Year 11 General Algebra: Linear And Non-Linear Relationships

Transposition Of Equations

10 practice questions 2 video lessons Theory + worked examples

Theory

Transposition means rearranging a formula so a chosen variable stands alone — the same as solving an equation, but the answer is another formula rather than a number. Use inverse operations on both sides, working from the outside in. Often combined with substitution: rearrange first, then plug in the values.

Transposition — also called rearranging or changing the subject — means manipulating a formula so that a chosen variable stands alone on one side of the equation. The same balance rules apply as for solving equations: whatever operation you do to one side, you must do to the other.

An inverse operation is the operation that undoes another. Addition and subtraction are inverses; multiplication and division are inverses; squaring and taking a square root are inverses.

The subject of a formula is the variable that stands alone on one side. For example, in $v = u + a t$ , the subject is $v$ . To make $t$ the subject, you rearrange to get $t = \frac{v - u}{a}$ .

The first diagram is the inverse-operations reference — the pairs of operations that undo each other. The second is the five-step workflow for transposing any formula.

Each operation is undone by its inverse. Apply the inverse to both sides.

Work from the outside in. Clear fractions, move terms, factor, divide, then take roots last.

There are no new formulas to memorise. What helps is recognising a handful of common transpositions that appear again and again in physics, geometry, and finance.

Original formula	Subject changed to	Rearranged form
$F = m a$	$a$	$a = \frac{F}{m}$
$v = u + a t$	$t$	$t = \frac{v - u}{a}$
$A = π r^{2}$	$r$	$r = \sqrt{\frac{A}{π}}$
$V = π r^{2} h$	$h$	$h = \frac{V}{π r^{2}}$
$I = \frac{P R T}{100}$	$R$	$R = \frac{100 I}{P T}$

r = \sqrt{\frac{A}{π}}

Square-root rule. In physical contexts (lengths, distances, periods, areas), keep only the positive square root. For example, c=a2+b2, not ±a2+b2.

Step-by-step transposition

Clear fractions or brackets if they're in the way. Multiply both sides by the denominator to clear a fraction. Distribute or factor to handle brackets.
Move terms not containing the target variable to the other side by applying their inverse operation.
If the target variable appears in more than one term, factor it out as a common factor.
Divide by the coefficient of the target so it stands alone.
Take a root at the very end if the target is still inside a square or square root. Keep the positive root in physical contexts.

Tip — work from the outside in

Look at what's being done to your target variable, then undo each operation in reverse order. The last thing applied to the variable is the first thing you undo.

EXAMPLE 1 — LINEAR, ONE STEP AT A TIME

Make

t

the subject of

v = u + a t

SOLUTION

Move $u$ across by subtracting it from both sides, then divide by $a$ .

$v$	$=$	$u + a t$
$v - u$	$=$	$a t$
$\frac{v - u}{a}$	$=$	$t$

Answer: $t = \frac{v - u}{a}$ .

t = \frac{v - u}{a}

EXAMPLE 2 — SQUARES AND ROOTS

Make

r

the subject of

A = π r^{2}

SOLUTION

Divide both sides by $π$ , then take the square root.

$A$	$=$	$π r^{2}$
$\frac{A}{π}$	$=$	$r^{2}$
$r$	$=$	$\sqrt{\frac{A}{π}}$

The radius is a length, so keep only the positive root.

Answer: $r = \sqrt{\frac{A}{π}}$ .

r = \sqrt{\frac{A}{π}}

EXAMPLE 3 — VARIABLE ON BOTH SIDES

Make

y

the subject of

3 y = 7 x + 2 y - 5

SOLUTION

Collect the $y$ terms on one side by subtracting $2 y$ from both sides.

$3 y$	$=$	$7 x + 2 y - 5$
$3 y - 2 y$	$=$	$7 x - 5$
$y$	$=$	$7 x - 5$

Answer: $y = 7 x - 5$ .

y = 7 x - 5

EXAMPLE 4 — REARRANGE, THEN SUBSTITUTE

The simple-interest formula is

I = \frac{P R T}{100}

. Find

R

when

I = $ 600

P = $ 4000

and

T = 3

years.

SOLUTION

First make $R$ the subject. Multiply both sides by $100$ , then divide by $P T$ .

$100 I$	$=$	$P R T$
$R$	$=$	$\frac{100 I}{P T}$

Now substitute the values.

$R$	$=$	$\frac{100 (600)}{4000 \times 3}$
	$=$	$\frac{60 000}{12 000}$
	$=$	$5$

Answer: the interest rate is $5 %$ per annum.

R = 5 %

Common pitfalls

Rearranging is not solving for a number. The answer to a transposition question is a formula, with the target variable alone on one side and everything else on the other. Stop when the variable is isolated — don't try to substitute values unless the question also asks for that.

Take the positive root in physical contexts. When the formula gives a length, distance, time, or area, write

c = \sqrt{a^{2} + b^{2}}

, not

\pm \sqrt{a^{2} + b^{2}}

. Negative lengths don't make sense.

Variable inside a bracket? Either expand the bracket first or divide out the factor in front. For

C = 2 π r (h + r)

, divide both sides by

2 π r

to get

h + r = \frac{C}{2 π r}

, then subtract

r

Don't divide before collecting like terms. If your target variable appears in more than one term, gather those terms first and factor the variable out, then divide. Dividing too early leaves a mess.

Do the same to both sides — every time. Multiplying just the right-hand side by something, or subtracting from only the left, breaks the formula. Every operation must be applied to both sides.

Frequently asked questions

What does transposition mean?

Transposition (also called rearranging or changing the subject) means manipulating a formula so that a chosen variable stands alone on one side of the equals sign. The same balance rules as for solving equations apply: whatever you do to one side, you must do to the other.

What's the difference between solving and transposing?

Solving an equation gives a numerical answer. Transposing gives another formula — the target variable is alone on one side and the other letters appear on the other side. No numbers are substituted in until you have the new formula.

How do I undo a square?

Take the square root of both sides. To undo a square root, square both sides. These are inverse operations. In most physical contexts (lengths, distances, periods), keep only the positive root.

What do I do if the variable I want appears in two places?

Collect all terms containing that variable on one side and everything else on the other. Then factor the variable out as a common factor and divide by what's left in the bracket.

How do I handle a variable inside a bracket?

Either expand the bracket first using the distributive law, or divide both sides by everything outside the bracket so the bracket stands alone. Choose whichever is simpler for that formula.

How does transposition help with substitution problems?

Many real problems give values for several variables and ask for the remaining one — but that variable isn't already the subject. Rearrange first to make it the subject, then substitute the values in. This avoids messy algebra with numbers in the middle of the manipulation.