Technical Algebra and Formula Rearrangement

Technical work often begins with a formula whose needed quantity is not already alone. This lesson develops a dependable method for rearranging literal equations, handling ratios and powers, and applying the result with units. It builds on the equation-solving ideas in the Algebra — Expressions, Equations, and Inequalities lesson and connects naturally to scientific notation and functions.

Equality Means “Same Value”

An equation remains true when the same reversible operation is applied to both sides. To isolate a target variable, undo the operations around it in reverse order. Treat the other letters as known numbers while rearranging.

For example, in F = ma, multiplication joins m and a. If m is nonzero, divide both sides by m to make a the subject:

a = F/m.

A restriction matters: division by zero is undefined, so this rearrangement assumes m ≠ 0. A formula may be algebraically correct but physically inappropriate for a particular situation; units and meaningful ranges provide a second check.

A Repeatable Isolation Method

1. State the variable to be isolated and note any denominators, roots, or possible zero values.
2. Clear fractions first when that makes the equation simpler.
3. Use inverse operations on both sides, one layer at a time.
4. Factor the target variable if it appears in more than one term.
5. Check by substituting the rearranged expression into the original equation and by checking units.

Worked example: a mechanical formula with a grouped denominator

For a belt drive, the rotational speed relationship is n₁d₁ = n₂d₂, where n is speed and d is pulley diameter. Isolate d₂.

n₁d₁ = n₂d₂

Divide both sides by n₂ (where n₂ ≠ 0):
d₂ = n₁d₁ / n₂

If a 1200 rpm motor drives a 60 mm pulley and the driven shaft is to run at 400 rpm, then

d₂ = (1200 rpm)(60 mm) / (400 rpm)
   = 180 mm

The rpm units cancel, leaving millimetres. The slower driven shaft needs the larger pulley, which is a useful reasonableness check.

Fractions, Products, and Powers

Parentheses identify a whole factor. In R = V/(I + I₀), multiplying by I + I₀ clears the entire denominator; multiplying only by I would be an error.

Worked example: isolate a current from a sensor model

A simplified current-sense model is V = V_off + GIR_s, where G is gain and R_s is a sense resistance. Solve for I.

V = Voff + GIRs
V − Voff = GIRs

Divide by the complete product GRs:
I = (V − Voff) / (GRs)

Let V = 2.50 V, V_off = 0.10 V, G = 20, and R_s = 0.050 Ω.

I = (2.50 V − 0.10 V) / [20(0.050 Ω)]
  = 2.40 V / 1.00 Ω
  = 2.40 A

Since an ohm is a volt per ampere, V/Ω = A. Keeping parentheses until the last numerical step prevents denominator mistakes.

Undoing a square

When a variable is squared, first isolate its square, then take a square root. In a physical context, choose the root consistent with the quantity. From kinetic energy E = ½mv², with m > 0:

2E = mv²
2E / m = v²
v = ±√(2E/m)

Mathematically there are two velocity directions. If v means speed, it is nonnegative, so v = √(2E/m).

Variables on Both Sides

When the target appears in several terms, collect those terms by factoring. This is common in formulas with losses, offsets, and gains.

Worked example: solve for a temperature coefficient

A resistance model is R = R₀(1 + αΔT). Isolate α.

R / R₀ = 1 + αΔT          (R₀ ≠ 0)
R / R₀ − 1 = αΔT
α = (R/R₀ − 1) / ΔT       (ΔT ≠ 0)

For R = 112.0 Ω, R₀ = 100.0 Ω, and ΔT = 30.0 °C:

α = (112.0/100.0 − 1) / 30.0 °C
  = 0.12 / 30.0 °C
  = 0.0040 °C⁻¹

The unit °C⁻¹ means “per degree Celsius.” A temperature difference has the same numerical size in degrees Celsius and kelvins.

Ratios, Proportions, and Scale

A ratio compares like kinds of quantities by division. A proportion says two ratios are equal. If a/b = c/d, with b and d nonzero, cross-multiplication gives ad = bc. It is not a shortcut to use unless two complete fractions are equal.

Worked example: scale a material mixture

A resin mix uses 100 g resin for every 35 g hardener. How much hardener corresponds to 260 g resin?

100 g resin / 35 g hardener = 260 g resin / h

100h = 35(260)
h = 91 g hardener

Alternatively, use the scale factor 260/100 = 2.6, then 35(2.6) = 91. Both quantities were multiplied by the same factor.

Scientific and Engineering Notation

Scientific notation writes a number as a × 10ⁿ, where 1 ≤ |a| < 10. Engineering notation uses exponents that are multiples of 3, matching prefixes such as milli (m, 10⁻³), micro (µ, 10⁻⁶), kilo (k, 10³), and mega (M, 10⁶).

Quantity	Scientific notation	Engineering form
0.000047 F	4.7 × 10−5 F	47 µF
3 300 000 Ω	3.3 × 106 Ω	3.3 MΩ
0.012 A	1.2 × 10−2 A	12 mA

For multiplication, multiply coefficients and add exponents. For division, divide coefficients and subtract exponents.

(6.8 × 10³ Ω)(2.2 × 10⁻⁶ F)
= (6.8 × 2.2) × 10^(3 + −6) s
= 14.96 × 10⁻³ s
= 14.96 ms

This is the time constant RC of a 6.8 kΩ resistor and a 2.2 µF capacitor. Notice that Ω·F = s.

Linear Interpolation Between Measurements

Interpolation estimates a value y between two known calibration points, assuming the relationship is linear over that interval. Between (x₁, y₁) and (x₂, y₂), with x₁ ≠ x₂:

y = y₁ + (x − x₁)(y₂ − y₁)/(x₂ − x₁).

Worked example: estimate a thermistor reading

A linearized calibration table gives 1.80 V at 20°C and 2.40 V at 50°C. Estimate the voltage at 32°C.

V = 1.80 V + (32 °C − 20 °C)(2.40 V − 1.80 V)/(50 °C − 20 °C)
  = 1.80 V + (12/30)(0.60 V)
  = 2.04 V

Interpolation stays between the measured endpoints. Using this line at, say, 90°C is extrapolation, which may be unreliable.

Piecewise Calibration

Many real instruments are only approximately linear. A piecewise model uses a different rule in each stated range. The boundary condition (< versus ≤) must assign every input to exactly one rule.

Suppose a load cell is calibrated by

F(V) = 250V N for 0 ≤ V ≤ 2.0 V, and F(V) = (180V + 140) N for 2.0 V < V ≤ 5.0 V.

At the join, the first rule gives 250(2.0) = 500 N and the second would give 180(2.0) + 140 = 500 N, so the model has no jump. At 3.2 V, select the second interval:

F(3.2 V) = [180(3.2) + 140] N
         = 716 N

The coefficients have implied units: 250 N/V and 180 N/V are sensitivities, while 140 N is an offset.

Common Mistakes and Checks

• Dropping parentheses: From V = V_off + GIR_s, the denominator is GR_s, not G + R_s.
• Dividing one term only: In (a + b)/c, both terms are divided: a/c + b/c.
• Ignoring a ±: Solving x² = 9 gives x = ±3 unless the context restricts x.
• Mixing prefixes: Convert 4.7 kΩ to 4700 Ω before combining it with 220 Ω.
• Using interpolation outside its interval: Label extrapolated estimates and justify why the trend should continue.
• Skipping units: Units should balance on each line; a mismatch often reveals a wrong operation or an unsuitable formula.

Practice Set

1. Rearrange P = VI to make V the subject. State a restriction.
2. Rearrange Q = mc(ΔT) to make ΔT the subject.
3. For V = IR, find I when V = 9.0 V and R = 1.5 kΩ.
4. Make x the subject of y = (3x − 5)/4.
5. Make r the subject of A = πr². Give the physically appropriate result for a radius.
6. A gear with 18 teeth drives a gear with 54 teeth. Using n₁N₁ = n₂N₂, find n₂ if n₁ = 900 rpm.
7. Write 0.0000033 A in scientific notation and engineering notation.
8. Calculate (4.0 × 10⁻³ m)(7.5 × 10² N).
9. A signal is 0.80 V at 10°C and 1.40 V at 40°C. Assuming linear behaviour, estimate it at 25°C.
10. For f(x) = 2x + 1 when x ≤ 3, and f(x) = 5x − 8 when x > 3, find f(3) and f(4). Is the model continuous at 3?
11. From R = R₀(1 + αΔT), find α for R = 105 Ω, R₀ = 100 Ω, and ΔT = 20°C.
12. A formula gives t = 2d/(u + v). Find d if t = 8.0 s, u = 3.0 m/s, and v = 5.0 m/s.

Answer Checks

1. V = P/I, provided I ≠ 0.
2. ΔT = Q/(mc), provided m and c are nonzero.
3. I = 9.0 V/(1.5 kΩ) = 9.0/1500 A = 0.0060 A = 6.0 mA.
4. 4y = 3x − 5, so 3x = 4y + 5 and x = (4y + 5)/3.
5. r² = A/π, so r = √(A/π). A radius is nonnegative.
6. n₂ = 900(18)/54 rpm = 300 rpm.
7. 3.3 × 10⁻⁶ A; 3.3 µA.
8. (4.0 × 7.5) × 10⁻³⁺² N·m = 30 × 10⁻¹ N·m = 3.0 N·m.
9. 0.80 V + (25 − 10)(1.40 − 0.80)/(40 − 10) V = 0.80 V + 0.30 V = 1.10 V.
10. f(3) = 2(3) + 1 = 7. f(4) = 5(4) − 8 = 12. The right-hand rule approaches 5(3) − 8 = 7, so it is continuous at 3.
11. α = (105/100 − 1)/20°C = 0.05/20°C = 0.0025 °C⁻¹.
12. Multiply by u + v: t(u + v) = 2d. Thus d = 8.0 s(3.0 + 5.0) m/s / 2 = 32 m.