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GuidePublished 15 Aug 20265 min readBy Kevin Joginliteral equationsformula rearrangementchanging the subjecttransposition
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KEVOS AILiteral Equations and Formula Rearrangement

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Literal Equations and Formula Rearrangement

Making a chosen symbol the subject of a formula, why the target must be isolated as a common factor, and the conditions that come with dividing by a variable.

Category Engineering / MathematicsStream EquationsLevel FoundationReading 5 minSource Week 3, page 8

What this page covers

  • Rearrange a formula to make a stated symbol the subject
  • Collect a target variable appearing in more than one term
  • State the conditions under which a rearrangement is valid
  • Check a rearrangement by substitution
On this page
  1. What changes and what does not
  2. The source's worked example
  3. The general procedure
  4. Further worked examples
  5. Conditions are part of the answer
  6. Common mistakes
  7. Frequently asked questions

What changes and what does not

A literal equation carries letters where a numeric equation carries numbers. The method is identical — apply inverse operations to both sides — but two things become more prominent.

Nothing collapses

3 + 5 becomes 8; a + b stays a + b. Every term must be carried along, so the working is longer and the bookkeeping matters more.

Conditions appear

Dividing by 4 is always safe. Dividing by 1 + rt requires it to be non-zero, and that condition belongs in the answer.

The target symbol is called the subject. Making P the subject means producing an expression of the form P = something with no P in it.

The source's worked example

Making P the subject of A = P + Prt

The source describes the context: A is the amount a sum P grows to at simple interest rate r over time t. Given A, r and t, find P.

P appears in both terms, so it cannot be isolated by transposing. It must be factored out first.

A = P + PrtA = P(1 + rt)A1 + rt = PP = A1 + rtSource example, Week 3, page 8

The source labels the key step directly: P is a monomial common factor. Recognising that is the whole of the problem; the rest is one division.

Note

The division requires 1 + rt ≠ 0. In this application r and t are non-negative, so 1 + rt ≥ 1 and the condition holds automatically — but it is worth saying why rather than ignoring it.

Check

Take P = 100, r = 0.05, t = 2. Then A = 100 + 100(0.1) = 110. The rearranged formula gives 1101.1 = 100. Correct.

The general procedure

  1. Clear fractions and radicals affecting the target, by multiplying through or squaring.
  2. Expand any brackets containing the target.
  3. Collect every term containing the target on one side, and everything else on the other.
  4. Factor the target out of that side.
  5. Divide by the bracket, and state the condition that it is non-zero.
  6. Check by substituting numbers into both forms.

Steps 3 and 4 are where the work is. If the target appears only once, they are trivial and the rearrangement is straightforward transposition.

Further worked examples

Target appearing once — make r the subject of A = π r2

A = π r2Aπ = r2r = √AπSource, Week 5, page 5, where the area of a circle is used to illustrate dependent and independent variables

The source adds the condition r > 0, which is the right call: algebraically r = ±√A/π, but a radius cannot be negative. Context, not algebra, discards the second root.

Target on both sides — make x the subject of ax + b = cx + d

ax + b = cx + dax - cx = d - bx(a - c) = d - bx = d - ba - c, a ≠ cThe condition a ≠ c is essential

If a = c the equation reduces to b = d, which is either an identity (if b = d) or has no solution (if not). Dividing by a - c without noticing would silently assume the general case.

Nested target — make y the subject of x + yx - y = k

x + y = k(x - y)x + y = kx - kyy + ky = kx - xy(1 + k) = x(k - 1)y = x(k - 1)1 + k, k ≠ -1Clear the fraction, expand, collect, factor, divide
Check

Take x = 3, k = 2. Then y = 3(1)3 = 1, and 3 + 13 - 1 = 2. Correct.

Conditions are part of the answer

A rearranged formula that omits its conditions is incomplete, and in an engineering context that omission can matter.

Where conditions come from
StepCondition generatedExample
Dividing by an expressionIt must be non-zeroP = A1 + rt needs 1 + rt ≠ 0
Taking an even rootThe radicand must be non-negativer = √A/π needs A ≥ 0
Choosing one sign of a rootJustified by contextr > 0 because a radius is positive
Multiplying by a denominatorIt must be non-zerox + yx - y = k needs x ≠ y
Watch out

Two of these are easy to lose. Squaring can introduce solutions that were never there; dividing by a variable can lose solutions that were. Both are covered in Radical Equations and Spurious Solutions.

Common mistakes

Errors and remedies
MistakeCorrectRemedy
Transposing when the target appears twiceCollect and factor firstP + Prt cannot be split by transposition
Dividing by a bracket that might vanishState the conditionAsk when the divisor is zero
Taking only the positive root without saying why± algebraically; context may discard oneName the reason, e.g. a radius is positive
Multiplying only some terms when clearing a fractionEvery term on both sidesCount terms before and after
Leaving the target on the rightWrite P = …Convention: the subject goes on the left

Frequently asked questions

What is a literal equation?

One whose coefficients are letters rather than numbers. Solving means isolating one chosen symbol in terms of the others — the same process as solving a numeric equation, with the arithmetic left undone.

What if the target appears twice?

Collect every term containing it on one side, factor it out, then divide by the bracket. That factoring step is the whole difficulty; without it the target cannot be isolated.

Do I need to state conditions?

Whenever you divide by something containing a variable, yes. Dividing by 1 + rt requires 1 + rt ≠ 0. In an applied context the quantity is often positive by its meaning, and saying so is enough.

Can I check a rearrangement?

Yes. Put numbers into the original, solve numerically, then check the rearranged version reproduces the same value. It takes a minute and catches sign errors reliably.

Related pages

  • Solving Linear Equations
  • The Field Laws of Real Number Arithmetic
  • Rational Algebraic Fractions
  • The Four Forms of a Straight Line

Source. Handwritten teaching notes, Week 3, page 8.

This page is an original exposition prepared for the KEVOS® knowledge library. It restates, reorganises and verifies the mathematics of the supplied teaching notes; it is not a reproduction of them. Numerical values taken from the notes are identified as source examples and are not presented as engineering standards.

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