Formulas and rearranging simple relationships

A formula names a relationship between quantities

A formula is an equation used to express a relationship through defined symbols. For a rectangle, A = L W means area A equals side length L multiplied by side length W. The adjoining symbols denote multiplication: L W = L × W. The formula describes many permitted rectangles; after choosing particular known values, it becomes a particular equation to evaluate or solve.

Define the symbols and the situation before using the rule. Here L and W are positive length quantities and A is a positive area. They are side labels, not a requirement that L be the longer side. The relationship concerns a rectangle; it does not give the area of every shape with two named lengths. A formula’s conditions are part of its meaning.

Decide what is known and what you want to find

The same relationship can answer different questions. If L = 3 m and W = 2 m are known, A = L W gives A = 6 m². If instead A = 6 m² and W = 2 m are known, the requested quantity is L. The rectangle relationship stays the same while the roles of the known quantities and the unknown change.

The subject of a formula is the quantity isolated on one side. A is the subject of A = L W. Making L the subject means writing an equivalent relationship with L isolated, including any necessary conditions. Naming the subject does not change a measured object or assign new dimensions to it; it changes which quantity the written rule computes from the others.

Separate substituting values from rearranging the rule

Substitution replaces known symbols with their values while retaining the formula’s structure. With A = 6 m² and W = 4 m, the original relation gives 6 m² = L × 4 m. Dividing both sides by 4 m then gives L = 3/2 m. You can solve this numerical equation after substitution.

Alternatively, first rearrange A = L W symbolically to L = A/W, requiring W ≠ 0, and then substitute L = (6 m²)/(4 m) = 3/2 m. Both routes express the same reasoning under the same conditions. Rearranging first is useful when you will repeat the calculation for several sets of known values. Keep fractions exact when the inputs are exact.

Apply reversible operations to both complete sides

A symbolic rearrangement should preserve all assignments of the quantities that satisfy the original relationship in its stated domain. Add or subtract the same compatible quantity on both sides, or multiply or divide both sides by the same nonzero quantity. The inverse operation must recover the original relationship without adding or losing admissible assignments.

From A = L W, division by W gives A/W = L only where W ≠ 0; swapping sides gives L = A/W. Multiplication by W reverses the step on that domain. Merely saying that W moves across the equals sign does not explain either the division or its condition. If the original relation permits W = 0, this quotient form cannot describe that branch, which must be considered separately.

Recover a rectangle side from area and the other side

For positive rectangle sides, A = L W can be written as L = A/W or W = A/L. The divisors are nonzero because the geometric domain requires positive sides. At exact values A = 6 m² and W = 4 m, L = (6 m²)/(4 m) = 3/2 m. Dividing an area unit by a length unit leaves a length unit.

Check the original area: (3/2 m) × (4 m) = 6 m². The result is positive and has the expected dimension. A zero or negative side is excluded from this rectangle model, even if some arithmetic statement with those values can be written. Dividing by a supplied zero width would be undefined, not a way to recover a rectangle length.

Undo a grouped perimeter rule in the right order

For a rectangle, P = 2(L + W), where P is perimeter and both sides are positive lengths. To make L the subject, divide both complete sides by two: P/2 = L + W. Then subtract W from both sides: L = P/2 − W. To make W the subject, use the corresponding form W = P/2 − L. The number two is dimensionless and nonzero.

Keep the grouping in the original rule and compatible length units in its rearrangement. At P = 13 m and L = 6 m, W = (13 m)/2 − 6 m = 1/2 m. Check: 2 × (6 m + 1/2 m) = 13 m. For a permitted rectangle the recovered side must be positive; a zero or negative arithmetic result means the supplied data do not describe a rectangle in this domain.

State the conditions of a rate relationship

In a constant-speed model, d = v t relates distance travelled d, nonnegative speed v and elapsed time t. Take d ≥ 0, v ≥ 0 and t ≥ 0. The formula gives distance from speed and time, but recovering speed by v = d/t requires t ≠ 0; in this context that means positive elapsed time. Recovering time by t = d/v requires v ≠ 0, which means positive speed here.

At exact distance d = 9 km and time t = 3/4 h, v = (9 km)/(3/4 h) = 12 km/h. The original check is (12 km/h) × (3/4 h) = 9 km. Keep the time unit compatible with the speed unit. If the speed changes during a journey, using one constant speed is an additional modeling choice; the appropriate whole-journey average speed uses total distance divided by positive elapsed time.

Handle a zero parameter through the original relationship

In d = v t, fixing t = 0 makes the right side zero for every permitted v. If d = 0, the equation is consistent with every permitted speed and does not identify a unique one. If d > 0, no permitted speed satisfies it. The quotient d/t is undefined at zero time; even 0/0 is not a computed speed. Inspect the original product instead of dividing by zero.

Likewise, if v = 0 and time is requested, d = 0 is consistent with every permitted t ≥ 0, while d > 0 has no solution in this model. Thus t = d/v applies only when v is nonzero. These all-solution or no-solution conclusions depend on which quantities are fixed and on the original domain. An inverse formula that requires a nonzero parameter must carry that condition with it.

Rearrange a scalar rule with a fixed multiplier and offset

For the scalar relationship y = a x + b, let a and b be fixed real parameters. If y is known and x is requested, subtract b from both sides: y − b = a x. When a ≠ 0, divide both sides by a to obtain x = (y − b)/a. The parentheses show that the whole difference is divided by a, rather than just the second term.

For a = −2, b = 3 and y = 2, x = (2 − 3)/(−2) = 1/2, checked by −2 × (1/2) + 3 = 2. If a = 0, the original rule is y = b regardless of x. With y fixed, every real x works when y = b, and no real x works when y ≠ b. Do not use the quotient form at a = 0 or invent a unique x from a relationship that does not determine one.

Check units, admissibility and the original formula

Quantities being added or subtracted must have compatible dimensions and units. For P = 2(L + W), convert mixed length units before adding numerical values: L = 150 cm = 3/2 m and W = 4 m give P = 2 × (3/2 m + 4 m) = 11 m. In A = L W the length units multiply to form area; in L = A/W the quotient must have a length unit.

Substitute the result into the original relationship, check its domain, and judge whether its size is sensible in the situation. Correct dimensions help detect mistakes but do not by themselves prove the arithmetic or model correct. The explorer’s inputs are exact examples. Measurements in a practical task can be approximate; preserve their stated precision and uncertainty instead of treating an exact formula as proof that its input measurements were exact.

Change the requested quantity while keeping the rectangle

Start with the rectangle L = 3 m and W = 2 m. Its area is 6 m² and perimeter is 10 m. Choose area or perimeter, then request the total, L or W. Total means area A for the area relationship and perimeter P for the perimeter relationship. Sides are reported in metres, area in square metres and perimeter in metres.

Four positive rectangle presets, two relationships and three requested quantities give 4 × 2 × 3 = 24 exact states. Changing the requested unknown retains the chosen rectangle: it changes which values are presented as known and which equivalent form computes the result. The diagram uses the same scale, 40 pixels per metre, for every preset. Side labels identify L and W even when W is the longer side.

For each state, read the symbolic relationship, known quantities, substitution or rearrangement and original-formula check. Fractional sides remain exact. No zero dimensions are selectable; the preceding nonzero and domain conditions explain why these rectangle inverses are allowed. Switching from area to perimeter changes the requested kind of total, not the geometric object.

Keep the relationship visible in every representation

Before calculating, define the quantities, name the known values and requested unknown, and state the domain. Rearrange only through reversible operations with their conditions, preserve grouping and units, and check the result in the original formula. If a zero parameter prevents a unique inverse, report the conditional all-solution or no-solution conclusion instead of a fabricated quotient.

A formula can also organize a table of permitted input and output pairs. That table must name its quantities and units, and sampled rows do not replace the full relationship. The next lesson, Coordinates, tables and plotting points, shows how to place such ordered pairs on labelled axes and connect a table with a diagram.

Worked examples

Try it yourself

1. For a positive rectangle, compare the tasks: find A from L = 3 m and W = 2 m; find L from A = 6 m² and W = 2 m. Does changing the subject change the rectangle relationship?

   Show the answer and reasoning

   The first task gives A = 6 m² using A = L W. The second gives L = 3 m using L = A/W. The relationship is the same; the known and requested quantities differ.

   A formula expresses a relation between defined quantities. Making another quantity the subject selects a useful equivalent form under the required conditions rather than changing the geometric object.

2. Find W when an exact rectangle has A = 3 m² and L = 6 m. Why does W = A/L require L ≠ 0? Would A = 3 m² with L = 0 m describe a rectangle in this lesson’s domain?

   Show the answer and reasoning

   W = (3 m²)/(6 m) = 1/2 m, checked by (6 m) × (1/2 m) = 3 m². The inverse divides by L, so L must be nonzero. Zero length is excluded from the positive rectangle domain and division by it is undefined.

   The area quotient has a length unit. A rearranged formula must carry its operation conditions and retain the original geometric domain.

3. An exact rectangle has L = 150 cm and W = 4 m. Find its perimeter in metres using P = 2(L + W). Why is 2 × (150 + 4) m not the stated calculation?

   Show the answer and reasoning

   L = 150 cm = 3/2 m, so P = 2 × (3/2 m + 4 m) = 11 m. The numerals 150 and four initially use different length units and cannot be added as if both were metres.

   Compatible dimensions alone are insufficient when numerical values use different units. Convert to a common unit, retain the grouping and check the result against the four boundary sides.

4. Suppose exact data P = 6 m and W = 4 m are offered for a positive rectangle. Use L = P/2 − W, check the arithmetic original relationship, and decide whether the data describe such a rectangle.

   Show the answer and reasoning

   The rearrangement gives L = 3 m − 4 m = −1 m. The arithmetic check 2 × (−1 m + 4 m) = 6 m succeeds, but the negative L is excluded. The data do not describe a rectangle with positive sides.

   Satisfying an arithmetic equality does not override the model’s domain. Check admissibility as well as the original formula.

5. In d = v t with d, v and t nonnegative, recover t for d = 9 km and v = 12 km/h. Then consider known v = 0: what times satisfy the model if d = 0, and if d > 0? What happens when speed is requested at t = 0 and d = 0?

   Show the answer and reasoning

   For positive v, t = d/v = 3/4 h. At v = 0 and d = 0, every permitted t ≥ 0 satisfies the product; at v = 0 and d > 0, none does. At t = 0 and d = 0, every permitted nonnegative v satisfies the product, so the model does not determine a unique speed and 0/0 is undefined.

   An inverse quotient requires a nonzero divisor. When it is zero, inspect the original relationship and report all permitted solutions or none instead of using division by zero.

6. For y = a x + b, take a = −2, b = 3 and known y = 2. Find x and check the original rule. If instead a = 0 and b = 3, which real x values work for known y = 3 and for known y = 2?

   Show the answer and reasoning

   For a = −2, x = (2 − 3)/(−2) = 1/2; the check is −2 × (1/2) + 3 = 2. With a = 0 and b = 3, every real x works for y = 3, while no real x works for y = 2.

   The whole difference y − b must be divided by nonzero a. When a = 0, the original rule is y = b, so fixed y determines consistency rather than a unique inverse input.

7. Use the explorer preset L = 3/2 m and W = 4 m. Find both totals. If you switch from requesting total area to requesting L, which quantities become known? Does requesting W or switching to perimeter change the chosen shape?

   Show the answer and reasoning

   A = 6 m² and P = 11 m. To recover L from area, the known quantities are A = 6 m² and W = 4 m, giving L = 3/2 m. Requesting W or switching the relationship retains the chosen rectangle; perimeter uses a length total instead of an area total.

   Keep the object distinct from the task. The explorer changes the given quantities and equivalent rule for the requested result while its fixed scale preserves comparisons between the presets.

What to carry forward

• A formula is an equation expressing a relationship between defined quantities; specify its domain and the known and requested values.
• Substitution uses known values, while rearrangement isolates a requested quantity through the same reversible operations on both sides.
• Keep nonzero conditions attached to inverse quotients and inspect zero parameters in the original relationship for all-solution or no-solution cases.
• Preserve grouping, compatible units and dimensions; an area total and a perimeter total are different kinds of quantity.
• Check the recovered value in the original formula and its context; correct arithmetic alone does not make an inadmissible geometric value valid.