Rearrange a formula: choose the unknown, keep the conditions
A formula describes a relationship. Rearranging it makes one chosen quantity the subject without changing that relationship on its allowed domain. Other letters represent quantities held fixed for this calculation; a letter is not automatically the unknown. This original workshop bridges high-school algebra and the formulas used in geometry and introductory science. It is practice, not an official assessment.
This continues the starter examples on isolating a variable, adding repeated variables, zero-coefficient cases and physical restrictions. Review denominator restrictions when you need them. Work symbolically first, then substitute numbers with consistent units. The two worked examples show the method; the six questions gradually remove the guidance.
One routine, with a check before each division
- Name the quantity to isolate and record restrictions from the original formula.
- Undo addition and subtraction around the target. Clear fractions by multiplying every term on both sides by a permitted common denominator.
- If the target occurs in several terms, collect those terms and factor the target out.
- Divide only after checking that the whole divisor is nonzero. If it can be zero, return to the equation before division and examine that case.
- Substitute the result into the original relationship. Check units and any positive-length or time conditions separately.
Why factor first? From px + q = rx + s, subtract rx and q to obtain (p − r)x = s − q. For p ≠ r, this gives x = (s − q)/(p − r). If p = r, the equation instead reduces to q = s: every real x works when that statement is true, and none works when it is false. Division by p − r would erase those cases.
Worked example 1: recover a trapezoid’s second base
A trapezoid has positive parallel base lengths a,b, perpendicular height h > 0, and area A. Its formula is A = (a + b)h/2. Isolate b, then find it when A = 54 cm², a = 5 cm and h = 6 cm.
Multiply both sides by 2, divide by the nonzero height, then subtract the entire first base:
The numerical result is b = 2(54)/6 − 5 = 13 cm. Substituting into the original formula gives (5 + 13)(6)/2 = 54 cm². Also, area divided by height has units of length, so the subtraction is meaningful.
Check the model: a positive second base requires 2A > ah. If the calculation gives zero or a negative value, the supplied measurements do not describe a trapezoid with two positive bases. Algebraic rearrangement does not remove that geometric condition.
Avoid this: (2A − a)/h is not the same formula. Subtract a only after dividing 2A by h. If writing one fraction, the correct numerator is 2A − ah.
Worked example 2: isolate acceleration, then test zero time
For straight-line motion with constant acceleration, displacement s, initial velocity u, elapsed time t, and acceleration a satisfy s = ut + at²/2. Here s,u,a are signed quantities relative to a chosen axis. Isolate a.
If s = 42 m, u = 3 m/s and t = 4 s, then a = 2(42 − 12)/16 = 3.75 m/s². The original equation checks: 3(4) + 3.75(4²)/2 = 12 + 30 = 42 m. Both s and ut are lengths; dividing their difference by time squared gives acceleration units.
What if t = 0? Do not use the divided formula. The original equation becomes s = 0. If s = 0, any real acceleration satisfies this single equation, so the equation cannot determine it. If s ≠ 0, no acceleration can satisfy it. In an elapsed-time application, a nonzero time is positive; the algebraic condition alone is just t ≠ 0.
Six questions: isolate, verify, then interpret
Write the restriction beside each rearranged formula. Open a hint if you need a first move; compare your reasoning with the full answer afterward.
1. A missing width
A rectangle has perimeter P = 2L + 2W and positive length L and width W. Isolate W. Find it for P = 46 cm, L = 15 cm, and state the condition for a positive width.
Hint 1
Subtract both length contributions before dividing by 2.
Answer 1 with reasoning
Subtracting 2L gives 2W = P − 2L, so W = (P − 2L)/2 = P/2 − L. Here W = 23 − 15 = 8 cm; checking gives 2(15) + 2(8) = 46 cm. Division by 2 needs no variable restriction. For the geometric model, W > 0 requires P > 2L, in addition to L > 0.
2. Reverse an affine formula
The numerical temperature readings C in degrees Celsius and F in degrees Fahrenheit obey C = 5(F − 32)/9. Isolate F, find it when C = 20, and explain why 9C/5 − 32 is incorrect.
Hint 2
First undo multiplication by 5/9. Then undo subtracting 32.
Answer 2 with reasoning
Multiply by 9/5 to get 9C/5 = F − 32, then add 32: F = 9C/5 + 32. For C = 20, F = 68; the check is 5(68 − 32)/9 = 20. Subtracting 32 again does not undo the original subtraction. No variable denominator or square root introduces a restriction in this algebraic conversion.
3. Choose a root that fits a radius
A cylinder has volume V = π r²h, with V > 0, h > 0, and radius r > 0. Isolate r. Use V = 200π cm³ and h = 8 cm. Explain what happens to the negative square root.
Hint 3
Divide by π h before taking square roots. Distinguish a real equation’s roots from the permitted geometric values.
Answer 3 with reasoning
Since h > 0, division is allowed: r² = V/(π h). The unrestricted real equation has the two roots , but a positive radius selects . Here , and π(5)²(8) = 200π cm³. The radicand has units of area, and its positive square root has units of length.
4. The target appears twice
Let y = (3x + 2)/(x − 1), with real x ≠ 1. Isolate x in terms of y. Decide whether y = 3 is possible, and find x when y = − 2.
Hint 4
Multiply by x − 1, collect the two terms containing x, then factor out x. Keep the original excluded value.
Answer 4 with reasoning
The original restriction permits y(x − 1) = 3x + 2. Thus yx − 3x = y + 2, or (y − 3)x = y + 2. For y ≠ 3,
The rearranged value therefore respects the original restriction. If y = 3, the equation before division would be 0x = 5, which is impossible. When y = − 2, x = 0; substitution gives (3(0) + 2)/(0 − 1) = − 2. The missing value 3 belongs to the output, whereas the original excluded value 1 belongs to the input.
5. Decide what a zero coefficient means
Here a is a fixed real parameter. Solve ax + 6 = 2x + 14 for all values of a. Then repeat for ax + 6 = 2x + 6. Explain why the case a = 2 has different outcomes.
Hint 5
Collect the x-terms into (a − 2)x. Examine the remaining equality at a = 2 before dividing.
Answer 5 with reasoning
The first equation gives (a − 2)x = 8. If a ≠ 2, its unique solution is x = 8/(a − 2). If a = 2, it gives 0 = 8, so there is no solution. The second gives (a − 2)x = 0. For a ≠ 2, its unique solution is x = 0. For a = 2, it gives 0 = 0, so every real x is a solution. A zero coefficient means return to the equation, not that the answer is automatically zero.
6. Transfer to a cylinder
A closed cylinder with r > 0 and h > 0 has total surface area S = 2π r² + 2π rh. Isolate h, state which values of S give a positive height for a fixed radius, and calculate h when r = 3 cm and S = 66π cm². A student proposes h = S/(2π r) − r². Diagnose the error.
Hint 6
Remove the two circular ends before dividing by the circumference. When splitting a fraction, divide every numerator term by the full denominator.
Answer 6 with reasoning
Positive height requires S > 2π r². With the supplied values, h = (66π − 18π)/(6π) = 8 cm. The check is 2π(3)² + 2π(3)(8) = 18π + 48π = 66π cm². The student failed to cancel 2π r²/(2π r) = r. Their subtraction also mixes a length with an area, which exposes the error without numerical substitution.
Retry without notes: name the target, identify a forbidden division, and verify the result in the original formula. For more geometric interpretation, continue to surface area and nets. For further work on variable denominators, return to rational equations. Use your own course outline to choose the appropriate level.
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