Rearrange formulas to highlight a quantity of interest using the reasoning of equation solving

Learning goal: Rearrange a formula to highlight a quantity of interest, isolating that quantity on one side using the same reasoning as in solving equations, for formulas in which the quantity is linear, for formulas in which it is squared, and for more complicated formulas in which it appears more than once or inside a fraction.

A formula is an equation with several variables. The formula for the area of a rectangle, A = lw, relates three quantities. Written this way it highlights A: given l and w, compute A. But a carpenter who knows the area and the length wants w. Rearranging the formula to w = A/l highlights the width instead. Nothing about the relationship has changed; only which quantity is written alone on one side. Solving a formula for a chosen variable uses exactly the moves used to solve an equation in one variable: add or subtract the same thing from both sides, multiply or divide both sides by the same nonzero quantity, take the square root of both sides when the variable is squared. The only difference is that the other letters are carried along instead of being numbers.

Quantities that appear linearly. Ohm's law states V = IR, voltage equals current times resistance. To highlight the resistance R, divide both sides by I: R = V/I, valid when I ≠ 0. With V = 12 volts and I = 0.5 amperes, R = 12/0.5 = 24 ohms. The perimeter formula P = 2l + 2w highlighted for w: subtract 2l from both sides, P − 2l = 2w, then divide by 2, w = (P − 2l)/2. The temperature formula F = (9/5)C + 32 highlighted for C: subtract 32, F − 32 = (9/5)C, then multiply by 5/9, C = (5/9)(F − 32). Check with a value: F = 212 gives C = (5/9)(180) = 100. The step students skip is subtracting 32 before multiplying; multiplying first, to get (5/9)F = C + (5/9)32, is not wrong but leaves the work unfinished and invites an error.

Quantities that appear squared: a fully worked model. The area of a circle is A = πr². To highlight the radius: divide both sides by π to get A/π = r², then take the square root of both sides, r = √(A/π). Only the positive root is kept, because a radius is a positive length; the equation r² = A/π has two solutions, but the situation allows one. With A = 50 square centimeters, r = √(50/π) ≈ √15.915 ≈ 3.99 centimeters. Check: π × 3.99² ≈ 3.1416 × 15.92 ≈ 50.0. The same reasoning solves the kinetic energy formula K = ½mv² for the speed: multiply by 2, 2K = mv²; divide by m, 2K/m = v²; take the root, v = √(2K/m), with m > 0 and K ≥ 0.

Formulas where the quantity appears more than once. The surface area of a closed cylinder is S = 2πr² + 2πrh. To highlight h, which appears once, subtract 2πr² and divide by 2πr: h = (S − 2πr²)/(2πr). With S = 150 and r = 3, h = (150 − 18π)/(6π) ≈ (150 − 56.55)/18.85 ≈ 4.96. Highlighting r in the same formula is harder because r appears in two terms with different powers; that is a quadratic in r and needs the quadratic formula, an example of the more complicated situations. The surface area of a closed rectangular box, S = 2lw + 2lh + 2wh, solved for h: the terms with h are 2lh + 2wh; move 2lw to the other side, S − 2lw = 2lh + 2wh; factor h out, S − 2lw = h(2l + 2w); divide, h = (S − 2lw)/(2l + 2w), valid when l + w ≠ 0. Factoring out the quantity of interest when it appears in several terms is the key move, and it is the same move used to solve ax + b = cx + d for x: ax − cx = d − b, x(a − c) = d − b, x = (d − b)/(a − c), provided a ≠ c.

Formulas with the quantity in a denominator or an exponent. The thin-lens formula 1/f = 1/u + 1/v relates a lens's focal length f, the distance u from the object to the lens and the distance v from the lens to the image; here it is used as a given algebraic relationship, with all three distances positive (the physics behind it and its sign conventions belong to a science course). Highlighted for v: subtract 1/u, 1/v = 1/f − 1/u = (u − f)/(fu), then take reciprocals of both sides, v = fu/(u − f), valid when u ≠ f and f, u ≠ 0. Compound growth A = P(1 + r)ᵗ highlighted for the starting amount: P = A/(1 + r)ᵗ; highlighted for the rate, (1 + r)ᵗ = A/P, so 1 + r = (A/P)^(1/t) and r = (A/P)^(1/t) − 1, valid for t ≠ 0, P ≠ 0 and A/P > 0 with 1 + r > 0, using the t-th root, the inverse of the t-th power in the same way the square root inverts squaring.

Stating the conditions. Every division carries a condition that the divisor is not zero, and every square root a condition that the quantity under it is not negative and, usually, that only the positive root fits the situation. A rearranged formula should be written with those conditions, as in R = V/I for I ≠ 0. They are not decoration: V = IR with I = 0 says V = 0 for any R, and R is genuinely undetermined there.

The misconception to remove. Students sometimes move a term across the equals sign by changing its sign even when it is a factor, writing V = IR as R = V − I. The move for a factor is division, not subtraction, because the inverse of multiplying by I is dividing by I. Testing with numbers exposes it: V = 12 and I = 0.5 give R = 24 by division, but V − I = 11.5 is not the resistance. A second error is taking the square root of each term separately in r² = A/π; the root is taken of each whole side.

Why "the same reasoning as solving equations" is the point. A student who can solve 3x + 7 = 22 by subtracting 7 and dividing by 3 can solve y = mx + b for x by subtracting b and dividing by m. The letters are different, the logic is identical, and the check is identical too: substitute the rearranged result back into the original formula and confirm it holds.

Worked linear quantity: Solve the distance formula d = rt for t, and use it. Divide both sides by r, valid when r ≠ 0: t = d/r. For a 150-mile trip at 60 miles per hour, t = 150/60 = 2.5 hours. Check in the original: 60 × 2.5 = 150. The same one-step division highlights r instead: r = d/t for t ≠ 0.

Worked squared quantity: Solve the cone volume formula V = (1/3)πr²h for r. Multiply both sides by 3: 3V = πr²h. Divide by πh (h ≠ 0): 3V/(πh) = r². Take the positive square root, since r is a length: r = √(3V/(πh)). With V = 100 and h = 12, r = √(300/(12π)) = √(25/π) ≈ √7.958 ≈ 2.82. Check: (1/3)π(2.82)²(12) ≈ (1/3)(3.1416)(7.95)(12) ≈ 99.9.

Worked quantity appearing twice: Solve the box surface-area formula S = 2lw + 2lh + 2wh for l. Collect the terms containing l on one side: 2lw + 2lh = S − 2wh. Factor: l(2w + 2h) = S − 2wh. Divide by 2w + 2h, which is positive for real dimensions: l = (S − 2wh)/(2w + 2h). Check with l = 2, w = 3, h = 4: S = 12 + 16 + 24 = 52, and (52 − 24)/(6 + 8) = 28/14 = 2.

Worked quantity in a denominator: Solve the average-speed relationship v = d/t for t and then the lens formula 1/f = 1/u + 1/v for u. First: multiply by t, vt = d, divide by v, t = d/v. Second, where v now means the image distance, not a speed: 1/u = 1/f − 1/v = (v − f)/(fv), so u = fv/(v − f), valid when v ≠ f. Check with f = 2, v = 6: 1/2 − 1/6 = 1/3, so u = 3; and fv/(v − f) = 12/4 = 3.

Representation task: Make a four-column table with the headings "formula", "quantity of interest", "rearranged formula" and "condition". Fill it for V = IR (for R), A = πr² (for r), P = 2l + 2w (for w), F = (9/5)C + 32 (for C), A = ½h(b₁ + b₂) (for b₁), S = 2lw + 2lh + 2wh (for h) and 1/f = 1/u + 1/v (for v). For two of the rows, write the step-by-step rearrangement in the margin and label each step with the equation-solving move it uses (subtract from both sides, divide both sides, take the square root of both sides, factor out the quantity).

Independent performance: Choose five formulas from science, geometry or finance that you can state correctly. For each, identify a quantity someone might need to compute, rearrange the formula to highlight it, state every condition, and check the rearrangement with a set of numbers substituted into both the original and the rearranged version. At least one formula must have the quantity squared, at least one must have it appearing in two terms, and at least one must have it inside a fraction. Success criteria: each step is a legitimate both-sides move; only whole sides are square-rooted; conditions such as nonzero divisors and positive roots are stated; and every numerical check agrees in both forms.