Multiplication and division as inverse operations, the multiplication and division properties of equality, dividing by a coefficient or equivalently multiplying by its reciprocal, the sign change caused by a negative coefficient, fractional coefficients, and modelling equal shares of a total.
Subject: Algebra 1 · 65 slides · symbolic lesson
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Title
Algebra 1 · Chapter 3 — Solving Linear Equations
Solving Equations Using Multiplication and 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. 3 Solving Linear Equations — Lesson 3.2 Solving Equations Using Multiplication and Division §3.2, pp. 138-143 — the lesson these objectives are drawn from
Warm-up
Lesson 3.1 used one pair of inverse operations. This lesson uses the other pair, and everything else is unchanged.
Discussion prompt
Lesson 3.1 solved x plus 6 equals 10 by subtracting 6 from both sides. What would you do to 6x equals 10, and why is it not subtraction?
Hint: Look at what is actually being done to the variable.
Answer:
\[ 6x = 10 \;\Longrightarrow\; x = \tfrac{10}{6} = \tfrac{5}{3} \]
Six is multiplying x rather than being added to it, so subtracting six would remove nothing. The inverse of multiplying by six is dividing by six, and applying that to both sides isolates x. The method is identical to Lesson 3.1 — name the operation, apply its inverse to both sides — with a different pair of inverses.
Concept
Multiplication and division are inverse operations. You can use multiplication to undo division and division to undo multiplication, applying the move to both sides exactly as before.
properties of equality — The rules permitting both sides of an equation to be multiplied by the same number, or divided by the same nonzero number, without changing the solution.
\[ \tfrac{x}{2} = 3 \;\Longrightarrow\; x = 6 \qquad 4x = 12 \;\Longrightarrow\; x = 3 \]
One condition is new: you may multiply by anything, but you may never divide by zero.
Figure (svg): A table showing an equation solved by multiplying and another by dividing
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.2 Solving Equations Using Multiplication and Division §3.2, pp. 138-138
Section
Section 1
Concept
When a number multiplies the variable, dividing both sides by that number isolates it. The coefficient and the divisor cancel by the inverse property of multiplication.
The divisor must not be zero, which is why an equation of the form zero times x equals something needs separate treatment.
Figure (svg): The equation 4x equals 1 solved by dividing both sides by 4
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.2 Solving Equations Using Multiplication and Division §3.2, pp. 138-138 — Example 1, Divide Each Side of an Equation
Picture it
The fours cancel on the left and the division stands on the right.
Figure (svg): The equation 4x equals 1 solved by dividing both sides by 4
The answer is a fraction, and that is perfectly normal. Lesson 1.4's mental-math method could not reach it, which is exactly why a written division is worth having.
Worked example
This is Example 1 from the textbook. The answer is a fraction rather than a whole number.
\[ \text{Solve } \; 4x = 1. \]
Name the operation on the variable
Why: Four is multiplying x, so the operation is multiplication.
\[ \text{multiplication by } 4 \]
Apply the inverse to both sides
Why: Divide each side by four, which undoes the multiplication.
\[ 4 x / 4 = \frac{1}{4} \]
Simplify the left side
Why: Four divided by four is one, and one times x is x, by the inverse and identity properties.
\[ x = \frac{1}{4} \]
State the solution
Why: One quarter, which is a perfectly ordinary answer.
\[ x = \frac{1}{4} \]
Figure (svg): The equation 4x equals 1 solved by dividing both sides by 4
\[ 4x = 1 \;\Longrightarrow\; x = \tfrac{1}{4} \]
Verify: substitute one quarter into the original
Why: Four times one quarter is one, which matches the right side, so the statement is true. A fractional solution checks exactly as readily as a whole-number one — there is nothing second-class about it.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.2 Solving Equations Using Multiplication and Division §3.2, pp. 138-138
Discrimination
Read what has been done to the variable in each equation.
Sort into buckets
Sort each equation by which pair of inverse operations it needs.
Worked example
Same move each time, with different coefficients.
\[ \text{Solve } \; 7x = 42, \quad 5y = -35, \quad 6n = 15. \]
Divide the first by 7
Why: Forty-two over seven is six.
\[ x = 6 \]
Divide the second by 5
Why: Negative thirty-five over five has opposite signs, so the quotient is negative seven.
\[ y = -7 \]
Divide the third by 6
Why: Fifteen over six simplifies to five halves.
\[ n = \frac{5}{2} \]
Note the three kinds of answer
Why: A whole number, a negative and a fraction — all ordinary outcomes of the same move.
\[ 6, -7, \frac{5}{2} \]
Figure (svg): The solution to Worked example three more divisions shown as a ladder of expressions, one row per algebraic move
\[ x = 6, \quad y = -7, \quad n = \tfrac{5}{2} \]
Verify: substitute each answer back
Why: Seven times six is forty-two; five times negative seven is negative thirty-five; six times five halves is fifteen. All three originals come out true, and the third confirms that a fraction is a real answer rather than an unfinished one.
Trap
\[ 4x = 12 \]
Subtract 4 from both sides, since Lesson 3.1 removed the number by subtracting
Why: The previous lesson's move is the one most recently practised, so it carries over.
\[ 4x - 4 = 8 \quad \text{(wrong and no progress)} \]
Subtracting removes an addition, and there is no addition here. The four is multiplying x, so nothing has been undone.
\[ \tfrac{4x}{4} = \tfrac{12}{4} \;\Longrightarrow\; x = 3 \]
Name the operation before choosing the inverse
Why: Multiplication is undone by division, not by subtraction. Reading the equation aloud — four times x — makes the operation obvious.
The whole method from Lesson 3.1 carries over unchanged. Only the pair of inverses is different, and identifying which pair applies is the first step every time.
Faded example
The move is chosen. Carry it out on both sides.
Fill in the blanks
4x = 1 \;\rightarrow\; \tfrac41/4 = \tfrac______} \;\rightarrow\; x = ___
Why: Dividing both sides by four cancels the coefficient on the left and leaves one quarter on the right. The two blanks are the same number because the property of equality requires the identical divisor on each side, and writing it explicitly on both is what keeps the balance visible.
Elimination
The equation is 7n equals 42.
Eliminate the wrong options
Which move isolates n in one step?
Survives elimination: A
Why: Seven is multiplying n, so dividing both sides by seven undoes it, giving n equal to six. Option D is worth noticing: it keeps the balance and is therefore a valid transformation, but it divides by the wrong number and so makes no progress towards isolating the variable.
Socratic
The cancellation is two properties from Chapter 2 doing their work.
Discussion prompt
Explain why dividing 4x by 4 leaves x, naming the two properties involved. Then say what the analogous argument was in Lesson 3.1.
Hint: One property is about a number and its reciprocal; the other is about one.
Answer:
\[ \tfrac{4x}{4} = \tfrac{1}{4}(4x) = \left(\tfrac{1}{4} \cdot 4\right)x = 1 \cdot x = x \]
The division rule turns dividing by four into multiplying by one quarter, the associative property regroups so that four and one quarter multiply first, the inverse property of multiplication makes that pair one, and the multiplicative identity leaves x alone.
In Lesson 3.1 the same shape of argument used the additive versions: associativity regrouped, the additive inverse gave zero, and the additive identity left the variable alone. The two lessons are the same argument with the two different pairs of properties.
Section
Section 2
Concept
When the variable is divided by a number, multiplying both sides by that number isolates it. This is the same relationship as before, used in the other direction.
\[ \tfrac{x}{2} = 3 \;\Longrightarrow\; 2 \cdot \tfrac{x}{2} = 2 \cdot 3 \;\Longrightarrow\; x = 6 \]
A fraction bar under a variable is a division, so an expression such as x over two is asking for a multiplication to undo it.
Figure (svg): A table showing an equation solved by multiplying and another by dividing
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.2 Solving Equations Using Multiplication and Division §3.2, pp. 138-138 — the Transforming Equations box
Picture it
One equation is undone by multiplying and one by dividing.
Figure (svg): A table showing an equation solved by multiplying and another by dividing
Deciding which row you are in takes a second: look at whether the number is multiplying the variable or dividing it, and do the opposite.
Worked example
The fraction bar is a division, so the inverse is a multiplication.
\[ \text{Solve } \; \tfrac{x}{2} = 3. \]
Name the operation on the variable
Why: x is being divided by two.
\[ \text{division by } 2 \]
Apply the inverse to both sides
Why: Multiply each side by two.
\[ 2 \cdot(\frac{x}{2}) = 2 \cdot 3 \]
Simplify the left side
Why: The two and the division by two cancel, leaving x.
\[ x = 6 \]
State the solution
Why: Six.
\[ x = 6 \]
Figure (svg): The solution to Worked example solve x over 2 equals 3 shown as a ladder of expressions, one row per algebraic move
\[ \tfrac{x}{2} = 3 \;\Longrightarrow\; x = 6 \]
Verify: substitute 6 into the original
Why: Six divided by two is three, matching the right side. The answer is larger than the right-hand side, which is what undoing a division should produce.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.2 Solving Equations Using Multiplication and Division §3.2, pp. 138-138
Sorting
Read what is being done to the variable, then do the opposite.
Sort into buckets
Sort each equation by the move that isolates the variable.
The three pairs use the same two numbers arranged differently and need opposite moves. Reading whether the number is above or below the fraction bar is the entire decision.
Worked example
Same move, with a negative and a decimal among them.
\[ \text{Solve } \; \tfrac{y}{5} = -4, \quad \tfrac{n}{3} = 7, \quad \tfrac{t}{10} = 0.6. \]
Multiply the first by 5
Why: Five times negative four is negative twenty.
\[ y = -20 \]
Multiply the second by 3
Why: Three times seven is twenty-one.
\[ n = 21 \]
Multiply the third by 10
Why: Ten times six tenths is six.
\[ t = 6 \]
Note the direction of every answer
Why: Each answer is larger in absolute value than the right-hand side, since a division was being undone.
Figure (svg): The solution to Worked example three more multiplications shown as a ladder of expressions, one row per algebraic move
\[ y = -20, \quad n = 21, \quad t = 6 \]
Verify: substitute each answer back
Why: Negative twenty over five is negative four; twenty-one over three is seven; six over ten is 0.6. All three check, and the negative one confirms that the sign is carried through the multiplication unchanged.
Error analysis
The student solved three equations. Two are wrong.
Annotate
On: \( \tfrac{x}{2} = 3 \;\rightarrow\; x = \tfrac{3}{2} \qquad 4x = 12 \;\rightarrow\; x = 3 \qquad \tfrac{y}{5} = -4 \;\rightarrow\; y = -\tfrac{4}{5} \)
Both errors are caught by a size check. Undoing a division must make the answer larger in absolute value, and both wrong answers came out smaller than the right-hand side.
Prediction
Predicting the direction is a check on the move as well as the arithmetic.
Predict first
In the equation x over 3 equals 12, will the solution be larger or smaller than 12?
Correct: Larger, because a division is being undone.
\[ \tfrac{x}{3} = 12 \;\Longrightarrow\; x = 36 \]
Why: The equation says that x divided by three is twelve, so x must be three times twelve, which is thirty-six. Undoing a division multiplies, and multiplying by a number above one increases the size. Predicting this before computing catches the commonest error here, which is dividing again instead of multiplying.
Faded example
The move is chosen. Carry it out on both sides.
Fill in the blanks
\tfrac22 = 3 \;\rightarrow\; 6 \cdot \tfrac______ = ___ \cdot 3 \;\rightarrow\; x = ___
Why: Multiplying both sides by two cancels the division on the left and doubles the right side to six. The first two blanks are the same number because the property of equality demands the identical factor on both sides, and writing it twice is what keeps the balance explicit.
Socratic
The move depends on reading the notation correctly.
Discussion prompt
Explain why x over 2 means x divided by 2 rather than x multiplied by 2, referring back to Lesson 1.1. Then say what the equation would look like if x really were multiplied by two, and how the required move would differ.
Hint: Lesson 1.1 listed the four operations and how algebra writes each.
Answer:
Lesson 1.1 established that a fraction bar is a division sign written vertically, so x over two is x divided by two. Multiplication by two would be written as 2x, with the number against the letter and no bar at all.
So x over two equals three needs a multiplication by two to undo it, giving six, while 2x equals three needs a division by two, giving three halves. The two answers differ by a factor of four, and the only thing distinguishing the equations on the page is whether the two sits above or below the line.
Section
Section 3
Concept
The multiplication property of equality says you may multiply both sides by any number. The division property says you may divide both sides by any nonzero number. That single word nonzero is the only asymmetry between them.
Figure (svg): The multiplication and division properties of equality stated in symbols
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.2 Solving Equations Using Multiplication and Division §3.2, pp. 138-139 — the properties of equality named in the lesson's key words
Picture it
Almost identical, and one carries a condition.
Figure (svg): The multiplication and division properties of equality stated in symbols
The condition on division comes straight from Lesson 2.8: zero has no reciprocal, so dividing by it is undefined rather than merely unhelpful.
Worked example
Every step in a solution is licensed by one of these properties.
\[ \text{Name the property used in each step of solving } \; 4x = 12. \]
Divide both sides by 4
Why: This is the division property of equality, and four is nonzero so the property applies.
Simplify the left side
Why: Four over four is one by the inverse property of multiplication.
Recognise 1 times x as x
Why: The multiplicative identity leaves x unchanged.
Read off the solution
Why: Twelve over four is three.
\[ x = 3 \]
Figure (svg): The solution to Worked example name the property used shown as a ladder of expressions, one row per algebraic move
\[ x = 3, \text{ by the division property of equality} \]
Verify: check that the divisor was allowed
Why: The divisor was four, which is not zero, so the division property applies. Had the coefficient been zero the equation would have read zero equals twelve, which has no solution at all — a case worth recognising rather than attempting to divide.
Elimination
Three of these are legal transformations of an equation.
Eliminate the wrong options
Which move is NOT permitted?
Survives elimination: A
Why: Division by zero is undefined, so it is not a transformation at all rather than merely a bad one. Note the asymmetry: multiplying by zero is permitted and merely destroys information, while dividing by zero is not permitted at any point.
Worked example
Legal moves are not always useful, and this is the extreme case.
\[ \text{What happens if you multiply both sides of } 4x = 12 \text{ by zero?} \]
Apply the multiplication property with c equal to zero
Why: The property permits any number, including zero.
\[ 0 \cdot 4 x = 0 \cdot 12 \]
Simplify both sides
Why: By the property of zero, both sides become zero.
\[ 0 = 0 \]
Judge the result
Why: The statement is true but says nothing about x — every number satisfies it.
State the lesson
Why: Multiplying by zero is legal and destroys all the information in the equation, so it is never a useful move.
Figure (svg): The solution to Worked example multiplying by zero shown as a ladder of expressions, one row per algebraic move
\[ 0 \cdot 4x = 0 \cdot 12 \;\Longrightarrow\; 0 = 0 \]
Verify: check whether the move was reversible
Why: It was not: from zero equals zero there is no way back to the original equation, because dividing by zero to undo it is undefined. That irreversibility is exactly why information was lost, and it is the reason every useful move in this chapter is reversible.
Trap
\[ 4x = 12x \]
Divide both sides by x to remove it
Why: Dividing by the coefficient worked before, and x looks like a coefficient here.
\[ 4 = 12 \quad \text{(false, and a solution has been lost)} \]
The division property requires a nonzero divisor, and x might be zero — in fact zero is the solution of this equation. Dividing by x threw it away.
\[ 4x = 12x \;\Longrightarrow\; 4x - 12x = 0 \;\Longrightarrow\; -8x = 0 \;\Longrightarrow\; x = 0 \]
Move the variable terms to one side rather than dividing by the variable
Why: Subtraction is always safe; division by something that might be zero is not.
Never divide both sides by an expression containing the variable unless you know it cannot be zero. Lesson 3.4 handles equations of this shape properly.
Sorting
For the equation 5x equals 20, judge each proposed move.
Sort into buckets
Sort each move by whether it is legal and whether it makes progress.
Four moves are legal and useless, one is legal and useful, and one cannot be done at all. Legality is the minimum bar and usefulness is the actual criterion.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| Multiplication property | Division property | |
|---|---|---|
| Statement | if a = b then ac = bc | if a = b then a/c = b/c |
| Condition on c | none — any number | c must not be zero |
| Why the condition | not needed | zero has no reciprocal |
The condition traces directly back to Lesson 2.8. Dividing by zero would mean multiplying by the reciprocal of zero, and no such number exists.
Socratic
It looks like an ordinary division and it is not.
Discussion prompt
Explain why dividing both sides of an equation by x can lose a solution, using the equation 4x equals 12x. Then give a safe alternative move that reaches the same answer.
Hint: Ask what value of x would make the divisor zero.
Answer:
Dividing by x is only permitted when x is not zero, and here zero is the solution — substituting it gives zero equals zero, which is true. So dividing by x quietly assumes away the very answer being looked for, and the resulting statement four equals twelve is simply false.
The safe move is to subtract twelve x from both sides, giving negative eight x equals zero, and then divide by negative eight, which is a definite nonzero number. That reaches x equals zero without ever dividing by something whose value is unknown, and it is the method Lesson 3.4 uses throughout.
Section
Section 4
Concept
A negative coefficient is divided by exactly like a positive one, and the sign rule from Lesson 2.8 decides the sign of the answer. A fractional coefficient is usually handled by multiplying by its reciprocal instead.
\[ -3x = 12 \;\Longrightarrow\; x = -4 \qquad \tfrac{2}{3}x = 8 \;\Longrightarrow\; x = 12 \]
Dividing by a fraction is legal but awkward; multiplying by its reciprocal is the same move in an easier form.
Figure (svg): Two columns contrasting dividing by a positive coefficient with dividing by a negative one
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.2 Solving Equations Using Multiplication and Division §3.2, pp. 138-139 — the Study Tip on being careful with signs
Picture it
The move is identical; only the sign of the divisor differs.
Figure (svg): Two columns contrasting dividing by a positive coefficient with dividing by a negative one
Dividing twelve by negative three gives negative four, since the two signs differ. The sign rule does the work, and there is nothing extra to remember about equations specifically.
Worked example
The sign of the answer follows from Lesson 2.8's quotient rule.
\[ \text{Solve } \; -3x = 12 \; \text{ and } \; -7y = -35. \]
Divide the first by the whole coefficient
Why: Divide both sides by negative three, sign included.
\[ 12 \div(-3) \]
Apply the quotient sign rule
Why: Opposite signs, so the quotient is negative four.
\[ x = -4 \]
Divide the second by negative 7
Why: Negative thirty-five over negative seven.
\[ -35 \div(-7) \]
Apply the sign rule again
Why: Same signs, so the quotient is positive five.
\[ y = 5 \]
Figure (svg): The solution to Worked example a negative coefficient shown as a ladder of expressions, one row per algebraic move
\[ -3x = 12 \;\Longrightarrow\; x = -4 \qquad -7y = -35 \;\Longrightarrow\; y = 5 \]
Verify: substitute both answers back
Why: Negative three times negative four is twelve; negative seven times five is negative thirty-five. Both originals come out true, and the second shows that a negative coefficient does not force a negative answer — the two signs together decide.
Prediction
The two signs together decide, exactly as in Lesson 2.8.
Predict first
In the equation negative 6x equals 24, will the solution be positive or negative?
Correct: Negative, since the signs differ.
\[ -6x = 24 \;\Longrightarrow\; x = \tfrac{24}{-6} = -4 \]
\[ \text{but } -6x = -24 \;\Longrightarrow\; x = 4 \]
Why: Twenty-four is positive and the coefficient negative six is negative, so the quotient has opposite signs and is negative four. Neither sign alone decides — it is the comparison of the two that does, which is why option C reaches the right answer by faulty reasoning and would fail on an equation with a negative right-hand side.
Worked example
Multiplying by the reciprocal is the same move as dividing, in an easier form.
\[ \text{Solve } \; \tfrac{2}{3}x = 8. \]
Identify the coefficient
Why: Two thirds is multiplying x.
\[ \text{coefficient } \frac{2}{3} \]
Choose the reciprocal rather than dividing
Why: Dividing by two thirds is the same as multiplying by three halves, and the multiplication is easier to write.
\[ \text{multiply by } \frac{3}{2} \]
Multiply both sides
Why: Three halves times two thirds is one on the left; three halves times eight is twelve on the right.
\[ x = 12 \]
State the solution
Why: Twelve.
\[ x = 12 \]
Figure (svg): An equation with a fractional coefficient solved by multiplying by the reciprocal
\[ \tfrac{2}{3}x = 8 \;\Longrightarrow\; x = 12 \]
Verify: substitute 12 into the original
Why: Two thirds of twelve is eight, matching the right side. The answer is larger than eight, which is what multiplying by a fraction below one requires — you need more than eight to end up with eight after taking two thirds of it.
Trap
\[ -3x = 12 \]
Divide both sides by 3 and put the minus sign back afterwards
Why: The digit looks like the coefficient and the sign feels like a separate decoration.
\[ x = 4 \text{ or } x = -4? \quad \text{unclear} \]
Dividing by three alone gives negative x equals four, which still is not solved — the variable carries a minus sign. Half the move was done.
\[ \tfrac{-3x}{-3} = \tfrac{12}{-3} \;\Longrightarrow\; x = -4 \]
Divide by the whole coefficient, minus sign included
Why: The sign belongs to the coefficient, so it goes into the divisor along with the digits.
Then the quotient sign rule from Lesson 2.8 does the rest: twelve over negative three has opposite signs, so the answer is negative four.
Sorting
Compare the sign of the coefficient with the sign of the right-hand side.
Sort into buckets
Sort each equation by the sign of its solution.
Three of the six have negative coefficients and they split evenly between the two columns. Only the comparison of the two signs decides, never either sign on its own.
Faded example
The reciprocal is the multiplier. Complete both sides.
Fill in the blanks
\tfrac3/23/2x = 8 \;\rightarrow\; 12 \cdot \tfrac______x = ___ \cdot 8 \;\rightarrow\; x = ___
Why: Multiplying by three halves cancels the two thirds on the left by the inverse property, and on the right three halves of eight is twelve. Choosing the reciprocal rather than dividing by a fraction is what keeps the arithmetic to a single multiplication.
Elimination
The equation is three quarters of x equals 9.
Eliminate the wrong options
Which move solves it most directly?
Survives elimination: A
Why: Multiplying by four thirds cancels the three quarters in one move and gives x equal to twelve. Options B and C reach the same answer by longer routes, which is worth noticing: several correct methods usually exist, and choosing among them is about effort rather than validity.
Section
Section 5
Concept
The commonest real situation producing a multiplication equation is a total divided into equal shares. The number of shares multiplies the size of one share to give the total.
Figure (svg): A pile of newspapers divided into equal bundles, modelled as a multiplication equation
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.2 Solving Equations Using Multiplication and Division §3.2, pp. 142-142 — the newspaper-bundle exercise the lesson opens with
Picture it
Five equal bundles making up one total weight.
Figure (svg): A pile of newspapers divided into equal bundles, modelled as a multiplication equation
One division recovers the weight of a single bundle. The same shape covers sharing a bill, cutting a length into equal pieces, and packing items into equal boxes.
Worked example
A pile of newspapers weighing 145 pounds is split into 5 equal bundles.
\[ \text{Find the weight } b \text{ of one bundle, given } \; 5b = 145. \]
Define the variable
Why: Let b be the weight of one bundle in pounds.
\[ b =\text{ pounds per bundle} \]
Write the verbal model
Why: The number of bundles times the weight of one bundle equals the total weight.
Translate and solve
Why: Five b equals 145; divide both sides by five.
\[ b = \frac{145}{5} \]
Compute and answer in words
Why: Twenty-nine pounds per bundle.
\[ 29\text{ pounds} \]
Figure (svg): A pile of newspapers divided into equal bundles, modelled as a multiplication equation
\[ 5b = 145 \;\Longrightarrow\; b = 29 \text{ pounds} \]
Verify: rebuild the total from the answer
Why: Five bundles at twenty-nine pounds each is 145 pounds, which is the total the problem gave. Recovering a number that was given rather than one you computed is the strongest available check.
Translation
Four situations. Let x be the unknown in each.
Match the pairs
Why: The first three all have the shape count times size equals total, and each is solved by one division. The fourth has a fractional coefficient and is solved by multiplying by the reciprocal, but its shape is identical — a multiple of the unknown equalling a known total.
Worked example
The same equation shape appears whenever a rate multiplies a quantity.
\[ \text{A car uses fuel at } 8 \text{ litres per } 100 \text{ km. It used } 46 \text{ litres. How far did it travel, in hundreds of km?} \]
Define the variable
Why: Let h be the number of hundreds of kilometres travelled.
\[ h =\text{ hundreds of } \text{km} \]
Write the verbal model
Why: The rate per hundred kilometres times the number of hundreds equals the total fuel used.
Translate
Why: Eight h equals forty-six.
\[ 8 h = 46 \]
Solve and answer
Why: Divide both sides by eight: 46 over 8 is 5.75, so the car travelled 575 kilometres.
\[ 575 \text{km} \]
Figure (svg): The solution to Worked example a rate rather than a share shown as a ladder of expressions, one row per algebraic move
\[ 8h = 46 \;\Longrightarrow\; h = 5.75, \text{ so } 575 \text{ km} \]
Verify: check the answer against the rate
Why: At eight litres per hundred kilometres, 575 kilometres uses 5.75 times eight, which is forty-six litres — the figure given. And the answer is a decimal rather than a whole number, which is entirely ordinary for a real measurement.
Trap
\[ 5b = 145 \;\rightarrow\; b = \tfrac{5}{145} \]
Divide the two numbers in the order they appear in the equation
Why: Both numbers are present and one has to go on top, and the leftmost is the obvious candidate.
Five over 145 is about 0.034 pounds, which would make five bundles weigh about one sixth of a pound rather than 145.
\[ b = \tfrac{145}{5} = 29 \]
Divide both sides by the coefficient, so the total goes on top
Why: The property of equality says divide each side by five, and the right side is 145 — so 145 is the number being divided.
A size check settles it instantly: one bundle must weigh less than the whole pile but not vastly less, and twenty-nine pounds is a fifth of 145 as it should be.
Elimination
A bill of 84 dollars is split equally between 6 people. Let x be each person's share.
Eliminate the wrong options
Which equation is correct?
Survives elimination: A
Why: Six shares of x dollars make up the 84 dollar total, so six x equals 84 and each share is fourteen dollars. Every wrong option fails a size check: a share of a bill between six people must be smaller than the bill and larger than nothing, and only one option produces such an answer.
Estimation
A rough answer first is a check on the exact one.
Predict first
A 145 pound pile is split into 5 equal bundles. Roughly how heavy is one bundle?
Correct: About 30 pounds.
\[ \tfrac{145}{5} = 29 \text{ pounds} \]
Why: One fifth of 145 is a little under thirty, since five thirties would be 150. The exact answer is twenty-nine. The three wrong options correspond to multiplying instead of dividing, dividing by fifty, and reporting the total — and an estimate rules out all three before any exact arithmetic.
Socratic
Choosing what the variable stands for shapes the whole equation.
Discussion prompt
In the newspaper problem you could let b be the weight of one bundle or let n be the number of bundles. Write the equation each choice produces, and say which is solvable from the information given and why.
Hint: Count how many quantities each version leaves unknown.
Answer:
\[ \text{letting } b \text{ be the bundle weight: } 5b = 145, \text{ solvable} \]
\[ \text{letting } n \text{ be the number of bundles: } n \cdot ? = 145, \text{ not solvable} \]
The second version leaves two unknowns — the number of bundles and the weight of each — because the problem told you the count and asked for the weight. Letting the letter stand for something you were given wastes the information and leaves the equation short.
The general rule is that the letter goes on the quantity the question asks for, and every other quantity should have a number against it. Counting the letters in the labels table, as Lesson 1.6 recommended, is what detects this before any solving is attempted.
Comparison
Fill the blanks from memory before you scroll back. The method is identical; only the pair of inverses differs.
Comparison matrix
| Lesson 3.1 | Lesson 3.2 | |
|---|---|---|
| Operation on the variable | added or subtracted | multiplied or divided |
| Inverse move | subtract or add | divide or multiply |
| Condition on the move | none | never divide by zero |
Only the bottom row differs in kind. Everything else about the method — name the operation, apply the inverse to both sides, check in the original — carries across unchanged.
Pattern
Whether the coefficient is a whole number, a negative or a fraction, the same five moves cover it.
Step four says sign included. Dividing by the digits alone leaves the variable carrying a minus sign, which means the equation is not yet solved.
Check
A multiplication to undo. Divide by the coefficient.
Check your understanding
Solve 9x equals 63.
Answer: A
Why: Nine multiplies x, so divide both sides by nine: sixty-three over nine is seven. Substituting confirms it, since nine sevens are sixty-three.
Check
A negative coefficient. Include the sign in the divisor.
Check your understanding
Solve negative 5y equals 40.
Answer: A
Why: Dividing both sides by negative five gives forty over negative five, and the two signs differ, so the quotient is negative eight. Substituting confirms it: negative five times negative eight is forty.
Check
A division to undo. Multiply by the denominator.
Check your understanding
Solve n divided by 4 equals negative 3.
Answer: A
Why: n is divided by four, so multiply both sides by four: four times negative three is negative twelve. Substituting confirms it, since negative twelve over four is negative three. Undoing a division makes the answer larger in absolute value.
Real world
A recipe for 6 servings needs 750 grams of flour. You want to make 10 servings, and separately you want to know the flour per serving.
Discussion prompt
Write and solve an equation for the flour per serving, then use it to find the flour needed for 10 servings. Say which of the two steps used a division property and which used a multiplication property, and why the second answer is larger than 750.
Hint: Find the per-serving amount first, then scale it up.
Answer:
\[ 6f = 750 \;\Longrightarrow\; f = 125 \text{ grams per serving} \]
\[ 10 \cdot 125 = 1250 \text{ grams for ten servings} \]
The first step divides both sides by six, using the division property of equality, and the second multiplies a known per-serving amount by ten, which is ordinary arithmetic rather than a property of equality — there is no equation being transformed there.
The second answer exceeds 750 because ten servings is more than six. Scaling a recipe is exactly the fixed-rate shape from Lesson 1.1, and the per-unit amount is what makes any scaling a single multiplication.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
To solve 6x equals 10, do you subtract 6 or divide by 6?
Correct: Divide by 6, because division undoes multiplication.
\[ \tfrac{6x}{6} = \tfrac{10}{6} \;\Longrightarrow\; x = \tfrac{5}{3} \]
\[ \text{check: } 6 \cdot \tfrac{5}{3} = 10 \;\checkmark \]
Why: The six is multiplying x, so only division undoes it. Subtracting six gives 6x minus 6 equals 4, which still has a coefficient on the variable and is no closer to a solution. The move is always determined by the operation actually present, and reading that operation correctly is the first step of every problem in this chapter.
Explain it
They have just learned to solve equations by adding and subtracting.
Discussion prompt
In no more than four sentences, explain when to divide rather than subtract, and how to tell which situation you are in. Then give them the one extra caution that applies to division and does not apply to subtraction.
Hint: The caution is about a particular number.
Answer:
A usable answer: look at how the number is attached to the letter. If there is a plus or minus sign between them, add or subtract; if the number is written against the letter or under it, multiply or divide. Six plus x needs a subtraction, and 6x needs a division.
The extra caution is never to divide by zero, and never to divide by anything whose value you do not know — an expression containing the letter might be zero, and dividing by it can throw away a solution. Subtraction has no such restriction, which is why moving terms is always safe when dividing is 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: Telling the types apart is fixed by looking for a plus or minus sign between the number and the letter. Multiply-or-divide is fixed by asking whether the number sits above or below a fraction bar. Negative coefficients are fixed by writing the whole coefficient, sign included, as the divisor. Fractional coefficients are fixed by multiplying by the reciprocal rather than dividing by a fraction. 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 make a two-column table of the four ways a number can be attached to a variable — added, subtracted, multiplying, dividing — and beside each write the inverse move. Underneath, solve four equations, one of each kind, writing the move on both sides explicitly and showing the substitution check against the original. In the middle, write the two properties of equality in symbols and circle the condition that appears in only one of them, with a note saying why. Near the bottom, solve one equation with a negative coefficient and one with a fractional coefficient, marking on each which sign rule or reciprocal you used. Finally, in the margin, write one equal-shares word problem, its variable definition, its equation and its answer in words.
The circled condition should be that the divisor cannot be zero. If you circled something on the multiplication side, look again at which of the two properties needs a restriction and why.
Recap
Five things, and the first one decides which of the two solving lessons you are in.
| If the question says | Your first move is |
|---|---|
| 4x = 12 | Divide both sides by 4 |
| x / 2 = 3 | Multiply both sides by 2 |
| -3x = 12 | Divide by -3, sign included |
| two thirds of x = 8 | Multiply both sides by 3/2 |
| Split equally between 5 | Write 5x = total, then divide |
Lesson 3.3 combines both lessons into one: equations needing more than one step, where an addition and a multiplication both have to be undone, and the order in which you undo them turns out to matter.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.2 Solving Equations Using Multiplication and Division §3.2, pp. 138-143 — everything on these slides traces back here
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