Solving a formula for one of its variables: rearranging with the same inverse operations used on numerical equations, the temperature and area formulas, using a rearranged formula to compute values, checking a rearrangement by its units, and knowing when rearranging first is worth the effort.
Subject: Algebra 1 · 65 slides · symbolic lesson
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Title
Algebra 1 · Chapter 3 — Solving Linear Equations
Formulas
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 3.7 Formulas §3.7, pp. 171-176 — the lesson these objectives are drawn from
Warm-up
Rearranging a formula uses no move you have not already made.
Discussion prompt
Solve 5x equals 20 for x. Now solve A equals lw for l. What is the same about the two, and what is different?
Hint: Look at what the move is, and then at what the answer looks like.
Answer:
\[ 5x = 20 \;\Longrightarrow\; x = 4 \qquad A = lw \;\Longrightarrow\; l = \tfrac{A}{w} \]
Both divide both sides by the coefficient of the variable being isolated. What differs is only the answer: the first collapses to a number because the other side was a number, and the second stays in letters because the other side was letters. The move is identical.
Concept
A formula is an algebraic equation relating two or more quantities. Solving it for one variable rearranges it to describe that quantity in terms of the others, using the same steps as any linear equation.
formula — An algebraic equation relating two or more quantities. Solving a formula for one of its variables produces an equivalent formula with that variable isolated.
Nothing is substituted and nothing is evaluated. The result is a formula rather than a number.
Figure (svg): The same formula shown solved for two different variables
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.7 Formulas §3.7, pp. 171-171
Section
Section 1
Concept
To solve a formula for a variable, transform it until that variable stands alone on one side. Every other letter stays where it is and is treated as a known quantity.
Treating the other letters as numbers is the whole trick: it turns an unfamiliar problem into an ordinary one.
Figure (svg): Two columns contrasting solving for a number with solving for a variable
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.7 Formulas §3.7, pp. 171-171 — the definition of a formula and the instruction to transform it
Picture it
The moves are the same; only the appearance of the answer differs.
Figure (svg): Two columns contrasting solving for a number with solving for a variable
If you can solve 5x equals 20, you can solve A equals lw. The second looks harder only because the answer is a fraction of letters rather than the number four.
Worked example
This is Example 3 part a from the textbook. One move is all it takes.
\[ \text{The area of a rectangle is } A = lw. \text{ Solve for } l. \]
Decide which letter to isolate
Why: The question asks for l, so w and A are treated as known.
Identify what has been done to l
Why: It has been multiplied by w.
Apply the inverse to both sides
Why: Divide each side by w.
\[ \frac{A}{w} = l w / w \]
Simplify
Why: The w's cancel on the right, leaving l alone.
\[ l = \frac{A}{w} \]
Figure (svg): The solution to Worked example solve the area formula for l shown as a ladder of expressions, one row per algebraic move
\[ A = lw \;\Longrightarrow\; l = \tfrac{A}{w} \]
Verify: substitute values into both forms
Why: With l equal to 7 and w equal to 5, the original gives an area of 35. The rearranged formula gives 35 over 5, which is 7 — the length we started from. The two formulas agree, which is what equivalence means.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.7 Formulas §3.7, pp. 172-172
Matching
Each formula is solved for a different one of its letters.
Match the pairs
Why: Each rearrangement divides by whichever letter is multiplying the one being isolated. The two formulas have the same shape — one quantity equal to a product of two others — so their four rearrangements follow the same pattern, and recognising that shape saves rederiving each one.
Worked example
The same formula rearranged for the other letter, to show that neither is special.
\[ \text{Solve } A = lw \text{ for } w. \]
Decide which letter to isolate
Why: This time w, so l and A are the known quantities.
Identify what has been done to w
Why: It has been multiplied by l.
Divide both sides by l
Why: The l's cancel on the right.
\[ w = \frac{A}{l} \]
Compare with the previous result
Why: The two rearrangements are mirror images, as the symmetry of the formula suggests.
Figure (svg): The solution to Worked example solve the same formula for w shown as a ladder of expressions, one row per algebraic move
\[ A = lw \;\Longrightarrow\; w = \tfrac{A}{l} \]
Verify: check with the same values
Why: With an area of 35 and a length of 7, the formula gives 35 over 7, which is 5 — the width. Both rearrangements recover the value they were solved for, which confirms neither move disturbed the relationship.
Trap
\[ A = lw \text{ with } A = 35 \text{ and } w = 5 \]
Put the numbers in first and then try to solve
Why: Numbers feel more concrete than letters, so getting them in early feels like progress.
\[ 35 = 5l \;\rightarrow\; l = 7 \]
This works for one set of values and has to be repeated in full for every other set. The formula was never rearranged, so nothing was gained for next time.
\[ A = lw \;\Longrightarrow\; l = \tfrac{A}{w} \;\Longrightarrow\; l = \tfrac{35}{5} = 7 \]
Rearrange once, then substitute as often as you like
Why: The rearranged formula answers the same question for every set of values without any further algebra.
For a single calculation either order works. For ten calculations the rearrangement is done once and the substitution ten times, which is far less work.
Sorting
Read which variable stands alone on one side.
Sort into buckets
Sort each formula by which variable it has been solved for.
Every formula of the shape one quantity equals a product has three arrangements, and two of them require a division. Knowing which arrangement you have been given tells you immediately whether any work is needed.
Elimination
The formula is d equals rt, and you want a formula for t.
Eliminate the wrong options
Which is correct?
Survives elimination: A
Why: The t is multiplied by r, so dividing both sides by r isolates it. The units confirm it: miles divided by miles per hour leaves hours, which is a time. Option B produces miles per hour divided by miles, which is a reciprocal time and describes nothing anyone measures.
Socratic
The other letters are unknown, and you still act as though they are known.
Discussion prompt
Explain why it is legitimate to treat w as a known number when solving A equals lw for l, even though w has no value. Then say what could go wrong if w happened to be zero.
Hint: Ask what the properties of equality require of a divisor.
Answer:
The properties of equality apply to any number, and w stands for a number whatever its value. So dividing both sides by w is exactly the same move as dividing by five would be, and it is legal for the same reason. Not knowing which number w is does not stop it being one.
The exception is zero: dividing by w assumes w is not zero, and the division property of equality requires that. For a rectangle a width of zero is not a real case, so the assumption is safe here — but in general a rearrangement that divides by a letter carries an unstated condition that the letter is not zero, and it is worth noticing when that condition might fail.
Section
Section 2
Concept
A formula needing two moves is rearranged in the usual order: undo the addition or subtraction first, then the multiplication or division. The presence of letters changes nothing about the sequence.
\[ C = \tfrac{5}{9}(F - 32) \;\Longrightarrow\; F = \tfrac{9}{5}C + 32 \]
Here the bracket is the outermost structure, so multiplying by the reciprocal comes first and the addition second.
Figure (svg): The Celsius formula rearranged step by step to give Fahrenheit
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.7 Formulas §3.7, pp. 171-171 — Example 1, Solve a Temperature Conversion Formula
Picture it
Four lines, one move each, with letters on the far side throughout.
Figure (svg): The Celsius formula rearranged step by step to give Fahrenheit
Multiplying by nine fifths cleared the fraction and the bracket in one move. Then adding thirty-two isolated F, and the answer is written the usual way round.
Worked example
This is Example 1 from the textbook. Celsius and Fahrenheit are related by the formula given.
\[ \text{Solve } \; C = \tfrac{5}{9}(F - 32) \; \text{ for } F. \]
Identify what is being done to F
Why: Thirty-two is subtracted, and the result is multiplied by five ninths. The multiplication is outermost.
\[ \text{outermost is } \times \frac{5}{9} \]
Undo the multiplication first
Why: Multiply each side by nine fifths, the reciprocal of five ninths.
\[ (\frac{9}{5}) C = F - 32 \]
Undo the subtraction
Why: Add thirty-two to each side.
\[ (\frac{9}{5}) C + 32 = F \]
Write the answer the usual way round
Why: The isolated variable conventionally goes on the left.
\[ F = (\frac{9}{5}) C + 32 \]
Figure (svg): The Celsius formula rearranged step by step to give Fahrenheit
\[ F = \tfrac{9}{5}C + 32 \]
Verify: test with a temperature you know
Why: Water freezes at 0 degrees Celsius and 32 degrees Fahrenheit. Substituting C equal to zero gives nine fifths of zero plus thirty-two, which is 32. And water boils at 100 Celsius: nine fifths of a hundred is 180, plus 32 is 212, which is correct. Two known pairs both check.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.7 Formulas §3.7, pp. 171-171
Ranking
Four moves for the temperature formula, in the right sequence.
Put in order
Why: Identifying the operations comes first, since it decides the order of the next two. The multiplication by five ninths is outermost — it wraps the whole bracket — so it is undone first, and the subtraction of thirty-two second. Rewriting with F on the left is cosmetic and comes last.
Worked example
Example 2 from the textbook. Two moves again, both multiplicative.
\[ \text{The area of a triangle is } A = \tfrac{1}{2}bh. \text{ Solve for } b. \]
Identify what has been done to b
Why: It has been multiplied by h and by one half.
\[ \times h\text{ and } \times \frac{1}{2} \]
Clear the fraction
Why: Multiply each side by two.
\[ 2 A = b h \]
Undo the multiplication by h
Why: Divide each side by h.
\[ 2 A / h = b \]
Write the answer
Why: b equals two A over h.
\[ b = 2 A / h \]
Figure (svg): A triangle with base b and height h, and the area formula rearranged for b
\[ A = \tfrac{1}{2}bh \;\Longrightarrow\; b = \tfrac{2A}{h} \]
Verify: substitute values into both forms
Why: A triangle with base 10 and height 6 has area one half times sixty, which is 30. The rearranged formula gives two times thirty over six, which is ten — the base we started from. Both forms describe the same triangle.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.7 Formulas §3.7, pp. 172-172
Error analysis
The student rearranged two formulas. One is wrong.
Annotate
On: \( C = \tfrac{5}{9}(F - 32) \;\rightarrow\; F = \tfrac{9}{5}C - 32 \qquad A = \tfrac{1}{2}bh \;\rightarrow\; b = \tfrac{2A}{h} \)
Testing a rearranged formula at a value you already know is the fastest possible check, and for the temperature formula there are two such values that everybody has.
Faded example
The first move is done. Complete the second.
Fill in the blanks
C = \tfrac+___(F - 32) \;\rightarrow\; \tfrac______C = F - 32 \;\rightarrow\; \tfrac______C ___ 32 = F
Why: Thirty-two was subtracted from F in the original, so adding thirty-two to both sides undoes it. Testing at C equal to zero confirms the sign: the formula must give thirty-two Fahrenheit, which it does only with a plus.
Elimination
The formula is C equals five ninths times the quantity F minus 32.
Eliminate the wrong options
What should you do first to solve for F?
Survives elimination: A
Why: Multiplying by the reciprocal clears the fraction and the bracket in one move, leaving F minus thirty-two on the right. Option C reaches the same answer by a longer route with fractions throughout, which is a good illustration that several correct routes exist and they differ in effort.
Socratic
Dividing by five ninths would also work.
Discussion prompt
Explain why multiplying by nine fifths is preferred to dividing by five ninths, and what it accomplishes to the bracket at the same time.
Hint: Think about what dividing by a fraction requires you to do anyway.
Answer:
Dividing by five ninths means multiplying by its reciprocal anyway, by the division rule from Lesson 2.8, so the two are the same move with one written more directly. Multiplying by nine fifths states the operation in the form you would actually carry out.
It also removes the bracket without any distributing, because the whole bracket was multiplied by five ninths and multiplying by the reciprocal returns it to itself. One move clears both the fraction and the grouping, which is why the alternative of distributing first is more work for the same answer.
Section
Section 3
Concept
The point of rearranging is to get a formula you can use directly. Once the variable you want is isolated, computing it for any set of values is a single substitution.
For one calculation the saving is small; for several it is the whole point.
Figure (svg): The rearranged formula used to compute a length from an area and a width
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.7 Formulas §3.7, pp. 172-172 — Example 3 part b, using the new formula
Picture it
The rearranged formula does the work.
Figure (svg): The rearranged formula used to compute a length from an area and a width
Thirty-five square feet divided by five feet gives seven feet. The substitution is a single division because the rearranging was done first.
Worked example
Example 3 part b from the textbook. The rearranged formula is already available.
\[ \text{A rectangle has area } 35 \text{ square feet and width } 5 \text{ feet. Find its length.} \]
Choose the rearranged formula
Why: Length equals area over width, which was derived earlier.
\[ l = \frac{A}{w} \]
Substitute the known values
Why: Thirty-five for A and five for w.
\[ l = \frac{35}{5} \]
Evaluate
Why: Thirty-five over five is seven.
\[ l = 7 \]
Attach the unit
Why: Square feet divided by feet leaves feet.
\[ 7\text{ feet} \]
Figure (svg): The rearranged formula used to compute a length from an area and a width
\[ l = \tfrac{A}{w} = \tfrac{35}{5} = 7 \text{ feet} \]
Verify: rebuild the area from the answer
Why: Seven feet by five feet gives thirty-five square feet, which is the area the problem gave. Recovering a given number is the strongest available check.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.7 Formulas §3.7, pp. 172-172
Elimination
You know a triangle's area and its base, and want its height.
Eliminate the wrong options
Which formula should you rearrange for?
Survives elimination: A
Why: The right side must contain only quantities you know, which here are A and b. Option D is worth checking on a triangle of base 10 and height 6: its area is 30, and the wrong formula gives 30 over 20, which is 1.5 rather than 6.
Worked example
Guided Practice 1 and 2. Rearrange for h, then use it.
\[ \text{Solve } A = \tfrac{1}{2}bh \text{ for } h, \text{ then find } h \text{ when } A = 25 \text{ square inches and } b = 10 \text{ inches.} \]
Rearrange for h
Why: Multiply both sides by two, then divide by b.
\[ h = 2 A / b \]
Substitute the values
Why: Two times twenty-five over ten.
\[ h = \frac{50}{10} \]
Evaluate
Why: Fifty over ten is five.
\[ h = 5 \]
Attach the unit
Why: Square inches divided by inches leaves inches.
\[ 5\text{ inches} \]
Figure (svg): The solution to Worked example find a triangle's height shown as a ladder of expressions, one row per algebraic move
\[ h = \tfrac{2A}{b} = \tfrac{50}{10} = 5 \text{ inches} \]
Verify: rebuild the area
Why: Half of ten times five is twenty-five square inches, which matches. And the rearranged formula for h is the mirror image of the one for b, as the symmetry of the original formula in b and h requires.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.7 Formulas §3.7, pp. 172-172
Trap
Asked for the length, given the area and the width, the student rearranges for w.
\[ w = \tfrac{A}{l} \]
Rearrange for whichever letter looks convenient
Why: Both rearrangements are one move and neither is obviously the wrong one.
The formula now needs l, which is the very thing being asked for. It cannot be used with the values given.
\[ l = \tfrac{A}{w} = \tfrac{35}{5} = 7 \]
Rearrange for the quantity the question asks for, so that every other letter has a value
Why: The point of a rearrangement is that the right side contains only known quantities.
A quick check before substituting: does every letter on the right have a number? If not, the formula was solved for the wrong variable.
Faded example
Substitute the known values and evaluate.
Fill in the blanks
h = \tfrac255 = \tfrac___})}___ = \tfrac______ = ___
Why: Substituting an area of twenty-five and a base of ten gives fifty over ten, which is five inches. Because the formula was rearranged first, the whole calculation is a single division rather than an equation to solve.
Prediction
For a single calculation the two orders take about the same effort.
Predict first
You need the length of twelve different rectangles, each given by its area and width. Is it worth rearranging first?
Correct: Yes — one rearrangement then twelve divisions.
\[ l = \tfrac{A}{w} \quad \text{used twelve times, versus twelve solves of } A = lw \]
Why: Substituting first means solving twelve separate equations, each with its own algebra. Rearranging first means one piece of algebra and then twelve substitutions, each a single division. The saving grows with the number of cases, which is exactly why formulas are published in several rearranged forms.
Socratic
Reference books often list the same relationship rearranged three ways.
Discussion prompt
Explain why a reference book might print d equals rt, r equals d over t, and t equals d over r as three separate entries, even though they say the same thing. Then say what skill this lesson gives you that makes such a list unnecessary.
Hint: Think about who is using the book and what they already know.
Answer:
Different readers arrive with different quantities known. Someone with a distance and a time wants the rate; someone with a rate and a time wants the distance. Printing all three saves each reader the algebra, which matters when the reader is in a hurry or is not confident rearranging.
The skill this lesson gives you is producing any of the three from any other in one move, which makes the list a convenience rather than a necessity. It also means you can rearrange formulas nobody has bothered to print in the form you need — which is most of them.
Section
Section 4
Concept
A rearranged formula can be checked two ways: substitute a set of values into both forms and confirm they agree, or check that the units on the right side produce the unit the left side should have.
The units check is faster and catches structural errors; the values check is slower and catches everything.
Figure (svg): A rearranged formula checked by its units
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.7 Formulas §3.7, pp. 172-172
Picture it
Square feet divided by feet leaves feet.
Figure (svg): A rearranged formula checked by its units
A rearrangement that produced feet squared or one over feet would be wrong, and the units say so without any values being chosen.
Worked example
Both checks on the same formula, to see what each one catches.
\[ \text{Check that } l = \tfrac{A}{w} \text{ is a correct rearrangement of } A = lw. \]
Choose values satisfying the original
Why: A rectangle 7 by 5 has area 35.
\[ l = 7, w = 5, A = 35 \]
Substitute into the rearrangement
Why: Thirty-five over five is seven, which is the length we started from.
\[ \text{recovers } l = 7 \]
Check the units
Why: Square feet divided by feet leaves feet, which is a length.
Compare what each check caught
Why: The values check confirms the whole formula; the units check confirms only its structure but costs almost nothing.
Figure (svg): A rearranged formula checked by its units
\[ \tfrac{35 \text{ sq ft}}{5 \text{ ft}} = 7 \text{ ft} \]
Verify: try a wrong rearrangement against both checks
Why: The wrong formula l equals A times w would give 175 with the same values, which the values check rejects, and its units would be feet cubed, which the units check also rejects. Both checks catch this one; the units check would miss an error that only changed a numerical coefficient.
Sorting
Work out the units of the right side and compare with the left.
Sort into buckets
Sort each proposed formula by whether its units are consistent.
Every wrong version here multiplied where it should have divided, and every one was caught by the units alone. The check costs nothing and requires no values at all.
Worked example
The units check is fast and structural, and here it is enough.
\[ \text{A student writes } t = dr \text{ as a rearrangement of } d = rt. \text{ Check it by units.} \]
Identify the units of each quantity
Why: Distance in miles, rate in miles per hour, time in hours.
\[ \text{mi}, \text{mi} / h, h \]
Compute the units of the right side
Why: Miles times miles per hour gives miles squared per hour.
\[ \text{mi} ^{2} / h \]
Compare with the left side
Why: Time is measured in hours, not miles squared per hour.
Conclude and correct
Why: The rearrangement is wrong; dividing rather than multiplying gives miles over miles per hour, which is hours.
\[ t = \frac{d}{r} \]
Figure (svg): The solution to Worked example a units check that catches an error shown as a ladder of expressions, one row per algebraic move
\[ t = \tfrac{d}{r}, \text{ since } \tfrac{\text{mi}}{\text{mi/h}} = \text{h} \]
Verify: confirm with values
Why: A journey of 100 miles at 50 miles per hour takes two hours. The correct formula gives 100 over 50, which is 2. The student's formula would give 5000, which is neither a time nor a plausible number for anything.
Trap
\[ C = \tfrac{5}{9}(F - 32) \;\rightarrow\; F = \tfrac{9}{5}C - 32 \]
Rearrange, write the answer, and move on
Why: The moves felt right at the time, and there is no number at the end to look suspicious.
At C equal to zero this gives negative thirty-two Fahrenheit. Water freezes at positive thirty-two, so the formula is wrong at the most familiar temperature there is.
\[ F = \tfrac{9}{5}C + 32 \quad \text{check at } C = 0: \; F = 32 \;\checkmark \]
Test the rearranged formula at a value you already know
Why: A rearrangement has no numerical answer to look wrong, so it needs a deliberate test.
Every well-known formula has at least one pair of values everybody knows. Freezing and boiling water check the temperature formula in about ten seconds.
Elimination
A student rearranges A equals half b h and writes b equals A over h, forgetting the factor of two.
Eliminate the wrong options
Which check exposes the mistake?
Survives elimination: A
Why: The missing factor of two does not affect the units, so only a values check catches it. A triangle with base 10 and height 6 has area 30, and the wrong formula gives 30 over 6, which is 5 rather than 10 — off by exactly the missing factor. Units checks catch structural errors and values checks catch everything.
Faded example
Work out the unit of the right side.
Fill in the blanks
In t = d / r the units are miles divided by miles per hour, which leaves hours.
Why: Miles divided by miles per hour cancels the miles and leaves hours, which is a time — exactly what t should be. The cancellation works because a rate carries a fraction of two units, and dividing by it flips that fraction over.
Socratic
Neither check is complete on its own.
Discussion prompt
Describe one error a units check would miss and one it would catch instantly. Then say why it is still worth doing given that a values check catches both.
Hint: Think about what a numerical coefficient does to the units.
Answer:
A units check misses any error in a numerical coefficient, since numbers have no units: writing b equals A over h instead of 2A over h passes the units check perfectly. It catches instantly any error that multiplies where it should divide, because that changes the unit of the answer entirely.
It is still worth doing because it takes about five seconds and needs no values chosen, so it can be run on every rearrangement as a matter of habit. A values check is more thorough and takes longer, so in practice you run the units check always and the values check when the formula matters.
Section
Section 5
Concept
A real problem hands you some quantities and asks for another. Rearranging the relevant formula so that the unknown is isolated turns the problem into a single substitution.
Figure (svg): The distance formula rearranged to give speed, applied to a journey to Mars
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.7 Formulas §3.7, pp. 174-174 — Example 5, estimating the Pathfinder spacecraft's average speed
Picture it
One formula, rearranged for whichever letter is unknown.
Figure (svg): The distance formula rearranged to give speed, applied to a journey to Mars
A spacecraft's journey is a distance and a duration, and the rate follows from dividing one by the other. The formula is the same one used for a car in Lesson 1.1.
Worked example
Example 5 in spirit. The spacecraft covered about 310 million miles in about 212 days.
\[ \text{Find the average speed in miles per day, using } d = rt. \]
Identify what is known and wanted
Why: Distance and time are known; the rate is wanted.
Rearrange the formula for r
Why: Divide both sides by t.
\[ r = \frac{d}{t} \]
Substitute the values
Why: Three hundred and ten million over two hundred and twelve.
\[ \frac{310000000}{212} \]
Evaluate and attach the unit
Why: About 1.46 million miles per day.
\[ 1460000 \text{mi} / d a y \]
Figure (svg): The distance formula rearranged to give speed, applied to a journey to Mars
\[ r = \tfrac{d}{t} \approx 1{,}460{,}000 \text{ miles per day} \]
Verify: check the units and the size
Why: Miles divided by days gives miles per day, which is a speed. And 1.46 million miles a day is about 61,000 miles an hour, which is the right order of magnitude for an interplanetary spacecraft — far faster than any aircraft and far slower than light.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.7 Formulas §3.7, pp. 174-174
Matching
Each situation gives two quantities and asks for a third.
Match the pairs
Why: In each case the wanted quantity is isolated and the two known ones appear on the right. Choosing the arrangement is a matter of reading the question rather than of algebra — the algebra is the one or two moves that produce it.
Worked example
Sometimes the harder decision is which formula to use at all.
\[ \text{A rectangular garden has perimeter } 36 \text{ m and width } 7 \text{ m. Find its length.} \]
Identify the quantities
Why: Perimeter and width are known; length is wanted.
Choose a formula relating all three
Why: Perimeter equals twice the length plus twice the width.
\[ P = 2 l + 2 w \]
Rearrange for l
Why: Subtract 2w from both sides, then divide by two.
\[ l = \frac{P - 2 w}{2} \]
Substitute and evaluate
Why: Thirty-six minus fourteen is twenty-two, over two is eleven.
\[ l = 11 m \]
Figure (svg): The solution to Worked example choosing which formula to rearrange shown as a ladder of expressions, one row per algebraic move
\[ l = \tfrac{P - 2w}{2} = \tfrac{36 - 14}{2} = 11 \text{ m} \]
Verify: rebuild the perimeter
Why: Twice eleven plus twice seven is twenty-two plus fourteen, which is thirty-six metres — the perimeter given. The area formula would have been the wrong choice here, since the area was neither given nor asked for.
Trap
Given the perimeter and the width, the student reaches for the area formula.
\[ A = lw \;\rightarrow\; l = \tfrac{A}{w} \]
Use the most familiar formula for the shape
Why: The area formula is the one most often met, so it comes to mind first.
The area was never given. The formula needs A, which is unknown, so it cannot be evaluated.
\[ P = 2l + 2w \;\rightarrow\; l = \tfrac{P - 2w}{2} \]
Choose a formula containing exactly the quantities you have and the one you want
Why: A usable formula has a value for every letter on the right side.
Before rearranging, list what you know and what you want, and pick a formula that mentions all of them and nothing else.
Missing information
A question can be perfectly well written and still be unanswerable.
Discussion prompt
A rectangular garden is 7 metres wide. Find its length. Say exactly what is missing, and name two different pieces of information that would each make the question answerable, together with the formula each would need.
Hint: The length has to be determined by something.
Answer:
Neither the area nor the perimeter has been given, so nothing determines the length — a 7-metre-wide garden can be any length at all.
\[ \text{given the area: } l = \tfrac{A}{w} \qquad \text{given the perimeter: } l = \tfrac{P - 2w}{2} \]
Either piece of information makes the question answerable, and each needs a different formula. Recognising which formula a given set of quantities calls for is as much part of the work as the rearranging is.
Estimation
A rough answer first is a check on the rearrangement as well as the arithmetic.
Predict first
A spacecraft covers about 300 million miles in about 200 days. Roughly what is its average speed in miles per day?
Correct: About 1.5 million miles per day.
\[ \tfrac{300{,}000{,}000}{200} = 1{,}500{,}000 \text{ miles per day} \]
Why: Three hundred million divided by two hundred is one and a half million. Option B multiplies instead of dividing, and the other two misplace the decimal point by three orders of magnitude — all three are ruled out by an estimate before any careful arithmetic is done.
Socratic
Both routes reach the answer, and one scales better.
Discussion prompt
You need the speed for twenty different journeys, each given by a distance and a time. Compare the work of rearranging once against solving twenty equations, and say what the rearranged formula is really giving you.
Hint: Count the algebraic steps in each approach.
Answer:
Solving twenty equations means twenty divisions plus twenty pieces of setting up, each with its own chance of error. Rearranging once means one division of both sides by t, and then twenty ordinary divisions with no algebra at all.
What the rearranged formula gives you is a procedure rather than an answer. It converts an algebra problem into an arithmetic one, permanently, and that conversion is why every science and engineering reference book is full of rearranged formulas rather than of worked examples.
Comparison
Fill the blanks from memory before you scroll back. The moves are identical.
Comparison matrix
| Numerical equation | Formula | |
|---|---|---|
| What the other side holds | numbers | letters |
| What the answer is | a number | another formula |
| How to check it | substitute the answer | substitute values, and check the units |
Only the middle row genuinely differs. The moves and the order are the same, and the check gains one extra tool because units are available.
Pattern
Whether the formula has two letters or five, the same five moves cover it.
Step one is worth stating explicitly. Rearranging for the wrong letter produces a correct formula that cannot be used with the values you have.
OpenStax Elementary Algebra 2e, §2.6 Solve a Formula for a Specific Variable §2.6
Check
One move. Divide by what multiplies the variable.
Check your understanding
Solve d equals rt for t.
Answer: A
Why: The t is multiplied by r, so dividing both sides by r isolates it. The units confirm it: miles divided by miles per hour leaves hours, which is a time.
Check
Two moves. Clear the fraction, then divide.
Check your understanding
Solve A equals one half b h for h.
Answer: A
Why: Multiplying both sides by two clears the half, giving 2A equals bh, and dividing by b isolates h. Testing on a triangle of base 10 and height 6, whose area is 30, gives sixty over ten, which is six.
Check
Use the rearranged formula.
Check your understanding
A triangle has area 24 square centimetres and base 8 centimetres. What is its height?
Answer: A
Why: Using h equals 2A over b gives forty-eight over eight, which is six centimetres. Checking rebuilds the area: half of eight times six is twenty-four square centimetres, as given.
Real world
A recipe's oven temperature is given as 180 degrees Celsius, and your oven is marked in Fahrenheit. Later you find another recipe calling for 425 degrees Fahrenheit and want the Celsius setting.
Discussion prompt
Use both arrangements of the temperature formula to answer the two questions, and say which arrangement each one needed. Then explain why having both forms available is more useful than being able to derive either on demand.
Hint: One question converts Celsius to Fahrenheit and the other goes the other way.
Answer:
\[ F = \tfrac{9}{5}(180) + 32 = 324 + 32 = 356 \text{ degrees Fahrenheit} \]
\[ C = \tfrac{5}{9}(425 - 32) = \tfrac{5}{9}(393) \approx 218 \text{ degrees Celsius} \]
The first needed the formula solved for F and the second the original form solved for C. Both are one rearrangement apart, and having both written down means neither conversion requires any algebra at the moment you need it.
That is exactly why kitchens keep conversion charts: the algebra is easy but doing it while cooking is not. Rearranging in advance converts a recurring algebra problem into a lookup, which is the practical value of this whole lesson.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
When you solve a formula for one of its variables, what does the answer look like?
Correct: Another formula, with the chosen variable isolated.
\[ A = lw \;\Longrightarrow\; l = \tfrac{A}{w} \]
Only afterwards, if values are given, does a substitution produce a number.
Why: Nothing is substituted during the rearrangement, so nothing collapses to a number. The other side stays in letters and the result is an equivalent formula describing the chosen quantity in terms of the others. That is what makes it reusable: one rearrangement serves every set of values, which is precisely the advantage over substituting first and solving each time.
Explain it
They can solve equations for x and have never rearranged a formula.
Discussion prompt
In no more than four sentences, explain how solving A equals lw for l is the same as solving 5x equals 20 for x. Then tell them the one thing to check that they could not check on a numerical equation.
Hint: The extra check is available because the letters mean something.
Answer:
A usable answer: in both cases something is multiplying the letter you want, so you divide both sides by it. In the numerical one the other side is twenty and collapses to four; in the formula the other side is A over w and stays in letters. The move is identical and only the appearance of the answer differs.
The extra check is the units. Area divided by width is square feet over feet, which gives feet — a length, which is what l should be. If the units had come out as feet cubed or one over feet, the rearrangement would be wrong, and you can see that without choosing a single number.
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: Treating letters as numbers is fixed by rewriting the formula once with a number in place of each other letter, solving that, and then copying the moves. Order is fixed by asking which operation is outermost, exactly as in Lesson 3.3. Choosing the variable is fixed by listing what you know and what you want before touching the formula. Checking is fixed by making the units check automatic and the values check deliberate. Pick yours and do five of that kind tonight rather than twenty mixed ones.
Connect it up
Do this on paper. It is worth more than rereading the slides.
Draw it
At the top of a page write one formula with three letters and rearrange it for each of its three variables in turn, showing every move. Beside each rearrangement write the units of the right side and confirm they give the unit of the left. In the middle, take the temperature formula and solve it for Fahrenheit, then test your answer at the freezing and boiling points of water. Near the bottom, invent a situation where you are given two of your three quantities and want the third, and use the matching rearrangement to answer it with a unit attached. Finally, in the margin, write one sentence saying why rearranging first beats substituting first when several sets of values are involved.
Your three rearrangements should all be one or two moves from the original and should all pass their units check. If one of them needs three or more moves, look again at whether you undid the operations in reverse order.
Recap
Five things, and the first one is what makes the rest ordinary.
| If the question says | Your first move is |
|---|---|
| Solve A = lw for l | Divide both sides by w |
| Solve for F | Undo the outermost operation on F first |
| Find a formula for the base | Isolate b, leaving A and h alone |
| Use the new formula to find | Substitute, then attach the unit |
| Is this rearrangement right | Check the units, then test known values |
Lesson 3.8 turns to ratios and rates, where two quantities are compared rather than combined, and where the unit of the answer carries as much meaning as its value.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.7 Formulas §3.7, pp. 171-176 — everything on these slides traces back here
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