number sold.
\[ 2 x \]
Write the total expenses
Why: Fifty cents each plus twelve dollars once.
\[ 0.50 x + 12 \]
Write and simplify the inequality
Why: Income minus expenses is at least three hundred.
\[ 1.50 x - 12 \ge 300 \]
Solve and interpret
Why: Add twelve, divide by one and a half.
\[ x \ge 208 \]
Figure (svg): The solution to Worked example the candle business shown as a ladder of expressions, one row per algebraic move
\[ 1.50x - 12 \ge 300 \;\Longrightarrow\; x \ge 208 \]
Verify: check that the boundary is a whole number here
Why: One and a half times 208 minus twelve is exactly 300, so 208 candles hits the target precisely and no rounding was needed. That the fly problem needed rounding and this one does not is a property of the numbers rather than of the method.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.3 Solving Multi-Step Inequalities §6.3, pp. 338-338
Trap
\[ x \ge 537.5 \]
Round to the nearest whole number, which is 538
Why: Rounding to the nearest is the automatic response to a decimal answer.
It happens to give the right answer here. Had the boundary been 537.2, the same habit would give 537 — and 537 flies produce a profit of 199.80, which misses the target.
Round in the direction the inequality requires, then check the boundary
Why: At least means the answer must satisfy the inequality, so round up to the first whole number that does.
Checking the two whole numbers either side of the boundary settles it in two lines and never depends on which way the decimal happened to fall.
Elimination
The solution is x greater than or equal to 537.5 flies.
Eliminate the wrong options
How many flies must be sold?
Survives elimination: A
Why: The question asks for the minimum, so the answer is the smallest whole number satisfying the inequality. Checking both candidates settles it: 537 gives 199.80 and 538 gives 200.20, so only 538 meets the target.
Hypothesis
Predict before you compute.
Predict first
If the desired profit rose from 200 dollars to 300, roughly how many more flies would be needed?
Correct: About 250 more, since each fly adds 40 cents of profit.
\[ 0.40x - 15 \ge 300 \;\Longrightarrow\; x \ge 787.5 \;\Longrightarrow\; 788 \text{ flies} \]
Why: The extra hundred dollars divided by forty cents per fly is 250 flies. Only the target changed, so only the right-hand side of the inequality moved, and the coefficient of x — which is the profit per fly — converts that change into a change in the number sold. The slope of the model is what turns one increase into the other, exactly as in Lesson 5.5.
Socratic
Both the materials and the book are expenses.
Discussion prompt
Explain why the twenty cents is multiplied by x and the fifteen dollars is not. Then say what the fifteen dollars corresponds to on the graph of the profit model.
Hint: Ask which cost depends on how many you make.
Answer:
The materials are bought per fly, so making twice as many costs twice as much in materials. The book is bought once whatever happens, so its cost does not depend on x at all — which is exactly the difference between a term multiplied by the variable and a constant term.
On the graph of profit against flies sold, the fifteen dollars is what makes the vertical intercept negative fifteen: with no flies sold you have spent fifteen dollars and earned nothing. The forty cents is the slope. So the two kinds of cost land in the two different roles of Lesson 4.7's slope-intercept form.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| One step (6.1, 6.2) | Multi-step (6.3) | |
|---|---|---|
| Operations needed | one | two or more |
| When to ask about reversing | once | at every step |
| Typical number of reversals | zero or one | still zero or one |
More steps does not mean more reversals. Most steps in a multi-step solve are additions or divisions by positives, and at most one usually turns the symbol.
Pattern
Whether the inequality has brackets, variables on both sides, or both, the same five moves cover it.
Step two is optional and it removes the only step at which the symbol can be lost, so it is worth the two seconds the comparison takes.
OpenStax Elementary Algebra 2e, §2.7 Solve Linear Inequalities §2.7
Check
Addition first, then division.
Check your understanding
Solve 4x - 7 <= 5.
Answer: A
Why: Adding seven gives 4x at most twelve, and dividing by four gives x at most three. Four is positive, so no reversal is needed. Testing zero gives negative seven, which is at most five, and zero is at most three — consistent.
Check
Ask the question at each step.
Check your understanding
How many steps reverse the symbol when solving -2x + 5 > 1?
Answer: A
Why: Subtracting five leaves negative 2x greater than negative four with no reversal, and dividing by negative two reverses it to x less than two. Only the division involves a negative multiplier.
Check
Compare the coefficients.
Check your understanding
For 3x - 4 < 7x + 8, which side avoids a reversal?
Answer: A
Why: Subtracting 3x from both sides leaves 4x on the right, a positive coefficient, so dividing by four needs no reversal. Collecting on the left would leave negative 4x and force one.
Real world
A stall at a market charges 45 dollars for the pitch. You sell jars of jam at 6 dollars each, and the ingredients cost 2 dollars 50 a jar.
Discussion prompt
Write and solve an inequality for the number of jars needed to make a profit of at least 100 dollars, and interpret the answer. Then say what the answer would be if the pitch fee doubled.
Hint: Profit is income minus expenses, and one cost is per jar while the other is not.
Answer:
\[ 6x - (2.5x + 45) \ge 100 \;\Longrightarrow\; 3.5x - 45 \ge 100 \;\Longrightarrow\; x \ge 41.43 \]
So at least 42 jars, since a fraction of a jar cannot be sold and 41 jars would give a profit of 98 dollars 50 — just short. Checking both whole numbers either side of the boundary is what confirms the rounding rather than guessing at it.
Doubling the pitch fee to 90 dollars gives 3.5x at least 190, so x at least 54.29, meaning 55 jars. The extra 45 dollars of fixed cost divided by the 3 dollars 50 profit per jar is about 13 more jars, which matches — the fixed cost and the per-jar profit play exactly the roles of intercept and slope from Lesson 4.7.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
Does distributing the -2 in -2(x - 1) < 2 reverse the inequality symbol?
Correct: No, since only one side is being rewritten.
\[ -2(x - 1) < 2 \;\Longrightarrow\; -2x + 2 < 2 \;\Longrightarrow\; -2x < 0 \;\Longrightarrow\; x > 0 \]
Why: Distributing rewrites the left side as an equivalent expression with the same value, so nothing being compared has changed and the comparison cannot change either. Reversing requires multiplying or dividing both sides. The reversal in this problem comes later, when dividing by negative two — so the symbol does turn once, just not here. Testing x equal to one gives zero, which is less than two, and only the correct route accepts it.
Explain it
They can solve one-step inequalities and are flipping the symbol once per problem rather than once per step.
Discussion prompt
In no more than four sentences, explain how the reversal rule applies in a multi-step problem. Then give them the trick that removes the question from most problems entirely.
Hint: Per step, not per problem.
Answer:
A usable answer: ask the question fresh at every line — is the number I am multiplying or dividing both sides by negative? Most steps are additions or divisions by positives, so the answer is usually no, and a typical problem reverses once or not at all. Distributing a negative inside brackets is not one of these steps, since it changes only one side.
The trick is that when the variable appears on both sides, collect it on whichever side already has the bigger coefficient. Then the number you end up dividing by is positive, and there is no reversal to remember at all.
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: Per-step decisions are fixed by writing the answer beside each line, even when it is no. Distributing is fixed by remembering that one side changing cannot change a comparison. The side to collect on is fixed by comparing the two coefficients before anything else. Rounding is fixed by testing both whole numbers either side of the boundary against the original requirement. 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
Down the left of a page solve three inequalities of increasing difficulty: a two-step one, one with brackets, and one with the variable on both sides. Beside every line of every solve, write the operation used and whether it reversed the symbol, including the lines where the answer is no. For the third problem, solve it a second time collecting the variable on the other side, and box both final answers to show they match, writing beside them which route needed a reversal. In the lower half, build a profit model of your own with one per-item cost and one fixed cost, solve it for a target of your choosing, and check the two whole numbers either side of your boundary against that target, marking which one is the answer. Finally, in the margin, write the one question that decides every reversal and the one exception to it.
Your two boxed answers must be identical. If they are not, substitute one test number into the original — whichever route rejects a genuine solution is the one where the symbol was lost.
Recap
Five things, and the second is where the symbols go missing.
| If the question says | Your first move is |
|---|---|
| Solve 2y - 5 < 7 | Add 5, then divide by 2 |
| There are brackets | Distribute; the symbol does not move |
| The variable is on both sides | Compare coefficients, collect on the larger |
| The answer is a decimal number of items | Test both whole numbers either side |
| A profit of at least | Income minus expenses, with >= |
Lesson 6.4 puts two inequalities together. When a quantity has to satisfy both a lower and an upper bound at once, the solution is a bounded interval rather than a ray, and the two conditions are solved side by side.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.3 Solving Multi-Step Inequalities §6.3, pp. 336-341 — everything on these slides traces back here
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