Equivalent equations and the balance model, inverse operations, transforming an equation by adding or subtracting the same number from both sides, isolating the variable, checking every solution in the original equation, and building one-step equations from described situations.
Subject: Algebra 1 · 65 slides · symbolic lesson
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Title
Algebra 1 · Chapter 3 — Solving Linear Equations
Solving Equations Using Addition and Subtraction
Objectives
Five outcomes, each one you can test yourself on with a pencil and no answer key.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.1 Solving Equations Using Addition and Subtraction §3.1, pp. 132-137 — the lesson these objectives are drawn from
Warm-up
Lesson 1.4 solved equations by reading them as questions. This lesson replaces that with something that scales.
Discussion prompt
Solve x plus 3 equals 5 in your head. Now try x plus 47 equals 132. Did the second one feel different, and if so, what changed?
Hint: The method that worked for the first probably stopped working for the second.
Answer:
\[ x + 3 = 5 \;\rightarrow\; x = 2 \qquad x + 47 = 132 \;\rightarrow\; x = 85 \]
The first is a recall question — three plus what makes five — and the second is not, because nobody has memorised what adds to a hundred and thirty-two. What you actually did for the second was subtract, and this lesson makes that subtraction an explicit written step so that it works whatever the numbers are.
Concept
An equation is two sides that must stay in balance. Applying the same operation to both sides produces a different-looking equation with exactly the same solution, and a chain of such moves ends with the variable alone.
equivalent equations — Equations that have the same solution or solutions. Transforming an equation legally always produces an equivalent one.
The goal of every transformation is to isolate the variable on one side.
Figure (svg): A balance scale with x plus 3 on the left and 5 on the right, then with 3 removed from each side leaving x and 2
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.1 Solving Equations Using Addition and Subtraction §3.1, pp. 132-132
Section
Section 1
Concept
Two equations are equivalent if they have the same solutions. You solve an equation by writing a chain of equivalent equations, each one simpler than the last, until the variable stands alone.
That is why the answer at the end of the chain is also the answer to the equation you started with.
Figure (svg): A balance scale with x plus 3 on the left and 5 on the right, then with 3 removed from each side leaving x and 2
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.1 Solving Equations Using Addition and Subtraction §3.1, pp. 132-132 — the balance diagram and the definition of equivalent equations
Picture it
Three units removed from each pan, and the scale is still level.
Figure (svg): A balance scale with x plus 3 on the left and 5 on the right, then with 3 removed from each side leaving x and 2
The right-hand scale is easier to read and says exactly the same thing about x. That is the whole strategy: make the equation simpler without changing what it claims.
Worked example
The picture and the algebra are the same argument written two ways.
\[ \text{Solve } \; x + 3 = 5 \; \text{ by keeping the equation in balance.} \]
Notice what has been done to x
Why: Three has been added to it, so the equation says x with three added makes five.
\[ 3\text{ added to } x \]
Subtract 3 from each side
Why: Removing the same amount from both pans keeps the scale level, so the new equation is equivalent.
\[ x + 3 - 3 = 5 - 3 \]
Simplify both sides
Why: On the left, adding three and subtracting three cancel by the inverse property; on the right, five minus three is two.
\[ x = 2 \]
State the solution
Why: The variable is alone, so the equation has been solved.
\[ x = 2 \]
Figure (svg): A balance scale with x plus 3 on the left and 5 on the right, then with 3 removed from each side leaving x and 2
\[ x + 3 = 5 \;\Longrightarrow\; x = 2 \]
Verify: substitute 2 into the original equation
Why: Two plus three is five, and the right side is five, so the statement is true and two is a solution. The check uses the original equation rather than any line in the middle, which is what makes it independent of the solving.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.1 Solving Equations Using Addition and Subtraction §3.1, pp. 132-132
Sorting
Two equations are equivalent when they have the same solutions.
Sort into buckets
Sort each pair by whether the two equations are equivalent.
The three wrong pairs all add when they should subtract, or vice versa. Substituting the claimed solution into the original equation catches every one of them in a few seconds.
Worked example
It is worth seeing once that the transformation does not lose or gain solutions.
\[ \text{Show that } x + 3 = 5 \text{ and } x = 2 \text{ have exactly the same solutions.} \]
Test a number that solves the second
Why: Two satisfies x equals 2, and substituting it into the first gives five equals five, which is true.
\[ 2\text{ works in both} \]
Test a number that does not
Why: Three satisfies neither: it gives six equals five in the first and three equals two in the second.
\[ 3\text{ fails both} \]
Explain why this always happens
Why: Subtracting three from both sides is reversible — adding it back returns the original — so no solution can be created or destroyed.
State the conclusion
Why: The two equations are equivalent, so solving the simpler one solves the harder one.
Figure (svg): The solution to Worked example why the solution survives shown as a ladder of expressions, one row per algebraic move
\[ x + 3 = 5 \;\Longleftrightarrow\; x = 2 \]
Verify: check the reverse direction
Why: Starting from x equals 2 and adding three to both sides gives x plus three equals five, the original. Being able to get back is exactly what reversible means, and it is the reason the chain of equations can be read in either direction.
Trap
\[ x + 3 = 5 \]
Subtract 3 from the left side to isolate x, and leave the right side alone
Why: The aim is to get rid of the three, and the three is on the left, so the left is where the work happens.
\[ x = 5 \quad \text{(wrong)} \]
The scale is no longer level. Removing weight from one pan and not the other changes what the equation claims, and five does not satisfy the original.
\[ x + 3 - 3 = 5 - 3 \;\Longrightarrow\; x = 2 \]
Apply every operation to both sides, without exception
Why: Balance is what preserves the solution, and it is broken the moment the two sides are treated differently.
Writing the operation on both sides explicitly, rather than mentally moving a term across, is the habit that prevents this. It costs one line and it is checkable.
Elimination
The equation is x plus 6 equals 10.
Eliminate the wrong options
Which move produces an equivalent equation?
Survives elimination: A
Why: Subtracting six from both sides keeps the balance and also isolates x, giving x equals 4. Options C and D are worth noticing separately: they are perfectly legal transformations that simply do not make progress, which shows that keeping the balance is necessary but not sufficient — the move also has to undo something.
Socratic
The balance picture is persuasive. It is worth saying why it is more than an analogy.
Discussion prompt
Explain why applying the same operation to both sides cannot create a new solution or destroy an existing one. Use the idea of reversibility in your answer.
Hint: Ask what would happen if you applied the opposite operation afterwards.
Answer:
If a number satisfies the original equation, then both sides are the same number for that value, and doing the same thing to two equal numbers leaves them equal — so it still satisfies the new equation. No solution is destroyed.
And no solution is created, because the move is reversible: adding three back to both sides returns the original equation, so any solution of the new one must also solve the old one. Both directions hold, which is exactly what equivalent means. In Chapter 9 you will meet a move that is not reversible — squaring both sides — and there solutions really can be created, which is why checking becomes compulsory.
Prediction
Predicting the direction is a check on the arithmetic.
Predict first
In the equation x plus 8 equals 3, will the solution be positive or negative?
Correct: Negative, since 8 is larger than 3.
\[ x + 8 = 3 \;\Longrightarrow\; x = 3 - 8 = -5 \]
Why: Adding eight to x has to reach only three, so x must start below zero. Solving confirms it: subtracting eight from both sides gives x equals negative five. A negative solution is an ordinary answer rather than a warning sign, and predicting its sign first catches a sign error immediately.
Section
Section 2
Concept
Addition and subtraction are inverse operations: each undoes the other. To isolate a variable, identify what has been done to it and apply the inverse to both sides.
This lesson uses the first two rows; Lesson 3.2 uses the last two.
| Operation | Its inverse |
|---|---|
| add 3 | subtract 3 |
| subtract 6 | add 6 |
| multiply by 4 | divide by 4 |
| divide by 2 | multiply by 2 |
Figure (svg): Pairs of inverse operations shown undoing each other
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.1 Solving Equations Using Addition and Subtraction §3.1, pp. 132-132 — the Transforming Equations box
Picture it
Every operation has exactly one partner that undoes it.
Figure (svg): Pairs of inverse operations shown undoing each other
The pairing comes straight from Chapter 2: adding the opposite undoes an addition, and multiplying by the reciprocal undoes a multiplication. Solving is those two facts used deliberately.
Worked example
The Transforming Equations table, worked one row at a time.
\[ \text{Solve } \; x - 3 = 5, \quad x + 6 = 10, \quad x + 8 = 3. \]
In the first, 3 has been subtracted from x
Why: The inverse of subtracting three is adding three, applied to both sides.
\[ x = 8 \]
In the second, 6 has been added to x
Why: The inverse is subtracting six from both sides.
\[ x = 4 \]
In the third, 8 has been added to x
Why: Subtract eight from both sides; three minus eight is negative five.
\[ x = -5 \]
Notice the pattern
Why: In every case the operation done to the variable was undone by its inverse, applied to both sides.
Figure (svg): A table of transformations showing original equations and their equivalents after adding or subtracting
\[ x = 8, \quad x = 4, \quad x = -5 \]
Verify: substitute each solution back
Why: Eight minus three is five; four plus six is ten; negative five plus eight is three. All three original equations come out true, and the third confirms that a negative solution is perfectly ordinary.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.1 Solving Equations Using Addition and Subtraction §3.1, pp. 132-132
Matching
Name what has been done to the variable, then its inverse.
Match the pairs
Why: In each case the move is the inverse of what the equation shows. An equation with a plus needs a subtraction and one with a minus needs an addition, which is exactly the opposite of what the eye wants to do. Saying the operation and its inverse aloud before writing anything is what breaks that instinct.
Worked example
Nothing about the method changes when the variable is on the other side.
\[ \text{Solve } \; 12 = y + 5. \]
Identify what has been done to the variable
Why: Five has been added to y, on the right-hand side.
\[ 5\text{ added to } y \]
Subtract 5 from both sides
Why: The side the variable happens to sit on does not affect the move.
\[ 12 - 5 = y + 5 - 5 \]
Simplify
Why: Seven on the left, and y alone on the right.
\[ 7 = y \]
Write the answer in the usual order
Why: y equals seven says the same thing as seven equals y.
\[ y = 7 \]
Figure (svg): The solution to Worked example the variable on the right shown as a ladder of expressions, one row per algebraic move
\[ 12 = y + 5 \;\Longrightarrow\; y = 7 \]
Verify: substitute 7 into the original
Why: Seven plus five is twelve, matching the left side. The equation was true with the variable on the right and it stays true after the answer is rewritten in the more familiar order, because an equal sign works in both directions.
Error analysis
The student solved three one-step equations. Two are wrong.
Annotate
On: \( x - 3 = 5 \;\rightarrow\; x = 2 \qquad x + 6 = 10 \;\rightarrow\; x = 4 \qquad x + 8 = 3 \;\rightarrow\; x = 11 \)
The reliable habit is to name the operation that has been done to the variable out loud, then name its inverse, then write the inverse on both sides. Three seconds, and it removes the guesswork.
Discrimination
Decide the move without solving.
Sort into buckets
Sort each equation by which move isolates the variable.
Faded example
The move is chosen. Complete both sides.
Fill in the blanks
x - 3 = 5 \;\rightarrow\; x - 3 + 3 = 5 + 3 \;\rightarrow\; x = 8
Why: Adding three to the left cancels the subtraction there, and the same three must be added to the right to keep the balance, giving eight. Writing the operation on both sides explicitly, rather than moving the term across, is what makes the two blanks obviously the same number.
Socratic
The cancellation on the left is a property from Chapter 2 doing its work.
Discussion prompt
Explain why adding three to x minus three leaves x alone, naming the two properties from Lesson 2.3 that make it happen. Then say what the equivalent argument would be for multiplication.
Hint: One property is about a number and its opposite; the other is about zero.
Answer:
\[ (x - 3) + 3 = x + (-3 + 3) = x + 0 = x \]
The associative property regroups so that the negative three and the three are added first, the inverse property makes that pair zero, and the identity property makes adding zero leave x unchanged. Three properties in one line, all from Lesson 2.3.
For multiplication the argument is identical with different properties: multiplying 4x by one quarter regroups as x times four times one quarter, the inverse property of multiplication makes that one, and the multiplicative identity leaves x alone. That is exactly what Lesson 3.2 will do.
Section
Section 3
Concept
Solving is finished when the equation reads variable equals number. Every legal move is judged by whether it brings you closer to that shape.
In a one-step equation a single move is enough; Lesson 3.3 chains several of them.
Figure (svg): The goal of solving shown as isolating the variable on one side with a number alone on the other
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.1 Solving Equations Using Addition and Subtraction §3.1, pp. 132-133
Picture it
This is what every solving process is aiming at.
Figure (svg): The goal of solving shown as isolating the variable on one side with a number alone on the other
Knowing the target shape is what lets you judge a move before making it. A move that leaves the variable no more alone than before is legal but useless.
Worked example
Several moves are legal. Only some of them help.
\[ \text{Solve } \; n + 14 = 9, \; \text{ and say why the move you chose was the right one.} \]
Identify what stands between n and being alone
Why: Fourteen is being added to it.
\[ \text{the } +14 \]
Apply the inverse to both sides
Why: Subtract fourteen from each side.
\[ n + 14 - 14 = 9 - 14 \]
Simplify both sides
Why: The left collapses to n; on the right, nine minus fourteen is negative five.
\[ n = -5 \]
Say why the move was right
Why: It removed the only thing standing between n and the equal sign, so one move was enough.
Figure (svg): The solution to Worked example choose the move that makes progress shown as a ladder of expressions, one row per algebraic move
\[ n + 14 = 9 \;\Longrightarrow\; n = -5 \]
Verify: substitute negative 5 into the original
Why: Negative five plus fourteen is nine, matching the right side. The answer is negative, which the sizes predicted — fourteen has to be added to reach only nine, so the starting value must be below zero.
Sorting
Every move listed is legal. Decide whether it helps.
Sort into buckets
For the equation x plus 6 equals 10, sort each move by whether it makes progress.
Five of six moves are legal and useless. Balance is necessary but not sufficient, and choosing the move that undoes the operation on the variable is what turns a legal move into a useful one.
Worked example
Sometimes the equation has to be tidied before the move is obvious.
\[ \text{Solve } \; x - (-4) = 10. \]
Simplify the left side first
Why: Subtracting negative four is adding four, by the rule from Lesson 2.4.
\[ x + 4 = 10 \]
Identify the operation on the variable
Why: Four is being added.
\[ \text{the } +4 \]
Apply the inverse to both sides
Why: Subtract four from each side.
\[ x = 6 \]
Note why the rewrite mattered
Why: Before the rewrite the equation looked like a subtraction, which would have suggested the wrong inverse.
Figure (svg): The solution to Worked example an equation needing a rewrite first shown as a ladder of expressions, one row per algebraic move
\[ x - (-4) = x + 4 = 10 \;\Longrightarrow\; x = 6 \]
Verify: substitute 6 into the original, not the rewritten form
Why: Six minus negative four is six plus four, which is ten. Checking against the original equation rather than the simplified one is what makes the check independent — an error in the rewrite would survive a check against the rewritten version.
Trap
\[ x + 6 = 10 \]
Add 6 to both sides, since adding the same thing to both sides is always allowed
Why: The rule about keeping the balance is remembered, and any balanced move satisfies it.
\[ x + 12 = 16 \]
Perfectly legal and completely useless. The variable is no more isolated than before, and the numbers are larger.
\[ x + 6 - 6 = 10 - 6 \;\Longrightarrow\; x = 4 \]
Choose the move that removes what stands between the variable and the equal sign
Why: Legality is the minimum requirement; usefulness is what decides which legal move to make.
Before making a move, say what it will remove. If the answer is nothing, choose a different one.
Ranking
Four moves, one correct sequence.
Put in order
Why: Identifying the operation comes first, because it determines which inverse to apply. Applying it to both sides is the transformation, simplifying makes the result readable, and the substitution check comes last since it needs a candidate answer. Skipping the first step is what produces the add-when-you-should-subtract error.
Faded example
The move is chosen. Complete it.
Fill in the blanks
n + 14 = 9 \;\rightarrow\; n = 9 - 14 = -5
Why: Subtracting fourteen from both sides leaves n on the left and nine minus fourteen on the right, which is negative five. The answer is negative because fourteen had to be added to reach only nine, so the starting value must have been below zero — a prediction worth making before the subtraction.
Edge cases
Some equations have zero as their solution, which is worth being comfortable with.
Discussion prompt
Solve x plus 7 equals 7 and x minus 7 equals negative 7. Say what is special about both answers, and why zero is a perfectly ordinary solution rather than a sign of trouble.
Hint: Solve them and compare the answers.
Answer:
\[ x + 7 = 7 \;\Longrightarrow\; x = 0 \qquad x - 7 = -7 \;\Longrightarrow\; x = 0 \]
Both have the solution zero, which happens whenever the two constants match. Substituting confirms it: zero plus seven is seven, and zero minus seven is negative seven.
Zero is an ordinary solution because it is an ordinary number — the identity property of addition says adding zero changes nothing, which is exactly what makes these equations true. An answer of zero is only worrying when it appears as a divisor, which is a completely different situation.
Section
Section 4
Concept
Every solution should be checked by substituting it into the equation you were given. Checking against a line you wrote yourself would confirm the arithmetic and miss any error made on the way there.
This is the routine from Lesson 1.4, and it is deliberately independent of how the answer was produced.
Figure (svg): A substitution check confirming that x equals 2 satisfies x plus 3 equals 5
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.1 Solving Equations Using Addition and Subtraction §3.1, pp. 133-133 — the instruction to check each solution in the original equation
Picture it
Three lines, ending in a verdict.
Figure (svg): A substitution check confirming that x equals 2 satisfies x plus 3 equals 5
The verdict is true or false with nothing in between. A check that comes out nearly right is a check that failed.
Worked example
The check is a separate calculation, not a re-reading of the solving.
\[ \text{Solve } \; y - 9 = -2 \; \text{ and check the solution.} \]
Identify the operation and its inverse
Why: Nine has been subtracted from y, so add nine to both sides.
\[ \text{add } 9 \]
Transform and simplify
Why: On the right, negative two plus nine is seven.
\[ y = 7 \]
Write the original equation again for the check
Why: Not the transformed version — the one that was given.
\[ y - 9 = -2 \]
Substitute and judge
Why: Seven minus nine is negative two, which matches the right side.
Figure (svg): The solution to Worked example solve and check shown as a ladder of expressions, one row per algebraic move
\[ y - 9 = -2 \;\Longrightarrow\; y = 7 \quad \text{and } 7 - 9 = -2 \;\checkmark \]
Verify: consider what a failed check would have meant
Why: If seven had given anything other than negative two, the error would be somewhere in the transformation rather than in the check, since the check is only substitution and arithmetic. That separation is what makes a failed check informative.
Elimination
A student has solved x minus 6 equals 4 and got x equals 10.
Eliminate the wrong options
Which check would actually test the answer?
Survives elimination: A
Why: Only the original equation is independent of the solving, so only it can catch an error in the transformation. Ten minus six is four, which is true, so the answer is confirmed — and had the student subtracted instead of added, the same check would have failed loudly.
Worked example
The check earns its keep when the solving went wrong.
\[ \text{A student solves } x + 8 = 3 \text{ and gets } x = 11. \text{ Check it.} \]
Write the original equation
Why: x plus eight equals three.
\[ x + 8 = 3 \]
Substitute 11
Why: Eleven plus eight is nineteen.
Judge
Why: Nineteen is not three, so the statement is false and eleven is not a solution.
Diagnose and correct
Why: The student added eight instead of subtracting it. Subtracting gives three minus eight, which is negative five.
\[ x = -5 \]
Figure (svg): The solution to Worked example a check that catches an error shown as a ladder of expressions, one row per algebraic move
\[ 11 + 8 = 19 \neq 3 \qquad -5 + 8 = 3 \;\checkmark \]
Verify: check the corrected answer
Why: Negative five plus eight is three, matching the right side exactly. The corrected solution passes the same check the wrong one failed, which is what confirms the diagnosis rather than merely replacing one guess with another.
Trap
\[ x + 8 = 3 \;\rightarrow\; x = 3 + 8 = 11 \]
Check by substituting 11 into the line x equals 3 plus 8
Why: That line is right there and the substitution is easy, so it looks like a check.
Eleven equals eleven, so the check passes — and the answer is still wrong. The line being checked against contains the very error the check was supposed to find.
\[ x + 8 = 3 \quad \text{check: } 11 + 8 = 19 \neq 3 \]
Always substitute into the equation you were given
Why: The original is the only line you did not write, so it is the only one that can be trusted as independent.
A check is only worth doing if it could fail. Checking against your own working can never fail, which makes it a ritual rather than a test.
Sorting
Substitute each claimed solution into its own original equation.
Sort into buckets
Sort each claim by whether the solution is correct.
Every failure here is the same error — using the visible operation instead of its inverse — and every one was exposed by a single substitution taking a few seconds.
Faded example
The solution has been found. Verify it.
Fill in the blanks
\text7 y - 9 = -2, \; \text-2 y = 7 \;\rightarrow\; ___ - 9 = ___ \;\checkmark
Why: Substituting seven for y gives seven minus nine, which is negative two, matching the right side of the original equation. The check is pure substitution and arithmetic, with no algebra in it at all, which is what makes it independent of how the seven was found.
Socratic
A failed check is more useful than a passed one.
Discussion prompt
A check fails. Name two different places the error could be, and describe how you would tell them apart. Then say why a check that passes is weaker evidence than one that fails is.
Hint: The check itself involves arithmetic too.
Answer:
The error could be in the solving or in the check itself, since the check involves its own arithmetic. Redoing just the substitution settles it: if the check comes out the same way twice, the error is upstream in the transformation, and the first thing to examine is whether the inverse operation was used.
A passed check is weaker evidence because a wrong answer can occasionally survive one — particularly if the same arithmetic slip is made twice, or if the equation happens to be satisfied by a nearby value. A failed check, by contrast, is conclusive: the claimed solution definitely does not satisfy the equation.
Section
Section 5
Concept
The commonest real situation producing a one-step equation is a total made of two parts, one of which is known. The translation skills from Lesson 1.5 and the plan from Lesson 1.6 supply the equation; this lesson supplies the solving.
Figure (svg): A bar model of a park's area split into a developed part and a wild part
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.1 Solving Equations Using Addition and Subtraction §3.1, pp. 136-137 — the city-park exercises the lesson opens with
Picture it
The known part and the unknown part together make the whole.
Figure (svg): A bar model of a park's area split into a developed part and a wild part
Whenever a situation says part of a total is something, the equation has this shape, and one subtraction finishes it.
Worked example
A park has 4210 acres in total, of which 180 acres are developed.
\[ \text{Find the undeveloped area } x, \text{ given } \; 180 + x = 4210. \]
Define the variable
Why: Let x be the undeveloped area in acres.
\[ x =\text{ undeveloped acres} \]
Write the verbal model
Why: The developed area plus the undeveloped area equals the total area.
Translate and solve
Why: One hundred and eighty plus x equals 4210; subtract 180 from both sides.
\[ x = 4210 - 180 \]
Compute and answer in words
Why: Four thousand and thirty acres are undeveloped.
\[ 4030\text{ acres} \]
Figure (svg): A bar model of a park's area split into a developed part and a wild part
\[ 180 + x = 4210 \;\Longrightarrow\; x = 4030 \text{ acres} \]
Verify: rebuild the total from the answer
Why: One hundred and eighty plus four thousand and thirty is 4210, which is the total the problem gave. Getting back 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: Two of these describe something being added and two something being taken away, and the equation records which. Notice that in three of the four the unknown is the starting quantity rather than the change or the total — reading which quantity the question asks for is what decides where the letter goes.
Worked example
The unknown is the starting value rather than the part.
\[ \text{A temperature falls } 17 \text{ degrees to reach } -5. \text{ Find the starting temperature.} \]
Define the variable
Why: Let t be the starting temperature in degrees.
\[ t =\text{ starting temperature} \]
Write the verbal model
Why: The starting temperature minus the fall equals the final temperature.
Translate
Why: t minus seventeen equals negative five.
\[ t - 17 = -5 \]
Solve and answer
Why: Add seventeen to both sides: negative five plus seventeen is twelve degrees.
\[ t = 12 \]
Figure (svg): The solution to Worked example a temperature drop shown as a ladder of expressions, one row per algebraic move
\[ t - 17 = -5 \;\Longrightarrow\; t = 12 \text{ degrees} \]
Verify: check the story against the answer
Why: Starting at twelve and falling seventeen degrees reaches negative five, which is what the problem described. And the answer is above zero while the final temperature is below it, which is exactly what a seventeen-degree fall should do.
Trap
\[ 180 + x = 4210 \;\Longrightarrow\; x = 4030 \]
Circle 4030 and move to the next question
Why: The algebra is finished, so the work feels finished.
Four thousand and thirty what? The question asked for an area, and a bare number is not an answer to it.
The undeveloped area is 4030 acres.
Look up what the letter stood for and answer in a sentence with the unit
Why: The definition written at the start is exactly what turns the number into an answer.
This is the same discipline as Lesson 1.6's plan, and it is the step examiners award a separate mark for.
Missing information
A question can be perfectly well written and still be unanswerable.
Discussion prompt
A park has 180 acres of developed land. How much is undeveloped? Say exactly what is missing and give two different answers depending on how the gap is filled.
Hint: Write the verbal model and see which quantity has no number.
Answer:
The total area is missing. The verbal model needs three quantities — developed, undeveloped and total — and only one has been given.
\[ \text{if the total is } 4210: \; x = 4030 \qquad \text{if the total is } 500: \; x = 320 \]
With one equation and two unknowns there is no unique answer, which is exactly the situation the labels stage from Lesson 1.6 detects by counting the letters. Two letters and one equation is a Chapter 7 problem rather than a Chapter 3 one.
Elimination
After spending 25 dollars you have 40 dollars left. Let x be what you started with.
Eliminate the wrong options
Which equation is correct?
Survives elimination: A
Why: Spending reduces the amount, so the starting amount minus twenty-five leaves forty, giving x equal to sixty-five. Checking against the story settles it: you must have started with more than you were left with, and three of the four options give an answer smaller than forty.
Socratic
Many one-step word problems can be done in your head.
Discussion prompt
Give one reason for writing the equation down even when the arithmetic is obvious, and describe the kind of problem where skipping it would cost you the answer entirely.
Hint: Think about a problem with three or four quantities in it.
Answer:
The equation records which quantity is which, so the answer can be reported correctly. In the spending problem the tempting mental answer is fifteen — the difference between the two numbers given — and writing the equation is what shows that the unknown is the starting amount rather than the difference.
It becomes essential the moment a problem has several quantities or the unknown appears more than once. Lesson 3.4 has equations with the variable on both sides, and no amount of mental arithmetic will produce those answers — the written equation is the only way in.
Comparison
Fill the blanks from memory before you scroll back. The move is always the opposite of what you can see.
Comparison matrix
| The equation shows | Your move | Example |
|---|---|---|
| a number added to the variable | subtract it from both sides | x + 6 = 10 gives x = 4 |
| a number subtracted from the variable | add it to both sides | x - 3 = 5 gives x = 8 |
| the variable on the right | the same move, either way round | 12 = y + 5 gives y = 7 |
The bottom row is worth noticing: the side the variable sits on never affects which move is needed, only where the answer ends up.
Pattern
Whether the equation comes from a page or from a word problem, the same five moves cover it.
Step two is the one that prevents the single commonest error in this lesson, which is performing the operation the equation displays rather than its inverse.
Check
An addition to undo. Name the inverse first.
Check your understanding
Solve x plus 12 equals 5.
Answer: A
Why: Twelve has been added to x, so subtract twelve from both sides: five minus twelve is negative seven. Substituting confirms it, since negative seven plus twelve is five. The answer is negative because twelve has to be added to reach only five.
Check
A subtraction to undo. The variable is on the right this time.
Check your understanding
Solve negative 3 equals y minus 8.
Answer: A
Why: Eight has been subtracted from y, so add eight to both sides: negative three plus eight is five. Substituting confirms it, since five minus eight is negative three. The variable sitting on the right changes nothing about which move is needed.
Check
A word problem. Decide which quantity is unknown.
Check your understanding
A hiker climbs 340 metres to reach an elevation of 1250 metres. What was the starting elevation?
Answer: A
Why: Letting e be the starting elevation, the equation is e plus 340 equals 1250, so e is 1250 minus 340, which is 910 metres. Checking against the story confirms it: starting at 910 and climbing 340 reaches 1250, and the starting elevation must be lower than the finish.
Real world
A bank account is overdrawn by 85 dollars. After a deposit the balance is 240 dollars.
Discussion prompt
Write an equation for the deposit using a signed starting balance, solve it, and say what the arithmetic in the final step is really recording. Then explain why the answer is larger than the final balance.
Hint: An overdraft is a negative balance, so the starting value carries a minus sign.
Answer:
\[ -85 + d = 240 \;\Longrightarrow\; d = 240 - (-85) = 240 + 85 = 325 \]
The deposit was 325 dollars. The final step subtracts a negative, which by Lesson 2.4 is an addition — and that is exactly right, because the deposit had to cover the overdraft before it could build a positive balance.
The answer exceeds the final balance because 85 dollars of it went to clearing the debt and only 240 remained. Whenever a starting value is negative, the change needed is larger than the finish, and the subtraction of a negative in the algebra records precisely that.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
To solve x plus 8 equals 3, do you add 8 or subtract 8 from both sides?
Correct: Subtract 8, because subtraction undoes addition.
\[ x + 8 - 8 = 3 - 8 \;\Longrightarrow\; x = -5 \]
\[ \text{check: } -5 + 8 = 3 \;\checkmark \]
Why: The move is always the inverse of what the equation displays, not a copy of it. Eight has been added to x, so subtracting eight from both sides removes it, giving x equal to negative five. Doing the visible operation instead is the single commonest error in this lesson, and it is caught immediately by substituting: eleven plus eight is nineteen rather than three.
Explain it
They can check whether a number is a solution and have never solved an equation with a written method.
Discussion prompt
In no more than four sentences, explain what solving an equation means and why doing the same thing to both sides is allowed. Then give them the one question to ask before every move, and say what error it prevents.
Hint: Your explanation should mention what the finished equation looks like.
Answer:
A usable answer: solving means rewriting the equation, over and over, until it says the letter equals a number. You are allowed to do anything to it as long as you do it to both sides, because an equation is a balance and treating the two sides equally keeps it level. The number you end up with also solves the equation you started with.
The question to ask before every move is what has been done to the letter, and then what undoes that. It prevents the commonest error, which is copying the operation you can see: an equation showing a plus needs a subtraction, and one showing a minus needs an addition.
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: Choosing the inverse is fixed by naming the operation on the variable out loud before writing anything. Both-sides is fixed by writing the operation on both sides explicitly instead of moving a term across. Negative solutions are fixed by predicting the sign before computing, from whether the constant added is larger than the target. Word problems are fixed by writing the variable definition as a full sentence first. 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 draw a balance scale twice: once holding an equation of your own choosing and once after you have applied a move to both pans, writing the equation under each. Underneath, make a two-column table of operations and their inverses, marking the two rows this lesson uses. In the middle, solve three one-step equations, one with a positive solution, one with a negative solution and one with the variable on the right, writing the operation on both sides explicitly in every case. Under each, show the substitution check against the original equation. Finally, in the margin, write one word problem of your own, its variable definition as a full sentence, its equation, and its answer in words with a unit.
Each of your three checks should substitute into the equation you were given, not into a line you wrote. If any check used one of your own lines, it could not have failed and so tested nothing.
Recap
Five things, and the second one is where nearly every mark in this lesson is won or lost.
| If the question says | Your first move is |
|---|---|
| Solve the equation | Name what has been done to the variable |
| x + 6 = 10 | Subtract 6 from both sides |
| x - 3 = 5 | Add 3 to both sides |
| Check your solution | Substitute into the original equation |
| What was the starting amount | Define the variable, then write the model |
Lesson 3.2 does the same job with the other pair of inverse operations: multiplication and division, undone by dividing and multiplying, with one extra caution about what happens when you divide by a negative number.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.1 Solving Equations Using Addition and Subtraction §3.1, pp. 132-137 — everything on these slides traces back here
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