HSA.CED.A.4: Rearranging Formulas to Highlight a Quantity of Interest
In plain English: HSA.CED.A.4 is the Common Core algebra standard that asks students to rearrange a formula to highlight a quantity of interest, such as rewriting V = IR as R = V/I. The key idea is that this uses the same inverse operations and properties of equality as solving an equation, but the result is an expression in the other variables. It is usually taught in Algebra I.
Rearrange formulas to highlight a quantity of interest, using the same reasoning as in solving equations. For example, rearrange Ohm's law V = IR to highlight resistance R.
Common Core State Standards for Mathematics · Domain: Creating Equations (CED) · Cluster: Create equations that describe numbers or relationships Also written as HSA-CED.A.4 or A-CED.4 · Official standard
In this lesson, students rewrite a formula so that a different variable stands alone, for example turning Ohm's law V = IR into R = V/I when the resistance is the quantity of interest. The key idea in the standard is that nothing new is needed: students use exactly the same inverse operations, in the same order, that they use to solve an equation such as 2x + 6 = 20. The only difference is that the result is an expression in the other letters instead of a single number.
Students rearrange formulas from science, geometry and finance, including formulas where the variable of interest has a coefficient, sits inside parentheses, appears in a fraction, is squared, or appears in two terms. They check every rearranged formula by substituting numbers, and they state restrictions such as "I cannot be 0" or "r must be positive" where the context requires them.
Learning Objectives
By the end of this lesson, students will be able to:
Solve a formula for a specified variable using inverse operations in the reverse order of how the variable was used
Explain each step of a rearrangement by comparing it with the same step in a numerical equation
Rearrange formulas in which the variable of interest has a coefficient, is inside parentheses or a fraction, is squared, or appears in more than one term
Check a rearranged formula by substituting known values
State restrictions on variables that come from division or from the context
Prior Knowledge Required
Students should already be comfortable with:
Solving multi-step linear equations in one variable 8.EE.C.7
Solving linear equations with coefficients represented by letters HSA.REI.B.3
Using the distributive property and combining like terms
Evaluating formulas for given values of the variables
Show the two tasks side by side and ask students to do the first before looking at the second.
Warm-Up Prompt
"(1) Solve 2x + 6 = 20. Write down each step. (2) The perimeter of a rectangle is P = 2l + 2w. A fence company knows the perimeter and the length of a yard and needs the width. Can you use the same steps to get w by itself?"
Many students will solve the first task quickly (x = 7). Have a volunteer list the steps: subtract 6, then divide by 2. Then work the second task together with the same steps: subtract 2l from both sides, then divide by 2, giving w = (P - 2l)/2. Name the idea: rearranging a formula is solving an equation where some of the numbers are letters.
Direct Instruction20 minutes
Teach a four-step routine:
Circle the quantity of interest: Decide which variable must end up alone and treat every other letter as if it were a number.
Trace how that variable was used: List the operations applied to it in order. In F = (9/5)C + 32, C is multiplied by 9/5 and then 32 is added.
Undo in reverse order: Apply inverse operations to both sides, last operation first. Collect terms first if the variable appears more than once.
Check and restrict: Substitute numbers that you know work in the original formula. Note any value that would cause division by zero or make no sense in the context.
Model each example below, writing a matching numerical equation next to the first two so students see that the steps are identical:
One step: Ohm's law
"An electrician knows the voltage V and the current I and needs the resistance R, where V = IR."
Equation: R = V/I (I ≠ 0); for V = 120 volts and I = 0.5 amps, R = 240 ohms
Two steps: perimeter
"Solve P = 2l + 2w for the width w."
Equation: w = (P - 2l)/2, which can also be written w = P/2 - l
Fraction coefficient: temperature
"Solve F = (9/5)C + 32 for C to convert Fahrenheit to Celsius."
Equation: C = (5/9)(F - 32); for F = 77, C = 25
Variable in a fraction: trapezoid
"Solve A = (1/2)h(b1 + b2) for the height h."
Equation: h = 2A/(b1 + b2); for A = 60, b1 = 6 and b2 = 9, h = 8
Squared variable: circle area
"Solve A = πr2 for the radius r."
Equation: r = √(A/π), taking the positive root because a radius is positive
Use Diagram 2 with the temperature example to show why the order matters: 32 was added last, so it is subtracted first. For the circle example, point out that the inverse of squaring is taking a square root, and that the negative root is discarded because of the context, not because of the algebra.
Guided Practice15 minutes
Pairs rearrange four formulas, one partner writing the steps and the other writing the matching numerical equation: d = rt for t, I = Prt for r, V = lwh for h, and 3x + 2y = 12 for y. After each, the pair checks the result by choosing numbers that satisfy the original formula (for example, l = 10, w = 6 and h = 4 give V = 240) and substituting them into the new one. Circulate and listen for errors such as dividing only one term by a coefficient. Debrief 3x + 2y = 12, which gives y = -(3/2)x + 6, and point out that this is slope-intercept form.
Independent Practice10 minutes
Students work alone on three formulas of increasing difficulty: D = m/V for V (the variable is in a denominator), S = 2πr2 + 2πrh for h (two terms, one step of subtraction first), and A = P + Prt for P (the variable appears in two terms, so students factor out P). For each, they write the rearranged formula, check it with numbers, and state any restriction. Students who finish early solve K = (1/2)mv2 for v.
Closure5-10 minutes
Exit ticket: "The cost of a club T-shirt order is C = 12n + 25 dollars for n shirts. (a) Solve for n. (b) Use your formula to find how many shirts $385 buys. (c) Write the equation 12n + 25 = 385 and solve it directly. Explain why (a) and (c) use the same steps." A correct ticket shows n = (C - 25)/12 and n = 30.
Differentiation Strategies
For Struggling Students
Have students write a numerical twin for each formula (for example, 20 = 2l + 2(4) next to P = 2l + 2w) and solve both side by side
Highlight the quantity of interest in one color and every other letter in another color
Start with one-step formulas (V = IR, d = rt) before two-step and fraction formulas
Provide the list of operations applied to the variable and ask students only to reverse it
For Advanced Students
Rearrange formulas where the variable appears in two terms, such as A = P + Prt for P or y = (x + 2)/(x - 1) for x
Solve the lens formula 1/f = 1/do + 1/di for di and state the restriction on the result
Write a spreadsheet formula that computes the width of a rectangle from its perimeter and length, and explain how it matches the rearranged formula
Assessment Guidance
What to Look For
Ask students to explain each step, not just produce the final formula. The standard asks for the same reasoning used in solving equations, so a student who can say "I subtracted 2l from both sides because 2l was added to 2w" has met it. Watch for three errors: dividing only one term by a coefficient (w = P - 2l/2), undoing operations in the wrong order (C = (5/9)F - 32), and leaving the variable of interest on both sides (P = A - Prt). A numerical check catches all three.
02
Classroom Activities
3 Activities
1
Formula Relay
20 minGroups of 3-4
Each group gets one formula card and a list of quantities of interest. Each student in turn rearranges the formula for the next variable on the list, and the group checks every version with one set of numbers.
Formula Cards
Simple interest I = Prt: solve for P, then r, then t. Check with P = 1,500, r = 0.04, t = 3 and I = 180.
Rectangular box V = lwh: solve for l, then w, then h. Check with l = 10, w = 6, h = 4 and V = 240.
Distance d = rt: solve for r, then t. Check with r = 60, t = 2.5 and d = 150.
Temperature F = (9/5)C + 32: solve for C. Check with C = 25 and F = 77.
Procedure
Student 1 rearranges for the first variable and passes the card; student 2 checks the work with the numbers before starting the next variable
If a check fails, the group stops and finds the error together
Debrief: Which rearrangement needed the most steps? Which used a step that none of the others did?
Modification for Distance Learning
Run the relay in a shared document with one row per student. Each student types the rearranged formula and the numerical check before the next student starts.
2
Find the Error
15 minPairs
Pairs receive six worked rearrangements, four of which contain an error. They find each error, explain it, and write the correct result.
Worked Rearrangements
P = 2l + 2w for w: "w = P - 2l/2" (error: only 2l was divided by 2; correct w = (P - 2l)/2)
F = (9/5)C + 32 for C: "C = (5/9)F - 32" (error: 32 must be subtracted before multiplying; correct C = (5/9)(F - 32))
A = P + Prt for P: "P = A - Prt" (error: P is still on the right side; correct P = A/(1 + rt))
D = m/V for V: "V = mD" (error: V was in the denominator; correct V = m/D)
V = IR for R: "R = V/I" (correct)
A = (1/2)bh for h: "h = 2A/b" (correct)
Discussion Questions
How could a numerical check have caught each error?
Which error comes from undoing operations in the wrong order?
Why does "P = A - Prt" not count as solving for P, even though every step is legal?
3
Formulas at Work
20 minIndividual then share
Each student chooses a formula used in a job or a hobby (a nurse's dosage rate, a cyclist's speed, a recipe scaled by servings, a loan's simple interest), rearranges it for a different quantity, and explains when a worker would need the new version.
Requirements
The original formula with every variable defined, including units
The rearranged formula with every step explained in words
A numerical check, and any restriction on the variables
One sentence describing a real situation in which the rearranged version is the one you would use
Gallery Walk Variation
Post the work around the room. Classmates choose one formula, test it with their own numbers, and leave a sticky note with their check.
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: Solving an Equation and Rearranging a Formula Use the Same Steps
The numerical equation 2x + 6 = 20 and the perimeter formula P = 2l + 2w are solved with the same two inverse operations in the same order. Display this side by side when introducing the standard.
Diagram 2: Undoing Operations in Reverse Order
To build F from C, multiply by 9/5 and then add 32. To get C back, undo the last operation first: subtract 32, then multiply by 5/9. The numbers 25 and 77 show that both directions agree.
04
Homework Assignment
~30 min
HSA.CED.A.4 Homework: Rearranging Formulas
Directions: For each problem, (a) solve the formula for the variable named, showing and explaining each step, (b) check your rearranged formula by substituting numbers, and (c) use it to answer the question. State any value a variable cannot take.
Part 1: One and Two Steps (Problems 1-3)
The distance a car can travel on a full tank is d = mg, where m is its fuel economy in miles per gallon and g is the number of gallons. Solve the formula for g. How many gallons does a car that gets 32 miles per gallon need for a 400-mile trip?
Newton's second law is F = ma, where F is the net force in newtons, m is the mass in kilograms and a is the acceleration in meters per second squared. Solve it for a. A net force of 3,000 newtons acts on a 1,200-kilogram car. What is its acceleration?
A gym membership costs C = 30m + 49 dollars for m months, including a $49 sign-up fee. Solve the formula for m. How many months of membership does $409 pay for?
Part 2: Multi-Step Formulas (Problems 4-6)
The volume of a cone is V = (1/3)πr²h. Solve it for h. A cone-shaped paper cup has a radius of 2 inches and holds 12π cubic inches. How tall is it?
The surface area of a cylinder is S = 2πr² + 2πrh. Solve it for h. A can has a radius of 5 cm and a surface area of 150π square centimeters. What is its height?
With sales tax, the total cost of an item is T = p + rp, where p is the price before tax and r is the tax rate as a decimal. Solve the formula for p. (Hint: p appears in two terms, so factor it out first.) A jacket costs $54 including 8% sales tax. What was the price before tax?
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
No Credit (0 pts)
Rearranged Formula
Correct, with the variable alone on one side
Correct steps, one algebra error
Incorrect or missing
Reasoning
Each step named as an inverse operation
Steps shown without explanation
No steps shown
Check
Numerical check shown and correct
Check attempted with an error
No check
Answer in Context
Correct value with units and any restriction stated
Correct value, units or restriction missing
No answer
05
Quiz: 20 Questions
Interactive, with answers
Instructions
Work through the questions in order. Your score updates as you answer, and Reset quiz clears everything so you or your students can try again.
Multiple choice: pick an option to check it. Short answer: write your answer, then reveal the model answer.
0 of 20 answered · 0 correct
Question 1 of 20 · Multiple Choice
Ohm's law states V = IR. Which formula gives the current I?
Answer: C
I is multiplied by R, so divide both sides by R: I = V/R. Check with V = 12, R = 4: I = 3, and 3 × 4 = 12. Choice A multiplies instead of dividing. Choice B divides in the wrong order.
Question 2 of 20 · Multiple Choice
Solve P = 2l + 2w for l.
Answer: B
Subtract 2w from both sides, then divide both sides by 2: l = (P - 2w)/2. Choice A forgets to divide by 2. Choice C divides only P by 2, not the whole side. Choice D adds 2w instead of subtracting it.
Question 3 of 20 · Multiple Choice
Solve y = (3/4)x - 6 for x.
Answer: D
The last operation applied to x was subtracting 6, so add 6 first, then multiply by 4/3: x = (4/3)(y + 6). Choice A undoes the operations in the wrong order. Choice B multiplies by 3/4 again instead of by its reciprocal. Choice C subtracts 6 instead of adding it.
Question 4 of 20 · Multiple Choice
The average of two test scores is A = (x + y)/2. Solve the formula for y, then find the score a student needs on the second test to average 85 if the first score was 78.
Answer: A
Multiply both sides by 2 to get 2A = x + y, then subtract x: y = 2A - x. With A = 85 and x = 78: y = 170 - 78 = 92. Check: (78 + 92)/2 = 85. Choice B subtracts 78 from 85 without first multiplying by 2. Choice C averages the two numbers given instead of solving for y. Choice D adds the two given numbers, 85 + 78, instead of solving for y.
Question 5 of 20 · Multiple Choice
Solve A = (1/2)bh for b.
Answer: C
Multiply both sides by 2 to get 2A = bh, then divide by h: b = 2A/h. Choice A divides by 2 instead of multiplying, so it undoes the 1/2 incorrectly. Choice B multiplies by h instead of dividing.
Question 6 of 20 · Multiple Choice
A driver travels 210 miles in 3.5 hours. Using d = rt solved for r, what is the average speed?
Answer: A
Solve for r: r = d/t = 210/3.5 = 60 miles per hour. Choice B multiplies d and t. Choice C subtracts t from d. Choice D divides t by d, which inverts the formula.
Question 7 of 20 · Multiple Choice
Solve 4x + 3y = 15 for y.
Answer: D
Subtract 4x: 3y = -4x + 15. Divide every term by 3: y = -(4/3)x + 5. Choice A loses the negative sign. Choice C divides only the x-term by 3, not the constant. Choice B forgets to divide by 3 at all.
Question 8 of 20 · Multiple Choice
The volume of a cylinder is V = πr2h. Solve for r, where r is the radius.
Answer: A
Divide by πh to get r2 = V/(πh), then take the positive square root because a radius is positive: r = √(V/(πh)). Choice B takes the square root before dividing by πh, so πh is not under the root. Choice C treats squaring as multiplying by 2. Choice D multiplies by πh instead of dividing.
Question 9 of 20 · Multiple Choice
Solve E = mc2 for m.
Answer: B
m is multiplied by c2, so divide both sides by c2: m = E/c2. Choice A subtracts, which undoes addition, not multiplication. Choice C multiplies again instead of dividing.
Question 10 of 20 · Multiple Choice
To solve ax + b = c for x, a student first subtracts b from both sides. Which equation results?
Answer: A
Subtracting b from both sides of ax + b = c gives ax = c - b, just as subtracting 5 from both sides of 3x + 5 = 20 gives 3x = 15. Choice B adds b to the right side instead of subtracting it. Choice D reverses the subtraction.
Question 11 of 20 · Multiple Choice
A box has volume V = lwh. Solve for h, then find h when V = 360 cubic inches, l = 12 inches and w = 5 inches.
Answer: C
h = V/(lw) = 360/(12 × 5) = 360/60 = 6 inches. Choice A divides by l only. Choice B multiplies all three numbers. Choice D subtracts l + w from V.
Question 12 of 20 · Multiple Choice
Pressure is P = F/A, where F is the force and A is the area it acts on. Which formula gives A?
Answer: D
Multiply both sides by A to get PA = F, then divide by P: A = F/P. Choice A multiplies F by P, a common error when the variable is in a denominator. Check with F = 200 newtons and A = 4 square meters: P = 50, and F/P = 4.
Question 13 of 20 · Multiple Choice
Solve ax + bx = c for x. Assume a + b ≠ 0.
Answer: B
Factor x out of the left side: x(a + b) = c. Divide both sides by (a + b): x = c/(a + b). Choice A is a true equation but not a solution, because x still appears on the right side. Choice C multiplies the coefficients instead of adding them. Choice D subtracts, which does not undo multiplication.
Question 14 of 20 · Multiple Choice
Solve I = Prt for t.
Answer: C
t is multiplied by the product Pr, so divide both sides by Pr: t = I/(Pr). Choice B inverts the fraction. Choice D subtracts, which does not undo multiplication.
Question 15 of 20 · Short Answer
The surface area of a rectangular box is S = 2lw + 2lh + 2wh. Solve for h.
h = (S - 2lw)/(2l + 2w) Subtract 2lw from both sides: S - 2lw = 2lh + 2wh. Factor h out of the right side: S - 2lw = h(2l + 2w). Divide both sides by 2l + 2w. Check: l = 5, w = 3 and h = 2 give S = 30 + 20 + 12 = 62, and (62 - 30)/(10 + 6) = 32/16 = 2.
Question 16 of 20 · Short Answer
Kinetic energy is K = (1/2)mv2. Solve for v, assuming v is positive. Then find v when K = 50 joules and m = 4 kilograms.
v = √(2K/m) Multiply both sides by 2: 2K = mv2. Divide by m: v2 = 2K/m. Take the positive square root. With K = 50 and m = 4: v = √(100/4) = √25 = 5 meters per second.
Question 17 of 20 · Short Answer
The area of a trapezoid is A = (1/2)h(b1 + b2). Solve for b2, then find b2 when A = 45, h = 5 and b1 = 7.
b2 = 2A/h - b1 Multiply by 2: 2A = h(b1 + b2). Divide by h: 2A/h = b1 + b2. Subtract b1. With the given values: b2 = 90/5 - 7 = 18 - 7 = 11.
Question 18 of 20 · Short Answer
Solve 5x - 7 = 18 for x and solve ax - b = c for x. Explain why the two solutions use the same steps.
x = 5 and x = (c + b)/a In both, add the constant to both sides (5x = 25 and ax = c + b), then divide both sides by the coefficient (x = 25/5 = 5 and x = (c + b)/a). The steps are the same because letters stand for numbers, so the same properties of equality apply. The second answer requires a ≠ 0.
Question 19 of 20 · Short Answer
The circumference of a circle is C = 2πr. Solve for r, then find the radius of a circular walking path around a pond with a circumference of 400 meters, to the nearest tenth of a meter.
r = C/(2π) Divide both sides by 2π. With C = 400: r = 400/(2π) ≈ 63.66, so the radius is about 63.7 meters.
Question 20 of 20 · Short Answer
A student solved y = mx + b for x and wrote x = y - b/m. Explain the error and give the correct formula.
Correct: x = (y - b)/m After subtracting b, the equation is y - b = mx, and the whole left side must be divided by m. The student's formula divides only b by m. A check shows the error: m = 2, b = 3 and x = 5 give y = 13, and (13 - 3)/2 = 5, but the student's formula gives 13 - 3/2 = 11.5. The correct formula requires m ≠ 0.
0 of 20 answered · 0 correct
06
Frequently Asked Questions
10 Questions
What does "highlight a quantity of interest" mean?
It means rewriting a formula so that the quantity you care about is alone on one side. The formula V = IR is convenient for finding voltage, but an electrician who needs the resistance wants R = V/I. Both formulas describe the same relationship; the rearranged one puts the quantity of interest in front.
Is rearranging a formula different from solving an equation?
No, and that is the point of the standard. Students use the same properties of equality and the same inverse operations in the same order. The only difference is that the answer is an expression in other variables, such as w = (P - 2l)/2, instead of a number such as x = 7. Placing a numerical equation next to the formula makes this visible.
How do I know which operation to undo first?
List the operations applied to the variable of interest in the order they happen, then undo them in reverse. In F = (9/5)C + 32, C is multiplied by 9/5 and then 32 is added, so you subtract 32 first and multiply by 5/9 second. Think of taking off shoes and socks: what went on last comes off first.
What if the variable appears in more than one term?
Collect those terms on one side and factor the variable out. In A = P + Prt, factor to get A = P(1 + rt), then divide by (1 + rt) to get P = A/(1 + rt). A result such as P = A - Prt is not finished, because P still appears on both sides.
What restrictions should students state?
Any variable you divide by cannot be zero: R = V/I requires I ≠ 0. When you take a square root, decide whether the negative root makes sense; a radius or a speed is positive, so r = √(A/π) uses only the positive root. Context also matters: a length, a mass or a time must be positive.
What are the common mistakes?
Dividing only one term by a coefficient, as in w = P - 2l/2
Undoing operations in the wrong order, as in C = (5/9)F - 32
Multiplying when the variable is in a denominator, as in V = mD for D = m/V
Leaving the variable of interest on both sides
Losing a negative sign when solving a linear equation such as 3x + 2y = 12 for y
How can students check a rearranged formula?
Choose numbers that satisfy the original formula, then substitute them into the new one. For P = 2l + 2w, the values l = 6 and w = 4 give P = 20. Substituting P = 20 and l = 6 into w = (P - 2l)/2 gives (20 - 12)/2 = 4, which matches. A wrong formula almost always fails this test.
Does HSA.CED.A.4 include formulas with squares or square roots?
Yes. The standard does not limit the type of formula. Courses usually begin with formulas that are linear in the variable of interest, such as d = rt and P = 2l + 2w, and add formulas such as A = πr2 or K = (1/2)mv2 once students can solve simple quadratic equations by taking square roots. This lesson includes both, with the squared cases marked in the examples.
Is HSA.CED.A.4 tested on the SAT?
Yes. Digital SAT math questions sometimes give a formula and ask which equation expresses one of its variables in terms of the others. These questions fall under the Algebra and Advanced Math domains, depending on the formula. The routine in this lesson, undoing operations in reverse order and checking with numbers, applies directly.
Where is this used outside math class?
Science courses use it constantly: students rearrange d = rt, D = m/V, V = IR and F = ma to find the quantity an experiment asks for. Spreadsheets, dosage calculations, unit conversions and loan calculations all rely on formulas solved for a particular quantity. Later in algebra, the same reasoning is used to find inverse functions (HSF.BF.B.4).
07
Related Standards
5 standards
These standards connect to HSA.CED.A.4: prerequisites to review first, parallel standards at the same level, and next steps that build on it.
