Rearranging Equations and Formulas

Introduction

In algebra, we often solve equations to find the value of an unknown variable.

For example:

3x + 5 = 20

Subtract 5 from both sides:

3x = 15

Divide both sides by 3:

x = 5

However, many equations contain more than one variable.

Consider the formula for the area of a rectangle:

A = lw

where:

• A = area

• l = length

• w = width

If we know the area and width, how can we calculate the length?

We rearrange the formula:

l = A/w

This process is called solving a literal equation.

A literal equation is an equation containing two or more variables, often representing a relationship between physical or mathematical quantities.

Rather than finding a numerical answer, we usually rearrange the equation to express one variable in terms of the others.

Solving literal equations is an essential skill in algebra, geometry, physics, chemistry, engineering, and many other areas of mathematics.

It allows us to:

• Rearrange formulas to calculate different quantities.

• Understand relationships between variables.

• Apply inverse operations.

• Solve problems involving several unknown quantities.

• Derive new formulas from existing equations.

• Verify that two formulas are equivalent.

What Is a Literal Equation?

A literal equation contains two or more variables.

Examples include:

A = lw

Area of a rectangle

P = 2l + 2w

Perimeter of a rectangle

d = vt

Distance travelled at constant speed

F = ma

Newton's Second Law

V = IR

Ohm's Law

C = 2πr

Circumference of a circle

Each equation describes a relationship between quantities.

Example

Consider:

d = vt

where:

• d = distance

• v = speed

• t = time

If distance and time are known, we may need to calculate speed.

Rearrange:

v = d/t

If distance and speed are known, we may need to calculate time.

Rearrange:

t = d/v

The same original equation can therefore be written in different equivalent forms.

Important Observation

Solving a literal equation usually means isolating a particular variable rather than calculating a numerical value immediately.

What Does It Mean to Isolate a Variable?

To isolate a variable means to rearrange an equation so that the required variable appears alone on one side.

For example:

A = bh

Suppose we want to isolate h.

Divide both sides by b:

A/b = h

Rewrite:

h = A/b

The variable h is now isolated.

Another Example

Consider:

y = mx + c

Suppose we want to isolate c.

Subtract mx from both sides:

y − mx = c

Therefore:

c = y − mx

Important Observation

The first step in solving a literal equation is identifying which variable must be isolated.

Identifying the Subject of a Formula

The subject of a formula is the variable that appears alone on one side of the equation.

Consider:

F = ma

The subject is F.

If we rearrange:

m = F/a

the subject becomes m.

If we rearrange:

a = F/m

the subject becomes a.

All three formulas describe the same relationship, provided the denominators are nonzero.

Practice Example

Identify the subject of each formula.

Important Observation

Changing the subject does not change the underlying mathematical relationship.

Understanding Inverse Operations

Inverse operations are mathematical operations that undo one another.

They are essential for rearranging equations.

Example: Addition and Subtraction

Consider:

x + 7 = 15

Subtract 7 from both sides:

x = 8

Subtraction undoes addition.

Example: Multiplication and Division

Consider:

5x = 30

Divide both sides by 5:

x = 6

Division undoes multiplication.

Example: Squaring and Square Roots

Consider:

x² = 49

Take square roots:

x = ±7

Both values satisfy the original equation.

Important Observation

Inverse operations allow us to undo the operations surrounding the variable we want to isolate.

The Balance Principle

An equation represents a balance between two equal quantities.

If we change one side, we must make an equivalent change to the other side to preserve equality.

Example

Consider:

x + 4 = 10

Subtract 4 from both sides:

x + 4 − 4 = 10 − 4

Simplify:

x = 6

The equation remains balanced because the same operation was performed on both sides.

Another Example

Consider:

3x = 18

Divide both sides by 3:

3x/3 = 18/3

Simplify:

x = 6

Important Observation

Every valid rearrangement must preserve equality between the two sides of the equation.

A General Strategy for Solving Literal Equations

A reliable approach is to follow a sequence of steps.

Step 1: Identify the Required Variable

Determine which variable must become the subject.

Step 2: Examine the Operations

Identify what is being added, subtracted, multiplied, divided, squared, or otherwise applied to the required variable.

Step 3: Use Inverse Operations

Undo those operations while preserving equality.

Step 4: Simplify

Combine terms, cancel common factors, or rewrite the expression into a clearer form.

Step 5: Verify

Substitute the rearranged formula into the original equation, or check using numerical values.

Step 6: Explain the Steps

State which operation was used and why it was appropriate.

Important Observation

A clear sequence of algebraic steps is often more reliable than attempting to rearrange a complicated equation mentally.

Isolating a Variable Using Addition and Subtraction

Worked Example

Make b the subject:

a = b + c

Subtract c from both sides:

a − c = b

Rewrite:

b = a − c

Explanation

The variable b was being increased by c.

Subtracting c undoes the addition.

Isolating a Variable in a Formula with Subtraction

Worked Example

Make x the subject:

y = x − a

Add a to both sides:

y + a = x

Therefore:

x = y + a

Explanation

Adding a undoes subtracting a.

Another Example

Make b the subject:

a = c − b

Subtract c from both sides:

a − c = −b

Multiply both sides by −1:

c − a = b

Therefore:

b = c − a

Important Observation

When the required variable has a negative sign, multiplying both sides by −1 can help isolate it.

Isolating a Variable Using Multiplication and Division

Worked Example

Make x the subject:

y = ax

Divide both sides by a:

y/a = x

Therefore:

x = y/a

This rearrangement assumes a ≠ 0.

Explanation

The variable x is multiplied by a.

Division by a undoes the multiplication.

Rearranging Formulas Involving Division

Worked Example

Make b the subject:

a = c/b

Multiply both sides by b:

ab = c

Divide both sides by a:

b = c/a

This assumes a ≠ 0 and b ≠ 0.

Explanation

The variable b originally appeared in the denominator.

Multiplying both sides by b removes it from the denominator.

Dividing by a then isolates b.

Important Observation

Multiplying both sides by a denominator is often a useful first step when the required variable appears in a fraction.

Rearranging a Formula with Two Operations

Worked Example

Make x the subject:

y = 3x + 7

First, subtract 7 from both sides:

y − 7 = 3x

Next, divide both sides by 3:

x = (y − 7)/3

Explanation

The original formula involves two operations on x:

• Multiplication by 3

• Addition of 7

We undo these operations in reverse order.

First undo the addition, then undo the multiplication.

Rearranging with a Negative Coefficient

Worked Example

Make x the subject:

y = 10 − 2x

Subtract 10 from both sides:

y − 10 = −2x

Divide both sides by −2:

x = (y − 10)/(−2)

Simplify:

x = (10 − y)/2

Verification

Start with:

x = (10 − y)/2

Multiply both sides by 2:

2x = 10 − y

Rearrange:

y = 10 − 2x

The original equation is recovered.

Important Observation

A negative coefficient must be handled carefully when dividing both sides.

Rearranging Equations with Brackets

Some literal equations contain brackets that group terms together.

Worked Example

Make x the subject:

y = a(x + b)

Divide both sides by a:

y/a = x + b

Subtract b:

x = y/a − b

This assumes a ≠ 0.

Alternative Method

Expand first:

y = ax + ab

Subtract ab:

y − ab = ax

Divide by a:

x = (y − ab)/a

Separate the fraction:

x = y/a − b

Both methods produce the same result.

Important Observation

When a bracket is multiplied by a coefficient, dividing first may be more efficient than expanding.

Rearranging Equations with Fractions

Worked Example

Make x the subject:

y = (x + a)/b

Multiply both sides by b:

by = x + a

Subtract a:

x = by − a

This assumes b ≠ 0.

Verification

Substitute x = by − a into the original formula:

y = [(by − a) + a]/b

Simplify the numerator:

y = by/b

Therefore:

y = y

The rearrangement is verified.

Rearranging Formulas with Multiple Terms

Worked Example

Make x the subject:

y = ax + bx

Both terms contain x.

Factor out x:

y = x(a + b)

Divide both sides by a + b:

x = y/(a + b)

This assumes a + b ≠ 0.

Explanation

Trying to divide individual terms separately would make the problem more complicated.

Factoring out the required variable creates a single product that can be rearranged.

Important Observation

If the required variable appears in more than one term, collecting or factoring the variable is often necessary.

Collecting Variables on One Side

Worked Example

Make x the subject:

ax + b = cx + d

Subtract cx from both sides:

ax − cx + b = d

Subtract b:

ax − cx = d − b

Factor out x:

x(a − c) = d − b

Divide by a − c:

x = (d − b)/(a − c)

This assumes a − c ≠ 0.

Explanation

The required variable originally appeared on both sides.

We collected all terms containing x on one side and all remaining terms on the other.

Important Observation

When a variable appears on both sides, it is often best to collect its terms before dividing.

Rearranging Geometry Formulas

Geometry provides many practical examples of literal equations.

Area of a Rectangle

The area formula is:

A = lw

Make l the subject:

Divide by w:

l = A/w

Make w the subject:

Divide by l:

w = A/l

These formulas assume the divisor is nonzero.

Worked Example

A rectangle has an area of 72 cm² and a width of 8 cm.

Find its length.

Use:

l = A/w

Substitute:

l = 72/8

l = 9 cm

Important Observation

Rearranging the formula allows us to calculate any one of the three quantities when the other two are known.

Rearranging the Perimeter of a Rectangle

The perimeter of a rectangle is:

P = 2l + 2w

Suppose we want to make w the subject.

Step 1: Subtract 2l

P − 2l = 2w

Step 2: Divide by 2

(P − 2l)/2 = w

Step 3: Simplify

w = P/2 − l

Worked Numerical Example

A rectangular field has perimeter 50 m and length 15 m.

Find its width.

w = P/2 − l

w = 50/2 − 15

= 25 − 15

w = 10 m

Verification

P = 2(15) + 2(10)

= 30 + 20

P = 50 m

The answer is correct.

Rearranging the Circumference Formula

The circumference of a circle is:

C = 2πr

where:

• C = circumference

• r = radius

• π ≈ 3.14159

Make r the subject.

Divide both sides by 2π:

r = C/(2π)

Worked Example

A circular track has circumference 62.8 m.

Calculate its approximate radius.

r = C/(2π)

r ≈ 62.8/(2 × 3.14)

r ≈ 62.8/6.28

r ≈ 10 m

Important Observation

The same formula can be used to calculate circumference or radius, depending on which quantity is unknown.

Rearranging the Area of a Triangle

The area of a triangle is:

A = ½bh

where:

• A = area

• b = base

• h = perpendicular height

Make h the Subject

Start:

A = ½bh

Multiply both sides by 2:

2A = bh

Divide both sides by b:

h = 2A/b

Make b the Subject

Start:

A = ½bh

Multiply both sides by 2:

2A = bh

Divide both sides by h:

b = 2A/h

Worked Example

A triangle has an area of 48 cm² and a base of 12 cm.

Find its height.

h = 2A/b

h = 2(48)/12

h = 96/12

h = 8 cm

Rearranging Physics Formulas

Physics frequently requires rearranging formulas to calculate different quantities.

Speed, Distance, and Time

The formula for constant speed is:

d = vt

where:

• d = distance

• v = speed

• t = time

Make v the subject:

v = d/t

Make t the subject:

t = d/v

Worked Example

A car travels 180 km at an average speed of 60 km/h.

Calculate the travel time.

Use:

t = d/v

Substitute:

t = 180/60

t = 3 hours

Important Observation

Rearranging a formula is useful only when the quantities and units are interpreted correctly.

Rearranging Newton's Second Law

Newton's Second Law is:

F = ma

where:

• F = resultant force in newtons (N)

• m = mass in kilograms (kg)

• a = acceleration in m/s²

Make m the Subject

Divide both sides by a:

m = F/a

Make a the Subject

Divide both sides by m:

a = F/m

Worked Example

A resultant force of 48 N produces an acceleration of 6 m/s².

Calculate the mass.

m = F/a

m = 48/6

m = 8 kg

Verification

F = ma

F = 8 × 6

F = 48 N

The calculation is correct.

Rearranging Ohm's Law

Ohm's Law is:

V = IR

where:

• V = potential difference in volts (V)

• I = electric current in amperes (A)

• R = resistance in ohms (Ω)

Make I the Subject

Divide by R:

I = V/R

Make R the Subject

Divide by I:

R = V/I

Worked Example

A circuit has a potential difference of 12 V and a current of 3 A.

Calculate the resistance.

R = V/I

R = 12/3

R = 4 Ω

Important Observation

Literal equations allow one relationship to be used in several different problem-solving situations.

Verifying Rearranged Formulas

Verification is an important part of algebra.

A rearranged formula should describe the same relationship as the original equation.

There are two common methods.

Method A: Algebraic Verification

Start with the rearranged formula and use algebra to recover the original equation.

Method B: Numerical Verification

Choose suitable numerical values and check that both forms agree.

Worked Example

Original formula:

y = 4x + 3

Rearranged formula:

x = (y − 3)/4

Choose x = 5.

Calculate y:

y = 4(5) + 3

y = 23

Now substitute into the rearranged formula:

x = (23 − 3)/4

x = 20/4

x = 5

The original value is recovered.

Important Observation

Numerical verification helps detect mistakes, but algebraic verification establishes the equivalence more generally.

Explaining Algebraic Steps Clearly

When solving literal equations, each step should have a mathematical reason.

Consider:

y = 5x − 8

Make x the subject.

This style makes the reasoning easy to follow.

Important Observation

Clear mathematical communication involves explaining both what operation was performed and why it was valid.

Common Mistakes in Basic Rearrangement

Performing an Operation on Only One Side

Incorrect:

y = x + 5

x = y + 5

Correct:

x = y − 5

Dividing Only One Term

Incorrect:

y = 3x + 6

x = y/3 − 6

Correct:

Subtract 6 first:

y − 6 = 3x

Then divide:

x = (y − 6)/3

Losing a Negative Sign

Incorrect:

y = 5 − 2x

x = (y − 5)/2

Correct:

x = (5 − y)/2

Forgetting Brackets

Incorrect:

x = y − 3/4

when the intended expression is:

x = (y − 3)/4

The brackets show that the entire numerator is divided by 4.

Cancelling Variables Incorrectly

Only common multiplicative factors can be cancelled.

Terms connected by addition or subtraction cannot simply be cancelled.

Dividing by Zero

When rearranging a formula, division by zero is not allowed.

For example:

x = y/a

requires a ≠ 0.

Important Observation

Many rearrangement errors can be avoided by writing each step clearly and checking that both sides remain equal.

Part 1 Practice

Identifying Variables

1. What is a literal equation?

2. Identify the subject of A = bh.

3. Identify the subject of v = d/t.

4. Explain what it means to isolate a variable.

5. What is the inverse operation of multiplication?

Basic Rearrangement

6. Make x the subject: y = x + 7.

7. Make x the subject: y = x − 4.

8. Make x the subject: y = 5x.

9. Make x the subject: y = x/3.

10. Make x the subject: y = 2x + 9.

Intermediate Rearrangement

11. Make x the subject: y = 7x − 5.

12. Make x the subject: y = 12 − 3x.

13. Make x the subject: y = a(x + b).

14. Make x the subject: y = (x + a)/b.

15. Make x the subject: y = ax + bx.

Formula Applications

16. Make l the subject: A = lw.

17. Make w the subject: P = 2l + 2w.

18. Make h the subject: A = ½bh.

19. Make a the subject: F = ma.

20. Make R the subject: V = IR.

Answers

Part 1 Summary

In this part, we learned that:

• Literal equations contain two or more variables.

• Solving a literal equation means making a specified variable the subject.

• Inverse operations undo addition, subtraction, multiplication, division, and powers.

• Every operation must preserve equality.

• The order of operations is important when isolating variables.

• Brackets and fractions must be handled carefully.

• Variables appearing in multiple terms may need to be collected or factored.

• Rearranged formulas can be verified algebraically or numerically.

• Literal equations are widely used in geometry, physics, and other sciences.

• Explaining each algebraic step helps communicate mathematical reasoning.

The most important principle is:

To isolate a variable, identify the operations acting on it and undo those operations while preserving equality.

Many literal equations require more than one of these techniques. The variable we want to isolate might:

• Appear in a denominator.

• Occur on both sides of an equation.

• Be contained inside brackets.

• Be raised to a power.

• Appear beneath a square root.

• Occur in several fractions.

• Require factoring before it can be isolated.

In this part, we will develop a systematic approach to these more challenging situations.

The central principle remains the same:

Every algebraic operation must preserve equality, and every division must be by a nonzero quantity.

Rearranging Equations with Variables in Denominators

When the required variable appears in the denominator of a fraction, multiplication is often the first useful inverse operation.

Worked Example: Isolating a Denominator

Make x the subject:

y = a/x

Multiply both sides by x:

xy = a

Divide both sides by y:

x = a/y

For this rearrangement, x and y must be nonzero.

Explanation

The original equation contains division by x.

Multiplying both sides by x removes x from the denominator.

Dividing by y then isolates x.

Verification

Start with:

x = a/y

Multiply by y:

xy = a

Divide by x:

y = a/x

The original equation is recovered.

Variables in Denominators with Additional Terms

Worked Example

Make x the subject:

y = a/x + b

Subtract b from both sides:

y − b = a/x

Multiply both sides by x:

x(y − b) = a

Divide by y − b:

x = a/(y − b)

This requires a ≠ 0 and y ≠ b for a valid solution of the original equation.

Numerical Example

Suppose:

y = 11, a = 24, b = 3

Then:

x = 24/(11 − 3)

x = 24/8

x = 3

Verify:

y = 24/3 + 3

y = 8 + 3

y = 11

Important Observation

When a fraction is combined with another term, first isolate the fraction before removing its denominator.

Rearranging Formulas with Two Fractions

Some equations contain several fractions.

A useful strategy is to multiply every term by a common denominator.

Worked Example

Make x the subject:

y = x/a + b/c

Subtract b/c:

y − b/c = x/a

Multiply both sides by a:

x = a(y − b/c)

Alternatively, distribute a:

x = ay − ab/c

Both forms are equivalent.

The original equation requires a ≠ 0 and c ≠ 0.

Explanation

Rather than combining every fraction immediately, we isolated the term containing x.

This made the rearrangement shorter.

Clearing Multiple Denominators

Worked Example

Make x the subject:

x/a + x/b = c

Assume a ≠ 0 and b ≠ 0.

Multiply every term by ab:

ab(x/a) + ab(x/b) = abc

Simplify:

bx + ax = abc

Factor out x:

x(a + b) = abc

Divide by a + b:

x = abc/(a + b)

This formula requires a + b ≠ 0.

Important Observation

Clearing denominators can turn a complicated fractional equation into a simpler equation involving multiplication and addition.

Rearranging Equations with Variables on Both Sides

When the required variable appears on both sides, collect all terms containing it before dividing.

Worked Example

Make x the subject:

ax + b = cx + d

Subtract cx:

ax − cx + b = d

Subtract b:

ax − cx = d − b

Factor out x:

x(a − c) = d − b

Divide:

x = (d − b)/(a − c)

This assumes a ≠ c.

Special Case

If a = c, the equation becomes:

ax + b = ax + d

Subtract ax:

b = d

If b = d, every value of x satisfies the equation.

If b ≠ d, no value of x satisfies the equation.

Important Observation

When a rearrangement requires division by an expression that could equal zero, the zero case must be considered separately.

Rearranging Equations with Brackets on Both Sides

Worked Example

Make x the subject:

a(x + b) = c(x + d)

Expand both sides:

ax + ab = cx + cd

Collect x terms:

ax − cx = cd − ab

Factor:

x(a − c) = cd − ab

Divide:

x = (cd − ab)/(a − c)

For a ≠ c, this is the required rearrangement.

Numerical Example

Let:

a = 5, b = 2, c = 3, d = 6

Then:

x = [3(6) − 5(2)]/(5 − 3)

= (18 − 10)/2

x = 4

Verify:

5(4 + 2) = 30

3(4 + 6) = 30

Both sides agree.

Rearranging Equations Containing Squared Variables

Sometimes the required variable is raised to the power of 2.

The inverse operation is taking a square root.

Worked Example

Make x the subject:

y = x²

Take square roots:

x = ±√y

For real solutions, y ≥ 0.

Why Are There Two Solutions?

Both positive and negative numbers can produce the same positive square.

For example:

5² = 25

(−5)² = 25

Therefore:

x² = 25

has two solutions:

x = 5 or x = −5

Important Observation

When solving a squared equation over the real numbers, both positive and negative square roots must be considered unless the context restricts the variable.

Rearranging a Formula with a Squared Variable

Worked Example

Make r the subject:

A = πr²

This is the formula for the area of a circle.

Divide by π:

A/π = r²

Take square roots:

r = ±√(A/π)

Since a radius cannot be negative:

r = √(A/π)

Numerical Example

A circular garden has an area of 78.5 m².

Calculate its radius.

r = √(A/π)

Using π ≈ 3.14:

r = √(78.5/3.14)

r = √25

r = 5 m

Important Observation

The mathematical rearrangement may produce two possible signs, but the physical meaning of the variable can restrict the acceptable answer.

Rearranging Formulas with Squared Binomials

Worked Example

Make x the subject:

y = (x + a)²

Take square roots:

x + a = ±√y

Subtract a:

x = −a ± √y

For real solutions, y ≥ 0.

Worked Example

Make x the subject:

y = 3(x − 2)² + 5

Subtract 5:

y − 5 = 3(x − 2)²

Divide by 3:

(y − 5)/3 = (x − 2)²

Take square roots:

x − 2 = ±√[(y − 5)/3]

Add 2:

x = 2 ± √[(y − 5)/3]

For real solutions, y ≥ 5.

Important Observation

When a squared expression contains several operations, undo the operations outside the square before taking square roots.

Rearranging Equations Involving Cubes

The inverse operation of cubing is taking a cube root.

Worked Example

Make x the subject:

y = x³

Take cube roots:

x = ∛y

Unlike square roots, real cube roots are defined for negative numbers.

For example:

∛(−27) = −3

because:

(−3)³ = −27

Another Example

Make x the subject:

y = 2x³ + 7

Subtract 7:

y − 7 = 2x³

Divide by 2:

(y − 7)/2 = x³

Take cube roots:

x = ∛[(y − 7)/2]

Important Observation

Cubing is one-to-one over the real numbers, so each real value of y gives exactly one real value of x.

Rearranging Equations Involving Square Roots

When the required variable appears beneath a square root, isolate the radical before squaring.

Worked Example

Make x the subject:

y = √(x + a)

Square both sides:

y² = x + a

Subtract a:

x = y² − a

The original equation requires y ≥ 0.

Explanation

Squaring removes the square root, but the nonnegative condition must be retained.

Square Roots with Additional Operations

Worked Example

Make x the subject:

y = 3√(x + 2) − 4

Add 4:

y + 4 = 3√(x + 2)

Divide by 3:

(y + 4)/3 = √(x + 2)

Square both sides:

[(y + 4)/3]² = x + 2

Subtract 2:

x = (y + 4)²/9 − 2

The original formula requires y ≥ −4.

Verification with Numbers

Let x = 7.

Original formula:

y = 3√(7 + 2) − 4

= 3(3) − 4

y = 5

Use the rearranged formula:

x = (5 + 4)²/9 − 2

= 81/9 − 2

= 9 − 2

x = 7

The rearrangement is verified.

Why Squaring Requires Care

Squaring both sides can introduce solutions that do not satisfy the original equation.

Consider:

√x = −3

Squaring both sides gives:

x = 9

However:

√9 = 3

not −3.

Therefore, the original equation has no real solution.

Important Observation

After squaring an equation, always check that the result satisfies the original equation and its sign restrictions.

Rearranging Formulas with Variables in Exponents

Sometimes the required variable appears as an exponent.

For example:

y = aˣ

If a > 0 and a ≠ 1, logarithms can isolate x.

Take logarithms:

ln y = x ln a

Divide by ln a:

x = ln y/ln a

This requires y > 0.

Numerical Example

Solve:

2ˣ = 32

Take logarithms:

x = ln 32/ln 2

x = 5

This agrees with:

2⁵ = 32

Important Observation

Logarithms are inverse operations for exponential relationships, just as roots are inverse operations for powers.

Rearranging the Formula for Simple Interest

Simple interest is calculated using:

I = Prt

where:

• I = interest

• P = principal

• r = annual interest rate expressed as a decimal

• t = time in years

Make P the Subject

Divide both sides by rt:

P = I/(rt)

Make r the Subject

Divide by Pt:

r = I/(Pt)

Make t the Subject

Divide by Pr:

t = I/(Pr)

Each formula requires a nonzero denominator.

Worked Example

An investment earns $360 in simple interest over 3 years at an annual rate of 4%.

Calculate the principal.

P = I/(rt)

P = 360/(0.04 × 3)

P = 360/0.12

P = $3,000

Verification

I = Prt

= 3000 × 0.04 × 3

I = $360

Rearranging the Density Formula

Density is defined as:

ρ = m/V

where:

• ρ = density

• m = mass

• V = volume

Make m the Subject

Multiply by V:

m = ρV

Make V the Subject

Start:

ρ = m/V

Multiply by V:

ρV = m

Divide by ρ:

V = m/ρ

This assumes ρ ≠ 0.

Worked Example

A metal sample has a mass of 540 g and density of 9 g/cm³.

Calculate its volume.

V = m/ρ

V = 540/9

V = 60 cm³

Important Observation

The same relationship can be rearranged to calculate density, mass, or volume.

Rearranging the Pressure Formula

Pressure is defined as:

P = F/A

where:

• P = pressure

• F = force

• A = area

Make F the Subject

Multiply both sides by A:

F = PA

Make A the Subject

Start:

P = F/A

Multiply by A:

PA = F

Divide by P:

A = F/P

This requires P ≠ 0.

Worked Example

A force of 800 N produces a pressure of 2000 Pa.

Calculate the contact area.

A = F/P

A = 800/2000

A = 0.4 m²

Rearranging the Kinetic Energy Formula

The kinetic energy of a moving object is:

Eₖ = ½mv²

Make m the Subject

Multiply by 2:

2Eₖ = mv²

Divide by v²:

m = 2Eₖ/v²

This requires v ≠ 0.

Make v the Subject

Start:

Eₖ = ½mv²

Multiply by 2:

2Eₖ = mv²

Divide by m:

2Eₖ/m = v²

Take square roots:

v = ±√(2Eₖ/m)

If v represents speed, it must be nonnegative:

v = √(2Eₖ/m)

This assumes m > 0 and Eₖ ≥ 0.

Worked Example

An object has mass 4 kg and kinetic energy 200 J.

Calculate its speed.

v = √(2Eₖ/m)

v = √[(2 × 200)/4]

= √100

v = 10 m/s

Important Observation

When a formula contains a squared variable, the rearrangement often involves a square root.

Rearranging Gravitational Potential Energy

Near Earth's surface, gravitational potential energy can be calculated using:

Eₚ = mgh

where:

• Eₚ = gravitational potential energy

• m = mass

• g = gravitational field strength

• h = height change

Make h the Subject

Divide both sides by mg:

h = Eₚ/(mg)

Make m the Subject

Divide by gh:

m = Eₚ/(gh)

Make g the Subject

Divide by mh:

g = Eₚ/(mh)

These formulas require nonzero denominators.

Worked Example

A 5 kg object gains 490 J of gravitational potential energy when lifted.

Using g = 9.8 N/kg, calculate the height.

h = Eₚ/(mg)

h = 490/(5 × 9.8)

h = 490/49

h = 10 m

Rearranging Temperature Conversion Formulas

Temperature conversion formulas are examples of literal equations involving several operations.

The relationship between Celsius and Fahrenheit is:

F = (9/5)C + 32

Make C the Subject

Subtract 32:

F − 32 = (9/5)C

Multiply by 5:

5(F − 32) = 9C

Divide by 9:

C = 5(F − 32)/9

Worked Example

Convert 95°F to Celsius.

C = 5(95 − 32)/9

= 5(63)/9

= 315/9

C = 35°C

Verification

F = (9/5)(35) + 32

= 63 + 32

F = 95°F

Rearranging the Equation of a Straight Line

The slope-intercept form of a linear equation is:

y = mx + c

where:

• m = gradient

• c = y-intercept

• x and y are coordinates

Make c the Subject

Subtract mx:

c = y − mx

Make m the Subject

Subtract c:

y − c = mx

Divide by x:

m = (y − c)/x

This requires x ≠ 0.

Make x the Subject

Subtract c:

y − c = mx

Divide by m:

x = (y − c)/m

This requires m ≠ 0.

Important Observation

Rearranging the same equation allows us to calculate different quantities depending on the information available.

Rearranging Formulas Involving Reciprocal Relationships

Some formulas involve sums of reciprocals.

An example is the equivalent resistance of two resistors connected in parallel:

1/R = 1/R₁ + 1/R₂

Suppose we want to make R the subject.

Step 1: Find a Common Denominator

1/R = (R₁ + R₂)/(R₁R₂)

Step 2: Multiply Both Sides by R

1 = R(R₁ + R₂)/(R₁R₂)

Step 3: Multiply by R₁R₂

R₁R₂ = R(R₁ + R₂)

Step 4: Divide

R = R₁R₂/(R₁ + R₂)

For positive resistances, the denominator is nonzero.

Important Observation

Expressions involving reciprocal quantities often become easier after combining fractions or clearing denominators.

Rearranging a Formula with a Variable in Two Denominators

Consider:

1/f = 1/u + 1/v

This relationship appears in the thin-lens equation under a common sign convention.

Make v the subject.

Step 1: Subtract 1/u

1/f − 1/u = 1/v

Step 2: Combine the Fractions

(u − f)/(fu) = 1/v

Step 3: Take Reciprocals

v = fu/(u − f)

This rearrangement assumes f, u, and v are nonzero and u ≠ f for a finite value of v.

Important Observation

When a variable appears as a reciprocal, isolating that reciprocal before inverting both sides is often effective.

Rearranging Equations with Nested Brackets

Worked Example

Make x the subject:

y = a[b(x + c) + d]

Assume a ≠ 0 and b ≠ 0.

Step 1: Divide by a

y/a = b(x + c) + d

Step 2: Subtract d

y/a − d = b(x + c)

Step 3: Divide by b

(y/a − d)/b = x + c

Step 4: Subtract c

x = (y/a − d)/b − c

Explanation

The operations applied to x were:

1. Add c.

2. Multiply by b.

3. Add d.

4. Multiply by a.

We reversed these operations in the opposite order.

Important Observation

When an equation contains nested operations, work from the outside inward to isolate the required variable.

Verifying Complex Rearrangements

A reliable verification method is to substitute the rearranged formula back into the original equation.

Worked Example

Original:

y = a(x + b)

Rearranged:

x = y/a − b

Substitute:

y = a[(y/a − b) + b]

Simplify inside the brackets:

y = a(y/a)

Cancel:

y = y

The original equation is satisfied for a ≠ 0.

Important Observation

A successful algebraic substitution confirms that the rearranged formula reproduces the original relationship under the stated conditions.

A Strategy for Multi-Step Literal Equations

When solving a challenging literal equation, use the following approach.

1. Identify the required variable. Decide exactly what must become the subject.

2. Inspect the equation. Determine whether the variable appears in fractions, brackets, powers, or multiple terms.

3. Simplify the structure. Expand, factor, or clear denominators when helpful.

4. Collect the variable terms. Bring terms containing the required variable together.

5. Apply inverse operations. Undo addition, subtraction, multiplication, division, powers, or roots.

6. Check restrictions. Avoid division by zero and account for the domains of roots and logarithms.

7. Verify the result. Substitute back into the original formula.

8. Explain each step. Use mathematical language to justify the rearrangement.

Common Mistakes in Advanced Rearrangement

Forgetting to apply an operation to every term

When multiplying an equation by a common denominator, every term must be multiplied.

Incorrectly cancelling terms

Cancellation is only valid for common multiplicative factors, not terms separated by addition or subtraction.

Forgetting the ± sign

When solving x² = a with a > 0, the solutions are x = ±√a.

Ignoring domain restrictions

Square roots, logarithms, and denominators have mathematical restrictions.

Squaring before isolating a radical

This often creates unnecessary complexity and increases the risk of errors.

Taking reciprocals incorrectly

If 1/x = a/b, then x = b/a, provided the relevant quantities are nonzero.

Failing to collect variable terms

When the required variable appears more than once, factoring it out may be necessary.

Assuming every equation has a unique solution

Some equations have no solutions, multiple solutions, or infinitely many solutions.

Not checking the physical meaning

A negative radius or negative mass may be mathematically possible in an intermediate calculation but invalid for the intended physical situation.

Part 2 Practice

Variables in Denominators

1. Make x the subject: y = a/x.

2. Make x the subject: y = a/x + b.

3. Make x the subject: x/a + x/b = c.

4. Make x the subject: y = (a + x)/b.

5. Make x the subject: y = a/(x + b).

Powers and Roots

6. Make x the subject: y = x² + a.

7. Make x the subject: y = 2(x − 3)².

8. Make x the subject: y = x³ − 5.

9. Make x the subject: y = √(x + 4).

10. Make x the subject: y = 2√(x − 1) + 3.

Applications

11. Make r the subject: A = πr².

12. Make v the subject: Eₖ = ½mv², where v represents speed.

13. Make h the subject: Eₚ = mgh.

14. Make C the subject: F = (9/5)C + 32.

15. Make R the subject: 1/R = 1/R₁ + 1/R₂.

Multi-Step Rearrangement

16. Make x the subject: ax + b = cx + d.

17. Make x the subject: a(x + b) = c(x + d).

18. Make x the subject: y = a[b(x + c) + d].

19. Make v the subject: 1/f = 1/u + 1/v.

20. Explain why rearranging y = x² to x = √y does not give every real solution.

Part 2 Answer Key

For questions involving division, the denominators must be nonzero. For questions involving real square roots, the quantity under the root must be nonnegative.

Part 2 Summary

In this part, we developed more advanced techniques for solving literal equations.

We learned how to:

• Isolate variables appearing in denominators.

• Clear fractions using common denominators.

• Collect variable terms from both sides.

• Rearrange equations containing nested brackets.

• Use square roots and cube roots as inverse operations.

• Recognise when positive and negative solutions are required.

• Rearrange equations containing radicals.

• Use logarithms to isolate variables in exponents.

• Rearrange formulas from geometry, physics, finance, and electricity.

• Recognise restrictions and special cases.

• Verify rearrangements algebraically and numerically.

• Explain each operation clearly.

The most important principle is that complex literal equations become manageable when we identify their structure and apply inverse operations systematically.