The four-step procedure covering every linear equation: simplify each side by distributing and combining, collect the variable terms on the side with the greater coefficient, isolate the variable, and check in the original. Includes brackets on both sides, fractional factors, and what it means when the variable terms cancel entirely.
Subject: Algebra 1 · 65 slides · symbolic lesson
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Title
Algebra 1 · Chapter 3 — Solving Linear Equations
More on Linear Equations
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 3.5 More on Linear Equations §3.5, pp. 157-162 — the lesson these objectives are drawn from
Warm-up
Every move you need has already appeared. This lesson assembles them into one procedure.
Discussion prompt
List the moves you have used to solve equations so far, in the order you would apply them. Then look at 4 times the quantity 1 minus x, plus 3x, equals negative 2 times the quantity x plus 1 — which move comes first, and why?
Hint: One move has to happen before you can even see what the coefficients are.
Answer:
\[ 4(1 - x) + 3x = -2(x + 1) \]
Distributing comes first, because until the brackets are gone the left side has no single coefficient to compare with anything. After distributing you get four minus four x plus three x on the left, which combines to four minus x — and only then does the comparison with the right side's negative two make sense.
Concept
Every linear equation yields to the same four steps: simplify each side, collect the variable terms on the side with the greater coefficient, isolate the variable with inverse operations, and check in the original equation.
identity — An equation that is true for every value of the variable. It appears when the variable terms cancel and the remaining numerical statement is true.
Nothing in this list is new. What is new is applying all four in sequence to equations complicated enough to need them.
Figure (svg): The four steps for solving any linear equation, shown as a numbered ladder
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.5 More on Linear Equations §3.5, pp. 157-157
Section
Section 1
Concept
The procedure is fixed and its order matters. The first two steps prepare the equation, the third solves it, and the fourth confirms the answer.
Steps one and two must come in that order, because a coefficient cannot be compared until the side is simplified.
Figure (svg): The four steps for solving any linear equation, shown as a numbered ladder
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.5 More on Linear Equations §3.5, pp. 157-157 — the Steps for Solving Linear Equations box
Picture it
Each rung says what to do and the note beside it says how.
Figure (svg): The four steps for solving any linear equation, shown as a numbered ladder
Step three is the whole of Lessons 3.1 to 3.3. The two lessons since then have added exactly one step each in front of it.
Worked example
This is Example 1 from the textbook. All four steps appear.
\[ \text{Solve } \; 4(1 - x) + 3x = -2(x + 1). \]
Distribute on both sides
Why: Four times one minus four times x on the left; negative two times x and negative two times one on the right.
\[ 4 - 4 x + 3 x = -2 x - 2 \]
Combine like terms on the left
Why: Negative four x and three x combine to negative x.
\[ 4 - x = -2 x - 2 \]
Collect the variable terms
Why: Negative one is greater than negative two, so collect on the left by adding two x to each side.
\[ 4 + x = -2 \]
Isolate the variable
Why: Subtract four from each side.
\[ x = -6 \]
Figure (svg): An equation with brackets on both sides, distributed before anything else happens
\[ 4(1 - x) + 3x = -2(x + 1) \;\Longrightarrow\; x = -6 \]
Verify: evaluate both sides of the original at negative 6
Why: The left side is four times seven plus negative eighteen, which is twenty-eight minus eighteen, or ten. The right side is negative two times negative five, which is ten. Both sides agree, which confirms every one of the four steps at once.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.5 More on Linear Equations §3.5, pp. 157-157
Ranking
Four steps, one correct sequence.
Put in order
Why: Simplifying comes first because the coefficients cannot be compared until each side is a single variable term and a constant. Collecting comes second and reduces the equation to the two-step form. Isolating is third, and the check is last since it needs an answer to test.
Worked example
Example 2 from the textbook. The same four steps, with more terms.
\[ \text{Solve } \; 3(4x - 1) - 6x = -8x + 24. \]
Distribute on the left
Why: Twelve x minus three, then the minus six x follows.
\[ 12 x - 3 - 6 x = -8 x + 24 \]
Combine like terms on the left
Why: Twelve x minus six x is six x.
\[ 6 x - 3 = -8 x + 24 \]
Collect the variable terms
Why: Six is greater than negative eight, so add eight x to each side.
\[ 14 x - 3 = 24 \]
Isolate the variable
Why: Add three to each side to get 14x equals 27, then divide by fourteen.
\[ x = \frac{27}{14} \]
Figure (svg): The solution to Worked example a longer chain shown as a ladder of expressions, one row per algebraic move
\[ 3(4x - 1) - 6x = -8x + 24 \;\Longrightarrow\; x = \tfrac{27}{14} \]
Verify: check the collection choice
Why: Six against negative eight: six is greater, so collecting on the left leaves a positive fourteen. Had the collection gone the other way the coefficient would have been negative fourteen and the final division would have been by a negative — correct, but with one more sign to manage.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.5 More on Linear Equations §3.5, pp. 158-158
Trap
\[ 4(1 - x) + 3x = -2(x + 1) \]
Add 2x to both sides straight away, since the right side has a -2x in it
Why: The 2x is visible inside the bracket, so it looks available to move.
The x inside the bracket has not yet been multiplied by negative two, so it is not a term of the right side. Moving it treats a partly-formed term as a finished one.
\[ 4 - 4x + 3x = -2x - 2 \;\Longrightarrow\; 4 - x = -2x - 2 \]
Distribute every bracket first, so that both sides really are sums of terms
Why: A bracket suspends its contents from the surrounding expression until the multiplication is carried out.
This is the same principle as Lesson 2.7: an expression's terms are not visible until the brackets are gone.
Sorting
Two of the four steps touch a single side and two touch both.
Sort into buckets
Sort each move by how many sides it must be applied to.
The switch happens between step two and step three. Everything in the preparation phase touches one side; everything in the solving phase touches both.
Elimination
The equation is 3 times the quantity 4x minus 1, minus 6x, equals negative 8x plus 24.
Eliminate the wrong options
What is the correct first move?
Survives elimination: A
Why: Distributing turns three times the bracket into twelve x minus three, after which the left side genuinely has terms that can be combined and a coefficient that can be compared. Options C and D both treat the factor three as though it were already a term of the side, which it is not.
Socratic
The four steps are listed in a particular sequence for a reason.
Discussion prompt
Explain why step two cannot come before step one, using an equation of your own or one from this lesson. Then say whether steps three and four could ever be swapped.
Hint: Ask what step two needs to know that step one supplies.
Answer:
Step two compares the coefficients of the two sides, and a side with a bracket or with two variable terms does not yet have a single coefficient. In 3 times the quantity 4x minus 1, minus 6x, the coefficient is six — a number that appears nowhere in the equation as written. Step one is what produces it.
Steps three and four cannot be swapped for a different reason: the check needs a candidate answer, and step three is what produces one. There is nothing to check before it. The order is not a convention in either case — each step depends on what the previous one supplies.
Section
Section 2
Concept
When both sides contain brackets, distribute on each side separately. Only when both sides are sums of plain terms can the like terms be combined and the coefficients compared.
\[ 4(1 - x) + 3x = -2(x + 1) \;\rightarrow\; 4 - 4x + 3x = -2x - 2 \]
A negative factor outside a bracket distributes with its sign, exactly as in Lesson 2.6.
Figure (svg): An equation with brackets on both sides, distributed before anything else happens
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.5 More on Linear Equations §3.5, pp. 157-157 — Example 1, Solve a More Complicated Equation
Picture it
Distribute, combine, collect, isolate — with the reason beside every line.
Figure (svg): An equation with brackets on both sides, distributed before anything else happens
Writing one move per line is what makes a five-line solution checkable. Compressed into two lines it would be unreadable and unfixable.
Worked example
Guided Practice 2. Two brackets, one with a negative factor in front.
\[ \text{Solve } \; 4x - (2 - x) = 3(x + 2). \]
Distribute the leading minus on the left
Why: A bare minus sign in front of a bracket is a factor of negative one, so it gives negative two plus x.
\[ 4 x - 2 + x \]
Distribute on the right
Why: Three x plus six.
\[ 3 x + 6 \]
Combine like terms on the left
Why: Four x and x make five x.
\[ 5 x - 2 = 3 x + 6 \]
Collect and isolate
Why: Five is greater than three, so subtract three x to get 2x minus two equals six, then 2x equals eight.
\[ x = 4 \]
Figure (svg): The solution to Worked example negative factors on both sides shown as a ladder of expressions, one row per algebraic move
\[ 4x - (2 - x) = 3(x + 2) \;\Longrightarrow\; x = 4 \]
Verify: evaluate both sides at 4
Why: The left side is sixteen minus the quantity two minus four, which is sixteen minus negative two, or eighteen. The right side is three times six, which is eighteen. Both agree, and the leading minus was distributed to both terms rather than only the first.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.5 More on Linear Equations §3.5, pp. 158-158
Prediction
Predicting the signs before distributing is a real check.
Predict first
Distributing negative 2 over the bracket x plus 1 gives which pair of terms?
Correct: -2x and -2, both negative.
\[ -2(x + 1) = -2x - 2 \]
\[ \text{but } -2(x - 1) = -2x + 2, \text{ one of each} \]
Why: The factor is negative and both terms inside the bracket are positive, so both products are negative. A negative factor over a sum always produces two negative terms, and any mixed-sign answer signals that the sign was left behind on one of them.
Worked example
Guided Practice 3. The right side has a bracket and a loose constant.
\[ \text{Solve } \; 2(4x - 2) = 2(x + 3) + 9. \]
Distribute on the left
Why: Eight x minus four.
\[ 8 x - 4 \]
Distribute on the right and keep the loose constant
Why: Two x plus six, then plus nine.
\[ 2 x + 6 + 9 \]
Combine like terms on the right
Why: Six and nine make fifteen.
\[ 8 x - 4 = 2 x + 15 \]
Collect and isolate
Why: Eight is greater than two, so subtract two x to get 6x minus four equals fifteen, then 6x equals nineteen.
\[ x = \frac{19}{6} \]
Figure (svg): The solution to Worked example brackets and a constant shown as a ladder of expressions, one row per algebraic move
\[ 2(4x - 2) = 2(x + 3) + 9 \;\Longrightarrow\; x = \tfrac{19}{6} \]
Verify: check the right side's combination
Why: Six and nine really are like terms, both being constants, so combining them to fifteen is legal and leaves the right side as a single variable term and a single constant. Only then is its coefficient of two ready to be compared.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.5 More on Linear Equations §3.5, pp. 158-158
Error analysis
The student distributed on both sides of two equations. One line is wrong.
Annotate
On: \( 4(1 - x) + 3x = -2(x + 1) \;\rightarrow\; 4 - 4x + 3x = -2x + 2 \qquad 4x - (2 - x) \;\rightarrow\; 4x - 2 + x \)
The error changes the answer from negative six to negative two thirds, and it survives every check except substituting into the original equation.
Faded example
Supply the two expansions.
Fill in the blanks
4(1 - x) + 3x = -2(x + 1) \;\rightarrow\; 4 - 4x + 3x = -2x - 2
Why: Four distributes over one minus x to give four minus four x, and negative two distributes over x plus one to give two negative terms. Doing both expansions before touching anything else is what makes the like terms on each side visible.
Sorting
Scan each side of each equation before deciding.
Sort into buckets
Sort each equation by where its brackets are.
The four-step procedure covers all three categories. Steps that have nothing to do simply pass, exactly as the order of operations does in Lesson 1.3.
Socratic
The x inside a bracket is visible. It is still not available.
Discussion prompt
In the equation 4 times the quantity 1 minus x, plus 3x, equals negative 2 times the quantity x plus 1, explain why the x inside the right-hand bracket cannot be moved before distributing. Say what it will actually become.
Hint: Ask what the term will be worth once the multiplication is carried out.
Answer:
The x inside the bracket has not yet been multiplied by negative two, so the right side does not contain a term x — it contains a term that will become negative two x once the distribution happens. Moving the x itself would move a quantity that is not there.
After distributing, the right side is negative two x minus two, and its variable term is negative two x. That is the term available to be collected, and it differs from the visible x by a factor of negative two. Distributing is what converts the contents of a bracket into terms of the equation.
Section
Section 3
Concept
A fractional factor is distributed like any other, and doing so often clears the fractions from the equation entirely. Multiplying each term by the fraction is usually easier than clearing denominators first.
\[ \tfrac{1}{4}(12x - 16) = 3x - 4 \]
Choose the fraction's denominator to divide the numbers inside where possible, and the products come out whole.
Figure (svg): A bracket multiplied by one quarter, distributed term by term
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.5 More on Linear Equations §3.5, pp. 158-158 — Example 3, with the one-quarter factor
Picture it
Each term inside is multiplied by the fraction.
Figure (svg): A bracket multiplied by one quarter, distributed term by term
Both products came out whole, because twelve and sixteen are both divisible by four. That is why the numbers inside such brackets are usually chosen to cooperate.
Worked example
This is Example 3 from the textbook.
\[ \text{Solve } \; \tfrac{1}{4}(12x - 16) = 10 - 3(x + 2). \]
Distribute the quarter on the left
Why: A quarter of twelve x is three x, and a quarter of sixteen is four.
\[ 3 x - 4 \]
Distribute the negative 3 on the right
Why: Negative three x minus six, then combined with the ten.
\[ 10 - 3 x - 6 \]
Combine like terms on the right
Why: Ten minus six is four.
\[ 3 x - 4 = 4 - 3 x \]
Collect and isolate
Why: Three is greater than negative three, so add three x to each side: six x minus four equals four, then six x equals eight.
\[ x = \frac{4}{3} \]
Figure (svg): The solution to Worked example a quarter outside a bracket shown as a ladder of expressions, one row per algebraic move
\[ \tfrac{1}{4}(12x - 16) = 10 - 3(x + 2) \;\Longrightarrow\; x = \tfrac{4}{3} \]
Verify: evaluate both sides at four thirds
Why: The left side is a quarter of sixteen minus sixteen, which is a quarter of zero — no: twelve times four thirds is sixteen, so the bracket is zero and the left side is zero. The right side is ten minus three times ten thirds, which is ten minus ten, also zero. Both sides are zero, so the answer checks.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.5 More on Linear Equations §3.5, pp. 158-158
Faded example
Multiply each term inside by the factor.
Fill in the blanks
\tfrac14(3y - 12) = ___y - ___
Why: A third of three y is one y, written simply as y, and a third of twelve is four. Both products came out whole because the numbers inside were chosen to be divisible by three — which is typical of textbook problems and worth expecting.
Worked example
Guided Practice 4. Same idea, with thirds.
\[ \text{Solve } \; \tfrac{1}{3}(3y - 12) = 6 - 2(y - 1). \]
Distribute the third on the left
Why: A third of three y is y, and a third of twelve is four.
\[ y - 4 \]
Distribute the negative 2 on the right
Why: Negative two y plus two, since a negative times a negative is positive.
\[ 6 - 2 y + 2 \]
Combine on the right
Why: Six and two make eight.
\[ y - 4 = 8 - 2 y \]
Collect and isolate
Why: One is greater than negative two, so add two y to each side: three y minus four equals eight, then three y equals twelve.
\[ y = 4 \]
Figure (svg): The solution to Worked example a third outside a bracket shown as a ladder of expressions, one row per algebraic move
\[ \tfrac{1}{3}(3y - 12) = 6 - 2(y - 1) \;\Longrightarrow\; y = 4 \]
Verify: evaluate both sides at 4
Why: The left side is a third of twelve minus twelve, which is a third of zero, or zero. The right side is six minus two times three, which is six minus six, also zero. Both sides agree at zero.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.5 More on Linear Equations §3.5, pp. 158-158
Trap
\[ \tfrac{1}{4}(12x - 16) \]
Multiply the 12x by a quarter and carry the 16 down unchanged
Why: The first product is easy and the second feels like a leftover.
\[ = 3x - 16 \quad \text{(wrong)} \]
The quarter reaches every term inside the bracket. A quarter of sixteen is four, not sixteen, and the error shifts the answer by twelve.
\[ \tfrac{1}{4}(12x - 16) = 3x - 4 \]
Multiply every term inside by the fraction, exactly as with a whole number
Why: A fraction is a factor like any other, and the distributive property does not treat it differently.
Substituting one value into both forms catches this immediately, which is the standard check for any distribution from Lesson 2.6.
Elimination
The expression is one half times the quantity 6x plus 10.
Eliminate the wrong options
Which is correct?
Survives elimination: A
Why: Half of six x is three x and half of ten is five, so both terms are halved. Substituting x equal to two settles it: the original bracket is twenty-two and half of that is eleven, and only the first option gives eleven.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| Approach | What you do | When it is easier |
|---|---|---|
| distribute the fraction | multiply each term inside by it | when the numbers inside divide evenly |
| multiply the whole equation | multiply both sides by the denominator | when several fractions appear |
Both are legal on any equation. Distributing is quicker when one fraction is present and the numbers cooperate; clearing denominators is quicker when fractions are scattered through the equation.
Socratic
A quarter outside a bracket containing 12 and 16 is not an accident.
Discussion prompt
Explain why textbook problems tend to put numbers divisible by the denominator inside such a bracket, and describe what you would do if they did not — say if the bracket held 13x minus 7 with a quarter outside.
Hint: Ask what the answer would look like in each case.
Answer:
Numbers divisible by the denominator make every product whole, so the equation stays free of fractions and the remaining steps are ordinary arithmetic. Problems are usually built that way so that the difficulty lies in the method rather than in the fraction handling.
If they did not cooperate, you would either distribute anyway and carry fractions through — thirteen quarters x minus seven quarters — or multiply both sides of the whole equation by four first, which clears the denominator before any distributing. The second is usually much easier, and it is the standard approach when Lesson 3.6 meets equations full of decimals and fractions.
Section
Section 4
Concept
Sometimes the collection step removes the variable from both sides at once. What remains is a numerical statement, and whether it is true or false decides the answer.
Figure (svg): Three possible endings to a linear equation: one solution, no solution, and every number a solution
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.5 More on Linear Equations §3.5, pp. 157-158 — the key word identity introduced in Lesson 3.4
Picture it
Only the first ends with a number.
Figure (svg): Three possible endings to a linear equation: one solution, no solution, and every number a solution
The two unusual endings are not errors. They are genuine answers to genuine questions, and recognising them is part of solving rather than a sign that something went wrong.
Worked example
The variable cancels and the remaining statement is false.
\[ \text{Solve } \; 2(x + 3) = 2x + 10. \]
Distribute on the left
Why: Two x plus six.
\[ 2 x + 6 = 2 x + 10 \]
Collect the variable terms
Why: Subtract two x from both sides.
\[ 6 = 10 \]
Judge the remaining statement
Why: Six is not ten, so the statement is false for every value of x.
State the conclusion
Why: No value of x makes the original equation true, so it has no solution.
Figure (svg): The solution to Worked example an equation with no solution shown as a ladder of expressions, one row per algebraic move
\[ 2(x + 3) = 2x + 10 \;\Longrightarrow\; 6 = 10, \text{ no solution} \]
Verify: test a value to confirm
Why: At x equal to 5 the left side is sixteen and the right side is twenty. At x equal to 100 the left is 206 and the right is 210. The two sides always differ by four, which is exactly what six equals ten was reporting.
Sorting
Simplify each and see what the collection step leaves behind.
Sort into buckets
Sort each equation by how many solutions it has.
The two identities were both a distributed expression set equal to its own undistributed form. That is the commonest way an identity is constructed, and spotting it can save all the working.
Worked example
The variable cancels and the remaining statement is true.
\[ \text{Solve } \; 3(x + 2) = 3x + 6. \]
Distribute on the left
Why: Three x plus six.
\[ 3 x + 6 = 3 x + 6 \]
Notice the two sides are identical
Why: Every term matches.
Collect the variable terms
Why: Subtract three x from both sides.
\[ 6 = 6 \]
Judge and conclude
Why: Six equals six is true for every value of x, so every number is a solution.
Figure (svg): Three possible endings to a linear equation: one solution, no solution, and every number a solution
\[ 3(x + 2) = 3x + 6 \;\Longrightarrow\; 6 = 6, \text{ an identity} \]
Verify: test two very different values
Why: At x equal to 0 both sides are six; at x equal to negative 20 both sides are negative fifty-four. The two sides agree everywhere, because the left side is simply the right side with the distribution not yet carried out.
Trap
\[ 2(x + 3) = 2x + 10 \;\rightarrow\; 6 = 10 \]
Write x equals 0, since no x appears in the final line
Why: The absence of a variable looks like an answer of nothing.
Substituting zero gives six on the left and ten on the right, which is false. Zero is not a solution, and neither is anything else.
The equation has no solution, because 6 equals 10 is false for every value of x.
Read the final numerical statement as a verdict rather than as an equation to solve
Why: The variable has already gone; what is left is a claim about numbers, and it is either true or false.
No solution and a solution of zero are completely different answers. Zero would satisfy the equation; no solution means nothing does.
Elimination
The collection step has removed the variable and left this statement.
Eliminate the wrong options
What should you conclude?
Survives elimination: A
Why: Six equals ten is false whatever x is, so no value of x makes the original equation true. Option D is worth taking seriously as a habit — it is always worth re-checking — but a false numerical statement is a genuine answer, and testing two values confirms it rather than a mistake.
Prediction
Some equations announce their answer before any work is done.
Predict first
What can you say about 4 times the quantity x minus 1, equals 4x minus 4, without solving?
Correct: Every number is a solution, since the sides are the same expression.
\[ 4(x - 1) = 4x - 4 \quad \text{for every } x \]
Why: Distributing the left side gives four x minus four, which is exactly the right side. The two sides are the same expression written differently, so the equation is true for every value of x. Recognising a distributed expression set equal to its undistributed form saves the entire calculation.
Socratic
Both are real answers, and they describe opposite situations.
Discussion prompt
Describe a real situation that would produce an equation with no solution, and one that would produce an identity. Use the comparison problems from Lesson 3.4 as your starting point.
Hint: Think about two plans with the same rate.
Answer:
No solution: two phone plans charging the same rate per minute but different monthly fees. Their costs differ by the fee difference at every level of usage, so they never cost the same and the equation setting them equal has no solution — which is a genuinely useful answer, since it says one plan is always cheaper.
An identity: the same plan described two different ways, say as ten pounds plus four pounds per gigabyte and as two lots of five pounds plus four pounds per gigabyte. The two expressions agree at every usage, so the equation is true always. Both outcomes tell you something real about the situation rather than signalling an error.
Section
Section 5
Concept
A comparison problem often produces an answer that is not a whole number. The algebra is finished, but the interpretation is not: a count has to be rounded, and which way depends on the situation.
Figure (svg): Two health club payment plans compared as expressions set equal
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.5 More on Linear Equations §3.5, pp. 160-160 — Example 4, comparing health club payment plans
Picture it
Pay per visit, against a membership plus a smaller per-visit charge.
Figure (svg): Two health club payment plans compared as expressions set equal
Twelve and a half visits is the exact crossing point, and nobody makes half a visit. The practical answer is that from the thirteenth visit onwards the membership is cheaper.
Worked example
Example 4 in spirit. Pay-per-visit costs 12 dollars a visit; membership costs 100 dollars plus 4 dollars a visit.
\[ \text{Solve } \; 12v = 100 + 4v \; \text{ for the number of visits } v. \]
Write an expression for each plan
Why: Twelve v for pay-per-visit; one hundred plus four v for membership.
\[ 12 v\text{ and } 100 + 4 v \]
Set them equal and collect
Why: Twelve is greater than four, so subtract four v from both sides.
\[ 8 v = 100 \]
Divide
Why: One hundred over eight is twelve and a half.
\[ v = 12.5 \]
Interpret the answer
Why: Nobody makes half a visit, so the plans cross between the twelfth and thirteenth visit.
\[ 12.5\text{ visits} \]
Figure (svg): Two health club payment plans compared as expressions set equal
\[ 12v = 100 + 4v \;\Longrightarrow\; v = 12.5 \text{ visits} \]
Verify: compare the costs at 12 and at 13 visits
Why: At twelve visits, pay-per-visit costs 144 and membership costs 148, so pay-per-visit is cheaper. At thirteen, they cost 156 and 152, so membership wins. The crossing really does fall between the two, exactly as 12.5 said.
Sorting
The situation decides, not the decimal.
Sort into buckets
Sort each answer by what the situation requires.
Three different treatments of a decimal, decided entirely by what the quantity means. The arithmetic is identical in all six cases.
Worked example
Rounding direction depends on the situation, not on the decimal.
\[ \text{A van holds } 9 \text{ boxes. Solve } \; 9n = 200 \; \text{ for the number of trips } n. \]
Solve the equation
Why: Two hundred over nine is about 22.2.
\[ n = 22.2 \]
Ask whether fractions make sense
Why: A trip is a whole event; you cannot make two tenths of a trip.
Decide the rounding direction
Why: Twenty-two trips would carry 198 boxes, leaving two behind, so a twenty-third trip is needed.
State the answer in words
Why: Twenty-three trips are required.
\[ 23\text{ trips} \]
Figure (svg): The solution to Worked example an answer that must round the other way shown as a ladder of expressions, one row per algebraic move
\[ 9n = 200 \;\Longrightarrow\; n \approx 22.2, \text{ so } 23 \text{ trips} \]
Verify: check both neighbouring whole numbers
Why: Twenty-two trips carry 198 boxes, which is not enough; twenty-three carry 207, which is enough with room to spare. Checking both neighbours is what settles the rounding direction, and it is a different question from the arithmetic.
Trap
\[ 9n = 200 \;\rightarrow\; n \approx 22.2 \]
Round 22.2 down to 22, since the decimal part is below a half
Why: Ordinary rounding rules say to round down below 0.5.
Twenty-two trips carry only 198 of the 200 boxes. The rounding rule answered a question about numbers rather than the question about boxes.
Twenty-three trips are needed, because twenty-two would leave two boxes behind.
Check both neighbouring whole numbers against the situation
Why: The rounding direction is decided by what the situation requires, not by the size of the decimal.
Some situations round up, some round down, and some genuinely allow fractions. Deciding which is part of answering the question rather than part of the arithmetic.
Elimination
Pay-per-visit costs 12 dollars a visit; membership costs 100 plus 4 a visit. They cross at 12.5 visits.
Eliminate the wrong options
You expect to visit about 20 times. Which is cheaper?
Survives elimination: A
Why: Past the crossing point the plan with the smaller per-visit rate wins, and membership charges four dollars a visit against twelve. At twenty visits membership costs 180 and pay-per-visit costs 240. The crossing point's job is exactly to divide the two regimes, so once it is known no further solving is needed.
Missing information
A question can be perfectly well written and still be unanswerable.
Discussion prompt
A health club charges 100 dollars for membership plus 4 dollars a visit. When does membership become cheaper? Say exactly what is missing, and give two different crossing points depending on how the gap is filled.
Hint: Cheaper than what?
Answer:
The alternative is missing. Membership can only be cheaper than something else, and no other plan has been described.
\[ \text{against } 12 \text{ per visit: } 12v = 100 + 4v \Rightarrow v = 12.5 \]
\[ \text{against } 8 \text{ per visit: } 8v = 100 + 4v \Rightarrow v = 25 \]
The crossing point doubles when the competing rate falls by a third. A comparison problem needs two expressions, and being handed only one is a signal that half the situation has not been described.
Socratic
It answers a question nobody asked, and that turns out to be the point.
Discussion prompt
The question was which plan is cheaper, and the equation answered when are they equal. Explain why solving the equality is the right way to answer the original question, and what you need to check afterwards.
Hint: Think about what happens on each side of the crossing point.
Answer:
The two costs can only swap over at a point where they are equal, so finding that point locates every place the answer changes. Between crossings one plan is uniformly cheaper, which means a single crossing divides the whole range into two simple regions.
What you check afterwards is which plan wins in each region, by testing one value on either side. At twelve visits pay-per-visit is cheaper and at thirteen membership is, which converts the equation's answer into the practical advice the question actually wanted. Chapter 6 will handle this directly with inequalities rather than by testing points.
Comparison
Fill the blanks from memory before you scroll back. Each lesson added one move to the front.
Comparison matrix
| Lesson | What it added | Example |
|---|---|---|
| 3.1 and 3.2 | one inverse operation | x + 6 = 10 |
| 3.3 | a second inverse operation, in reverse order | 3x - 7 = 8 |
| 3.4 | collecting variable terms | 7x - 19 = 2x + 55 |
| 3.5 | distributing brackets on both sides | 4(1 - x) + 3x = -2(x + 1) |
Reading upwards, each row's equation becomes the next row's after one move. That is why the four-step procedure is the same list every time, with more of it in play.
Pattern
This is the complete procedure for every linear equation in the book.
Steps one and two prepare the equation and touch one side at a time; steps three and four transform it and must touch both. If the variable cancels during step three, read the remaining numerical statement as the answer.
OpenStax Elementary Algebra 2e, §2.4 Use a General Strategy to Solve Linear Equations §2.4
Check
Brackets on both sides. Distribute before anything else.
Check your understanding
Solve 2 times the quantity x plus 4, equals 3 times the quantity x minus 1.
Answer: A
Why: Distributing gives 2x plus 8 equals 3x minus 3. Three is greater than two, so collect on the right by subtracting 2x, giving 8 equals x minus 3, and adding three gives eleven. Both sides check at thirty.
Check
A fractional factor. Distribute it to every term.
Check your understanding
Solve one half times the quantity 8x minus 6, equals x plus 5.
Answer: A
Why: Half of 8x is 4x and half of 6 is 3, so the left side is 4x minus 3. Collecting on the left gives 3x minus 3 equals 5, then 3x equals 8, so x is eight thirds. Both sides check at about 7.67.
Check
The variable cancels. Read the remaining statement.
Check your understanding
Solve 5 times the quantity x plus 2, equals 5x plus 7.
Answer: A
Why: Distributing gives 5x plus 10 equals 5x plus 7, and subtracting 5x leaves 10 equals 7, which is false. No value of x makes the original true, so the equation has no solution. Testing any value confirms it: the two sides always differ by three.
Real world
Two removal quotes. Firm A charges 150 dollars plus 40 dollars an hour. Firm B charges 90 dollars plus 55 dollars an hour, but gives a 10 percent discount off the whole bill.
Discussion prompt
Write an expression for each firm's cost after h hours, remembering the discount applies to firm B's whole bill, and find the number of hours at which they cost the same. Then say which firm is cheaper for a three-hour job and which for a ten-hour one.
Hint: Ninety percent of a bill is 0.9 times it, and the 0.9 distributes over both of firm B's terms.
Answer:
\[ 150 + 40h = 0.9(90 + 55h) = 81 + 49.5h \]
\[ 150 + 40h = 81 + 49.5h \;\Longrightarrow\; 69 = 9.5h \;\Longrightarrow\; h \approx 7.26 \text{ hours} \]
At three hours firm A costs 270 and firm B costs 229.50, so B is cheaper. At ten hours A costs 550 and B costs 576, so A is cheaper. The crossing at about 7.26 hours divides the two cases.
The distribution of the 0.9 over both of firm B's terms is what makes this a Lesson 3.5 problem rather than a Lesson 3.4 one. Applying the discount to only the hourly rate would have shifted the crossing point by more than an hour.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
If the variable terms cancel and you are left with 5 equals 5, what is the answer?
Correct: Every number is a solution.
\[ 3(x + 2) = 3x + 6 \;\rightarrow\; 6 = 6, \text{ an identity} \]
\[ 2(x + 3) = 2x + 10 \;\rightarrow\; 6 = 10, \text{ no solution} \]
Why: Five equals five is true regardless of x, so every value satisfies the original equation and it is an identity. The commonest wrong answers read the surviving number as a value of x, or treat the disappearance of the variable as meaning nothing works — but a true remaining statement means everything works, and a false one means nothing does. The verdict comes from whether the statement is true, not from which numbers appear in it.
Explain it
They can solve equations with variables on both sides and have not met brackets on both sides.
Discussion prompt
In no more than four sentences, give them the four-step procedure and say which steps go on one side and which on both. Then tell them what to do if all the letters disappear during the solving.
Hint: The last part has two possible outcomes.
Answer:
A usable answer: first tidy each side on its own, multiplying out brackets and adding up like terms. Then get all the letters onto whichever side has the bigger number in front. Then undo the plus or minus and after that the times or divide, doing each of those to both sides. Finally check by working out both sides of the equation you were given.
If all the letters vanish, look at the numbers that are left. If the statement is true, like six equals six, then every number works. If it is false, like six equals ten, then nothing works. Neither is a mistake — both are real answers.
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: Negative factors are fixed by predicting the sign pattern before multiplying — a negative over a sum gives two negatives. Simplifying first is fixed by refusing to look at the equal sign until each side is a single variable term and a constant. Fractional factors are fixed by writing both products explicitly before compressing. A cancelled variable is fixed by reading the remaining statement as true or false rather than as an equation. 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 write the four steps as a numbered ladder, and beside each write whether it touches one side or both. To the right, solve one equation with brackets on both sides in full, one move per line with the reason beside it, and mark which line is which step. Below that, solve one equation whose variable terms cancel and write the verdict in words, then do the same for one that cancels the other way. Near the bottom, write one comparison problem whose answer is not a whole number, and say what the fractional answer means in the situation. Finally, in the margin, write the one sentence that decides between no solution and every number.
Your marginal sentence should be about whether the remaining numerical statement is true or false. If it mentions the value of x, look again at what has actually been left behind.
Recap
Five things, and the first one is the complete procedure for every linear equation in the book.
| If the question says | Your first move is |
|---|---|
| Solve the equation | Simplify each side before looking at the equal sign |
| 4(1 - x) + 3x = -2(x + 1) | Distribute both brackets |
| one quarter of a bracket | Multiply every term inside by the quarter |
| you are left with 6 = 10 | Read it as false, so no solution |
| the answer is 12.5 visits | Test 12 and 13 against the situation |
Lesson 3.6 handles equations whose numbers are decimals or fractions, using one extra move at the front — multiply the whole equation by a common denominator — so that the four steps you now have can run on whole numbers.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.5 More on Linear Equations §3.5, pp. 157-162 — everything on these slides traces back here
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