Rearranging Equations and Formulas

1. Solving Literal Equations

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.

2. Rearranging Scientific Formulae

Introduction

Scientific formulae describe mathematical relationships between measurable quantities.

Scientists use formulae to calculate quantities such as speed, acceleration, force, energy, pressure, density, and electrical resistance.

For example, Newton's Second Law is written:

F = ma

where:

• F = resultant force in newtons (N)

• m = mass in kilograms (kg)

• a = acceleration in metres per second squared (m/s²)

This formula allows us to calculate force when mass and acceleration are known.

However, what happens if we know the force and acceleration but need to calculate the mass?

We must rearrange the formula:

m = F/a

Similarly, if we need to calculate acceleration:

a = F/m

All three formulae describe the same scientific relationship, provided the required divisions are defined.

The ability to rearrange scientific formulae is essential because scientists rarely receive information in exactly the form needed for a calculation.

In this topic, we will learn how to rearrange scientific and geometric formulae, isolate variables in multi-step equations, substitute numerical values, and verify our results.

Why Do Scientists Rearrange Formulae?

Scientific formulae are often written in a conventional form, but the unknown quantity can change from one problem to another.

Consider:

v = d/t

where:

• v = speed

• d = distance

• t = time

This formula calculates speed.

However, the same relationship can be rearranged to calculate distance:

d = vt

Or time:

t = d/v

Example

A car travels 150 km at a constant average speed of 75 km/h.

How long does the journey take?

Use:

t = d/v

Substitute:

t = 150/75

t = 2 hours

Important Observation

Rearranging a formula allows us to calculate different quantities without memorising a separate equation for every situation.

Understanding the Subject of a Formula

The subject of a formula is the variable isolated on one side of the equation.

For example:

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.

Identifying the Subject

Important Observation

Before rearranging a formula, always identify the variable that needs to become the subject.

The Balance Principle

An equation represents a balance between two equal quantities.

If an operation is performed on one side, an equivalent operation must be performed on the other side.

Consider:

3x + 4 = 19

Subtract 4 from both sides:

3x = 15

Divide both sides by 3:

x = 5

The same principle applies to scientific formulae.

Example

F = ma

To isolate m, divide both sides by a:

F/a = ma/a

Simplify:

m = F/a

This division requires a ≠ 0.

Important Observation

Every rearrangement must preserve the equality of the original formula.

Using Inverse Operations

Inverse operations undo one another.

Example

Make x the subject:

y = 4x + 7

Subtract 7:

y − 7 = 4x

Divide by 4:

x = (y − 7)/4

Explanation

The operations applied to x were:

• Multiply by 4.

• Add 7.

We reverse these operations in the opposite order:

• Subtract 7.

• Divide by 4.

Important Observation

When isolating a variable, undo the operations surrounding it in reverse order.

A General Strategy for Rearranging Scientific Formulae

A reliable approach is to follow these steps.

1. Identify the unknown quantity. Determine which variable must be isolated.

2. Write the original formula. Include the correct symbols and mathematical operations.

3. Identify the operations. Determine what is being added, subtracted, multiplied, divided, squared, or otherwise applied.

4. Apply inverse operations. Perform equivalent operations on both sides.

5. Simplify the rearranged formula. Collect terms or cancel common factors where appropriate.

6. Check restrictions. Avoid division by zero and invalid square roots.

7. Substitute known values. Use consistent units.

8. Verify the result. Substitute the answer into the original formula.

Rearranging the Speed Formula

Speed is calculated using:

v = d/t

where:

• v = speed in m/s

• d = distance in m

• t = time in s

Make d the Subject

Start:

v = d/t

Multiply both sides by t:

vt = d

Therefore:

d = vt

Make t the Subject

Start:

v = d/t

Multiply by t:

vt = d

Divide by v:

t = d/v

This requires v ≠ 0.

Worked Example

A cyclist travels at a constant speed of 6 m/s for 25 seconds.

Calculate the distance travelled.

Use:

d = vt

Substitute:

d = 6 × 25

d = 150 m

Verification

v = d/t

v = 150/25

v = 6 m/s

The calculation is correct.

Rearranging the Acceleration Formula

Average acceleration is calculated using:

a = (v − u)/t

where:

• a = acceleration in m/s²

• v = final velocity in m/s

• u = initial velocity in m/s

• t = time in s

Make v the Subject

Start:

a = (v − u)/t

Multiply both sides by t:

at = v − u

Add u:

v = u + at

Make u the Subject

Start:

a = (v − u)/t

Multiply by t:

at = v − u

Subtract v:

at − v = −u

Multiply by −1:

u = v − at

Make t the Subject

Start:

a = (v − u)/t

Multiply by t:

at = v − u

Divide by a:

t = (v − u)/a

This requires a ≠ 0.

Worked Example

A vehicle accelerates uniformly from 4 m/s to 20 m/s over 8 seconds.

Calculate its acceleration.

a = (v − u)/t

a = (20 − 4)/8

a = 16/8

a = 2 m/s²

Verification

v = u + at

v = 4 + 2(8)

v = 20 m/s

The result is correct.

Rearranging Newton's Second Law

Newton's Second Law is:

F = ma

where:

• F = resultant force in N

• m = mass in kg

• a = acceleration in m/s²

Make m the Subject

Divide by a:

m = F/a

Make a the Subject

Divide by m:

a = F/m

Worked Example

A resultant force of 45 N produces an acceleration of 3 m/s².

Calculate the mass.

m = F/a

m = 45/3

m = 15 kg

Verification

F = ma

F = 15 × 3

F = 45 N

Important Observation

Rearranging a formula before substituting values makes calculations easier to organise and verify.

Rearranging the Weight Formula

Weight is the gravitational force acting on an object.

The formula is:

W = mg

where:

• W = weight in N

• m = mass in kg

• g = gravitational field strength in N/kg

Make m the Subject

Divide by g:

m = W/g

Make g the Subject

Divide by m:

g = W/m

Worked Example

An object has a weight of 196 N on Earth.

Using g = 9.8 N/kg, calculate its mass.

m = W/g

m = 196/9.8

m = 20 kg

Verification

W = mg

W = 20 × 9.8

W = 196 N

Important Observation

Mass and weight are different quantities, but the weight formula allows us to calculate one from the other when gravitational field strength is known.

Rearranging the Density Formula

Density is defined as mass per unit volume.

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 requires ρ ≠ 0.

Worked Example

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

Calculate its volume.

V = m/ρ

V = 540/9

V = 60 cm³

Verification

ρ = m/V

ρ = 540/60

ρ = 9 g/cm³

Rearranging the Pressure Formula

Pressure is force per unit area.

P = F/A

where:

• P = pressure in pascals (Pa)

• F = perpendicular force in newtons (N)

• A = area in m²

Make F the Subject

Multiply 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 600 N produces a pressure of 1500 Pa.

Calculate the area.

A = F/P

A = 600/1500

A = 0.4 m²

Verification

P = F/A

P = 600/0.4

P = 1500 Pa

Rearranging the Work Done Formula

For a constant force acting in the direction of displacement:

E = Fd

where:

• E = work done or energy transferred in joules (J)

• F = force in newtons (N)

• d = displacement in metres (m)

Make F the Subject

Divide by d:

F = E/d

Make d the Subject

Divide by F:

d = E/F

Worked Example

A force transfers 360 J of energy while moving an object 12 m in the direction of the force.

Calculate the force.

F = E/d

F = 360/12

F = 30 N

Verification

E = Fd

E = 30 × 12

E = 360 J

Important Observation

This formula applies when the force is constant and acts in the direction of the displacement. If the force acts at an angle, a more general formula is required.

Rearranging the Power Formula

Power is the rate of energy transfer.

P = E/t

where:

• P = power in watts (W)

• E = energy transferred in joules (J)

• t = time in seconds (s)

Make E the Subject

Multiply by t:

E = Pt

Make t the Subject

Multiply by t:

Pt = E

Divide by P:

t = E/P

This requires P ≠ 0.

Worked Example

An electrical device operates at 200 W for 30 seconds.

Calculate the energy transferred.

E = Pt

E = 200 × 30

E = 6000 J

Verification

P = E/t

P = 6000/30

P = 200 W

Rearranging the Kinetic Energy Formula

The kinetic energy of a moving object is:

Eₖ = ½mv²

where:

• Eₖ = kinetic energy in J

• m = mass in kg

• v = speed in m/s

This formula requires more advanced rearrangement because the speed is squared.

Make m the Subject

Start:

Eₖ = ½mv²

Multiply both sides 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 the nonnegative square root because v represents speed:

v = √(2Eₖ/m)

This assumes m > 0.

Worked Example

A 4 kg object has kinetic energy of 200 J.

Calculate its speed.

v = √(2Eₖ/m)

v = √[(2 × 200)/4]

= √100

v = 10 m/s

Verification

Eₖ = ½mv²

Eₖ = ½(4)(10²)

= 2 × 100

Eₖ = 200 J

Important Observation

When a variable is squared, the final rearrangement often requires a square root.

Rearranging Gravitational Potential Energy

Near Earth's surface, the change in gravitational potential energy is:

Eₚ = mgh

where:

• Eₚ = gravitational potential energy change in J

• m = mass in kg

• g = gravitational field strength in N/kg

• h = vertical height change in m

Make m the Subject

Divide by gh:

m = Eₚ/(gh)

Make g the Subject

Divide by mh:

g = Eₚ/(mh)

Make h the Subject

Divide by mg:

h = Eₚ/(mg)

Worked Example

A 6 kg object gains 588 J of gravitational potential energy.

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

h = Eₚ/(mg)

h = 588/(6 × 9.8)

h = 588/58.8

h = 10 m

Verification

Eₚ = mgh

Eₚ = 6 × 9.8 × 10

Eₚ = 588 J

Rearranging Ohm's Law

Ohm's Law is:

V = IR

where:

• V = potential difference in volts (V)

• I = 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 resistor has a potential difference of 18 V and a current of 2 A.

Calculate its resistance.

R = V/I

R = 18/2

R = 9 Ω

Verification

V = IR

V = 2 × 9

V = 18 V

Important Observation

Rearranging Ohm's Law allows scientists to calculate voltage, current, or resistance from the other two quantities.

Rearranging Electrical Power

Electrical power can be calculated using:

P = VI

where:

• P = power in watts (W)

• V = potential difference in volts (V)

• I = current in amperes (A)

Make V the Subject

Divide by I:

V = P/I

Make I the Subject

Divide by V:

I = P/V

Worked Example

An electrical appliance uses 1200 W of power at a potential difference of 240 V.

Calculate the current.

I = P/V

I = 1200/240

I = 5 A

Verification

P = VI

P = 240 × 5

P = 1200 W

Rearranging Wave Speed

Wave speed is calculated using:

v = fλ

where:

• v = wave speed in m/s

• f = frequency in hertz (Hz)

• λ = wavelength in metres (m)

Make f the Subject

Divide by λ:

f = v/λ

Make λ the Subject

Divide by f:

λ = v/f

Worked Example

A wave travels at 12 m/s and has a frequency of 4 Hz.

Calculate its wavelength.

λ = v/f

λ = 12/4

λ = 3 m

Verification

v = fλ

v = 4 × 3

v = 12 m/s

Rearranging Momentum

Momentum is calculated using:

p = mv

where:

• p = momentum in kg·m/s

• m = mass in kg

• v = velocity in m/s

Make m the Subject

Divide by v:

m = p/v

Make v the Subject

Divide by m:

v = p/m

Worked Example

A 5 kg object has momentum of 40 kg·m/s.

Calculate its velocity.

v = p/m

v = 40/5

v = 8 m/s

Verification

p = mv

p = 5 × 8

p = 40 kg·m/s

Important Observation

Momentum is a vector quantity, so its direction must also be considered in more advanced problems.

Using Units to Check Rearranged Formulae

Scientific calculations should produce answers with appropriate units.

Units can provide a useful check on rearranged formulas.

Example: Speed

v = d/t

Units:

m/s

Correct unit for speed.

Example: Force

F = ma

Units:

kg × m/s²

This is equivalent to:

N

Example: Pressure

P = F/A

Units:

N/m²

This is equivalent to:

Pa

Example: Energy

E = Fd

Units:

N × m

This is equivalent to:

J

Example: Power

P = E/t

Units:

J/s

This is equivalent to:

W

Important Observation

Dimensional analysis can reveal mistakes in scientific formulae, although correct units alone do not guarantee that a formula is correct.

A Worked Example Combining Rearrangement and Unit Conversion

A vehicle travels 2.4 km in 120 seconds.

Calculate its average speed in m/s.

Step 1: Identify the Formula

v = d/t

Step 2: Convert Units

2.4 km = 2400 m

Step 3: Substitute

v = 2400/120

v = 20 m/s

Step 4: Verify

d = vt

d = 20 × 120

d = 2400 m

Important Observation

Always convert quantities into compatible units before substituting them into a scientific formula.

Common Mistakes in Scientific Rearrangement

Rearranging Without Identifying the Required Variable

Before starting, determine which quantity the problem asks you to calculate.

Applying Operations to Only One Side

Every operation must preserve equality.

Dividing Only Part of an Expression

Incorrect:

y = 3x + 6

x = y/3 − 6

Correct:

x = (y − 6)/3

Losing Negative Signs

Incorrect:

y = 10 − 2x

x = (y − 10)/2

Correct:

x = (10 − y)/2

Forgetting Square Roots

Incorrect:

Eₖ = ½mv²

v = 2Eₖ/m

Correct:

v = √(2Eₖ/m)

when v represents speed and m > 0.

Using Inconsistent Units

For example, substituting distance in kilometres and time in seconds into a calculation intended to produce m/s.

Forgetting Restrictions

Division by zero is not permitted.

Failing to Verify

Substitute the calculated value into the original formula to check the result.

Part 1 Practice

Basic Scientific Formulae

1. Make d the subject: v = d/t.

2. Make t the subject: v = d/t.

3. Make m the subject: F = ma.

4. Make a the subject: F = ma.

5. Make g the subject: W = mg.

Intermediate Rearrangement

6. Make V the subject: ρ = m/V.

7. Make A the subject: P = F/A.

8. Make t the subject: P = E/t.

9. Make R the subject: V = IR.

10. Make λ the subject: v = fλ.

Multi-Step Formulae

11. Make v the subject: a = (v − u)/t.

12. Make u the subject: a = (v − u)/t.

13. Make m the subject: Eₖ = ½mv².

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

15. Make h the subject: Eₚ = mgh.

Numerical Applications

16. A car travels 300 m in 15 s. Calculate its average speed.

17. A resultant force of 72 N produces an acceleration of 6 m/s². Calculate the mass.

18. A 5 kg object has a weight of 49 N. Calculate gravitational field strength.

19. A resistor has a potential difference of 24 V and a current of 3 A. Calculate resistance.

20. A wave travels at 18 m/s and has a frequency of 6 Hz. Calculate wavelength.

Part 1 Answer Key

Part 1 Summary

In this part, we learned how to:

• Identify the required variable in a scientific formula.

• Apply inverse operations to isolate variables.

• Rearrange formulas involving multiplication and division.

• Rearrange multi-step formulas.

• Use square roots when isolating squared variables.

• Substitute numerical values into rearranged formulas.

• Verify calculations using the original equations.

• Check scientific units.

• Apply formula rearrangement to motion, forces, energy, pressure, density, electricity, and waves.

The most important principle is:

Rearrange the formula first, substitute values second, and verify the result using the original relationship.

Introduction

In Part 1, we explored how to rearrange scientific formulae involving speed, acceleration, force, density, pressure, energy, electricity, and waves.

We also learned that a reliable calculation follows a clear sequence:

Identify the unknown → Rearrange the formula → Substitute values → Calculate → Verify

In this part, we will extend these skills to geometric formulae and more challenging scientific equations.

We will investigate how to:

• Rearrange formulae involving perimeter, area, surface area, and volume.

• Isolate variables raised to powers.

• Work with square roots and cube roots.

• Rearrange equations containing brackets and fractions.

• Collect variables that appear more than once.

• Solve multi-step scientific problems.

• Verify rearranged formulae algebraically and numerically.

The goal is to understand not only how to rearrange a formula, but also why each algebraic operation is valid.

Rearranging the Perimeter of a Rectangle

The perimeter of a rectangle is the total distance around its boundary.

P = 2l + 2w

where:

• P = perimeter

• l = length

• w = width

Making Length the Subject

Start with:

P = 2l + 2w

Subtract 2w from both sides:

P − 2w = 2l

Divide both sides by 2:

(P − 2w)/2 = l

Therefore:

l = P/2 − w

Making Width the Subject

Start with:

P = 2l + 2w

Subtract 2l:

P − 2l = 2w

Divide by 2:

w = P/2 − l

Worked Example

A rectangular field has a perimeter of 90 m and a width of 18 m.

Calculate its length.

l = P/2 − w

l = 90/2 − 18

l = 45 − 18

l = 27 m

Verification

P = 2l + 2w

P = 2(27) + 2(18)

= 54 + 36

P = 90 m

The calculation is correct.

Important Observation

When a variable appears in a formula involving addition and multiplication, undo addition or subtraction before dividing.

Rearranging the Area of a Rectangle

The area of a rectangle is:

A = lw

Making Length the Subject

Divide by w:

l = A/w

Making Width the Subject

Divide by l:

w = A/l

These rearrangements require nonzero divisors.

Worked Example

A rectangular laboratory floor has an area of 84 m² and a length of 12 m.

Calculate its width.

w = A/l

w = 84/12

w = 7 m

Verification

A = lw

A = 12 × 7

A = 84 m²

Important Observation

The same area formula can calculate any one of three quantities when the other two are known.

Rearranging the Area of a Triangle

The area of a triangle is:

A = ½bh

where:

• A = area

• b = base

• h = perpendicular height

Making Height the Subject

Start:

A = ½bh

Multiply both sides by 2:

2A = bh

Divide by b:

h = 2A/b

Making Base the Subject

Start:

A = ½bh

Multiply by 2:

2A = bh

Divide by h:

b = 2A/h

Worked Example

A triangular sail has an area of 24 m² and a base of 6 m.

Calculate its height.

h = 2A/b

h = 2(24)/6

h = 48/6

h = 8 m

Verification

A = ½bh

A = ½(6)(8)

A = 24 m²

Important Observation

Fractions in formulae can often be removed by multiplying both sides by the denominator.

Rearranging the Area of a Trapezium

The area of a trapezium is:

A = ½(a + b)h

where:

• a = length of one parallel side

• b = length of the other parallel side

• h = perpendicular height

Making Height the Subject

Start:

A = ½(a + b)h

Multiply by 2:

2A = (a + b)h

Divide by a + b:

h = 2A/(a + b)

This requires a + b ≠ 0.

Making a the Subject

Start:

A = ½(a + b)h

Multiply by 2:

2A = (a + b)h

Divide by h:

2A/h = a + b

Subtract b:

a = 2A/h − b

This requires h ≠ 0.

Worked Example

A trapezium has an area of 70 cm².

Its parallel sides are 8 cm and 12 cm.

Calculate its height.

h = 2A/(a + b)

h = 2(70)/(8 + 12)

h = 140/20

h = 7 cm

Verification

A = ½(8 + 12)(7)

= ½(20)(7)

A = 70 cm²

Important Observation

When several variables are grouped inside brackets, treat the entire bracket as one expression until it is appropriate to simplify it.

Rearranging the Circumference of a Circle

The circumference of a circle is:

C = 2πr

where:

• C = circumference

• r = radius

• π ≈ 3.14159

Making Radius the Subject

Divide both sides by 2π:

r = C/(2π)

Worked Example

A circular running track has a circumference of 125.6 m.

Calculate its approximate radius.

Using π ≈ 3.14:

r = 125.6/(2 × 3.14)

r = 125.6/6.28

r = 20 m

Verification

C = 2πr

C ≈ 2(3.14)(20)

C ≈ 125.6 m

Rearranging the Area of a Circle

The area of a circle is:

A = πr²

The radius is squared, so rearranging requires a square root.

Making Radius the Subject

Start:

A = πr²

Divide by π:

A/π = r²

Take square roots:

r = ±√(A/π)

Since radius is nonnegative:

r = √(A/π)

Worked Example

A circular pond has an area of 153.86 m².

Using π ≈ 3.14, calculate its radius.

r = √(A/π)

r = √(153.86/3.14)

r = √49

r = 7 m

Verification

A = πr²

A ≈ 3.14 × 7²

= 3.14 × 49

A ≈ 153.86 m²

Important Observation

A squared variable requires a square root, and the physical meaning of the variable determines whether negative values are appropriate.

Rearranging the Volume of a Rectangular Prism

The volume of a rectangular prism is:

V = lwh

where:

• V = volume

• l = length

• w = width

• h = height

Making Height the Subject

Divide by lw:

h = V/(lw)

Making Length the Subject

Divide by wh:

l = V/(wh)

Making Width the Subject

Divide by lh:

w = V/(lh)

Worked Example

A storage container has a volume of 2400 cm³.

Its length is 20 cm and its width is 12 cm.

Calculate its height.

h = V/(lw)

h = 2400/(20 × 12)

h = 2400/240

h = 10 cm

Verification

V = lwh

V = 20 × 12 × 10

V = 2400 cm³

Rearranging the Volume of a Cylinder

The volume of a cylinder is:

V = πr²h

where:

• V = volume

• r = radius

• h = height

Making Height the Subject

Divide by πr²:

h = V/(πr²)

This requires r ≠ 0.

Making Radius the Subject

Start:

V = πr²h

Divide by πh:

V/(πh) = r²

Take square roots:

r = ±√[V/(πh)]

Since radius is nonnegative:

r = √[V/(πh)]

For a physical cylinder, h > 0.

Worked Example

A cylindrical container has a volume of 1256 cm³ and a height of 10 cm.

Calculate its radius using π ≈ 3.14.

r = √[V/(πh)]

r = √[1256/(3.14 × 10)]

= √40

r ≈ 6.32 cm

Verification

V = πr²h

Using the unrounded value r² = 40:

V = 3.14 × 40 × 10

V = 1256 cm³

Important Observation

When a variable is squared and multiplied by other quantities, divide first and take square roots afterward.

Rearranging the Volume of a Sphere

The volume of a sphere is:

V = (4/3)πr³

The radius is cubed, so we must use a cube root.

Making Radius the Subject

Start:

V = (4/3)πr³

Multiply both sides by 3:

3V = 4πr³

Divide by 4π:

3V/(4π) = r³

Take cube roots:

r = ∛[3V/(4π)]

Worked Example

A spherical object has a volume of approximately 904.32 cm³.

Using π ≈ 3.14, calculate its radius.

r = ∛[3V/(4π)]

r = ∛[(3 × 904.32)/(4 × 3.14)]

= ∛(2712.96/12.56)

= ∛216

r = 6 cm

Verification

V = (4/3)πr³

V ≈ (4/3)(3.14)(6³)

= (4/3)(3.14)(216)

V ≈ 904.32 cm³

Important Observation

The inverse operation depends on the power of the required variable.

• x² requires a square root.

• x³ requires a cube root.

Rearranging the Surface Area of a Cylinder

The total surface area of a closed cylinder is:

S = 2πr² + 2πrh

where:

• S = total surface area

• r = radius

• h = height

Suppose we want to make h the subject.

Step 1: Subtract 2πr²

S − 2πr² = 2πrh

Step 2: Divide by 2πr

h = (S − 2πr²)/(2πr)

For r ≠ 0, this can be simplified:

h = S/(2πr) − r

Worked Example

A closed cylinder has:

• Surface area = 376.8 cm²

• Radius = 5 cm

Using π ≈ 3.14, calculate its height.

h = S/(2πr) − r

h = 376.8/[2(3.14)(5)] − 5

= 376.8/31.4 − 5

= 12 − 5

h = 7 cm

Verification

S = 2πr² + 2πrh

S = 2(3.14)(5²) + 2(3.14)(5)(7)

= 157 + 219.8

S = 376.8 cm²

Important Observation

Some geometry formulae contain the required variable in only one term, even when the overall equation has several terms.

Multi-Step Rearrangement with Brackets

Scientific equations sometimes contain several operations surrounding the required variable.

Worked Example

Make x the subject:

y = a(x + b) + c

Step 1: Subtract c

y − c = a(x + b)

Step 2: Divide by a

(y − c)/a = x + b

Step 3: Subtract b

x = (y − c)/a − b

This requires a ≠ 0.

Numerical Example

Suppose:

y = 29, a = 4, b = 3, c = 5

Then:

x = (29 − 5)/4 − 3

= 24/4 − 3

= 6 − 3

x = 3

Verification

y = 4(3 + 3) + 5

= 24 + 5

y = 29

Rearranging Formulae with Nested Brackets

Worked Example

Make x the subject:

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

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

This requires a ≠ 0 and b ≠ 0.

Explanation

The operations originally applied to x were:

• Add c.

• Multiply by b.

• Add d.

• Multiply by a.

To isolate x, undo them in reverse order.

Important Observation

Nested expressions are often easiest to rearrange by working from the outermost operation inward.

Rearranging Formulae with Variables on Both Sides

Sometimes the required variable appears on both sides of the equation.

Worked Example

Make x the subject:

ax + b = cx + d

Subtract cx:

ax − cx + b = d

Subtract b:

ax − cx = d − b

Factor x:

x(a − c) = d − b

Divide by a − c:

x = (d − b)/(a − c)

This requires a ≠ c.

Numerical Example

Let:

a = 5, b = 7, c = 2, d = 19

Then:

x = (19 − 7)/(5 − 2)

= 12/3

x = 4

Verification

Left side:

5(4) + 7 = 27

Right side:

2(4) + 19 = 27

Both sides are equal.

Important Observation

When a variable appears in several terms, collect those terms and factor out the variable before dividing.

Rearranging Formulae with Fractions

Worked Example

Make x the subject:

y = (ax + b)/c

Multiply by c:

cy = ax + b

Subtract b:

cy − b = ax

Divide by a:

x = (cy − b)/a

This requires a ≠ 0 and c ≠ 0.

Numerical Example

Suppose:

y = 7, a = 3, b = 5, c = 2

Then:

x = (2 × 7 − 5)/3

= (14 − 5)/3

= 9/3

x = 3

Verification

y = (3 × 3 + 5)/2

= 14/2

y = 7

Clearing Multiple Denominators

Some formulae contain several fractions.

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

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 x:

x(a + b) = abc

Divide by a + b:

x = abc/(a + b)

This requires a + b ≠ 0.

Important Observation

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

Rearranging a Formula with a Variable in the Denominator

Worked Example

Make x the subject:

y = a/(x + b)

Step 1: Multiply by x + b

y(x + b) = a

Step 2: Divide by y

x + b = a/y

Step 3: Subtract b

x = a/y − b

This assumes a ≠ 0 and y ≠ 0.

Verification

Substitute x = a/y − b into the original formula:

y = a/[(a/y − b) + b]

Simplify:

y = a/(a/y)

Therefore:

y = y

The rearrangement is verified.

Rearranging Formulae with Square Roots

Some scientific equations contain square roots.

Worked Example

Make x the subject:

y = √(x + a)

Square both sides:

y² = x + a

Subtract a:

x = y² − a

The original formula requires y ≥ 0.

Another 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)²/9 = x + 2

Subtract 2:

x = (y + 4)²/9 − 2

The original formula requires y ≥ −4.

Important Observation

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

Why Restrictions Matter

Rearranging an equation can introduce expressions that are undefined for certain values.

For example:

y = a/x

The original formula requires:

x ≠ 0

Rearranging gives:

x = a/y

This requires y ≠ 0.

If a = 0, the original equation becomes y = 0 for every nonzero x, so the rearrangement cannot uniquely determine x.

Another Example

y = x²

Rearranging gives:

x = ±√y

For real solutions:

y ≥ 0

If y > 0, there are two solutions.

If y = 0, there is one solution.

If y < 0, there are no real solutions.

Important Observation

A complete rearrangement should include restrictions whenever division, roots, or other domain-sensitive operations are involved.

Verifying Rearranged Formulae Algebraically

Numerical substitution is useful, but algebraic verification provides a more general check.

Worked Example

Original:

y = a(x + b)

Rearranged:

x = y/a − b

Substitute into the original formula:

y = a[(y/a − b) + b]

Simplify inside the brackets:

y = a(y/a)

Cancel:

y = y

The rearrangement is verified for a ≠ 0.

Important Observation

Algebraic verification shows that the rearranged formula reproduces the original relationship for all permitted values.

Verifying Scientific Formulae Using Units

Dimensional analysis provides another useful check.

Example: Cylinder Volume

V = πr²h

Units:

cm² × cm = cm³

Correct.

Example: Density

ρ = m/V

Units:

g/cm³

Correct.

Example: Speed

v = d/t

Units:

m/s

Correct.

Example: Force

F = ma

Units:

kg × m/s² = N

Correct.

Important Observation

Dimensional analysis can detect many incorrect rearrangements, but matching units does not prove that a formula is correct.

Common Mistakes in Multi-Step Rearrangement

Dividing Before Collecting Terms

Incorrect:

ax + bx = c

x = c/a + c/b

Correct:

x(a + b) = c

x = c/(a + b)

Forgetting Brackets

Incorrect:

x = y − c/a − b

when the intended formula is:

x = (y − c)/a − b

Losing Negative Signs

Incorrect:

y = 10 − 3x

x = (y − 10)/3

Correct:

x = (10 − y)/3

Forgetting Square Roots

Incorrect:

A = πr²

r = A/π

Correct:

r = √(A/π)

for a nonnegative radius.

Cancelling Terms Instead of Factors

Incorrect:

(a + x)/a = x

Correct:

(a + x)/a = 1 + x/a

for a ≠ 0.

Ignoring Restrictions

Dividing by an expression that may equal zero can remove valid cases or introduce undefined results.

Failing to Verify

Substituting into the original formula can help identify algebraic mistakes.

Part 2 Practice

Geometry Formulae

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

2. Make w the subject: A = lw.

3. Make h the subject: A = ½bh.

4. Make h the subject: A = ½(a + b)h.

5. Make r the subject: C = 2πr.

Area and Volume

6. Make r the subject: A = πr².

7. Make h the subject: V = lwh.

8. Make r the subject: V = πr²h.

9. Make h the subject: V = πr²h.

10. Make r the subject: V = (4/3)πr³.

Multi-Step Rearrangement

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

12. Make x the subject: y = (ax + b)/c.

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

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

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

Numerical Applications

16. A rectangle has perimeter 84 m and width 15 m. Calculate its length.

17. A triangle has area 60 cm² and base 12 cm. Calculate its height.

18. A circle has area 78.5 cm². Calculate its radius using π ≈ 3.14.

19. A rectangular prism has volume 3600 cm³, length 20 cm, and width 15 cm. Calculate its height.

20. A cylinder has volume 628 cm³ and radius 5 cm. Calculate its height using π ≈ 3.14.

Part 2 Answer Key

Part 2 Summary

In this part, we learned how to:

• Rearrange perimeter and area formulae.

• Isolate variables in volume and surface area equations.

• Use square roots and cube roots when rearranging powers.

• Solve multi-step equations involving brackets.

• Collect variables appearing on both sides.

• Clear fractions using common denominators.

• Rearrange formulae containing variables in denominators.

• Recognise restrictions and special cases.

• Verify rearrangements algebraically.

• Check scientific formulae using units.

The most important principle is:

Identify the structure of the formula, isolate the required variable systematically, and verify that the rearranged expression is equivalent to the original equation.

Introduction

Scientific formulae are powerful because one equation can describe several related quantities.

For example:

F = ma

can be rearranged to calculate force, mass, or acceleration.

Similarly:

V = πr²h

can be used to calculate the volume, height, or radius of a cylinder.

In Parts 1 and 2, we developed the algebraic skills required to rearrange scientific and geometric formulae.

In this final part, we will apply these techniques to realistic investigations, interpret numerical results, and develop reliable methods for checking our work.

A successful scientific calculation should demonstrate four things:

• The correct formula was selected.

• The required variable was isolated correctly.

• Values and units were substituted accurately.

• The final result was verified and interpreted.

Application: Investigating the Motion of a Vehicle

A student investigates the motion of a small electric vehicle.

The vehicle starts from rest and accelerates uniformly for 8 seconds.

Its final velocity is 24 m/s.

The student wants to calculate its acceleration.

Identifying the Formula

Average acceleration is:

a = (v − u)/t

where:

• a = acceleration in m/s²

• v = final velocity in m/s

• u = initial velocity in m/s

• t = time in seconds

Substituting Values

The vehicle starts from rest:

u = 0 m/s

Final velocity:

v = 24 m/s

Time:

t = 8 s

Therefore:

a = (24 − 0)/8

a = 3 m/s²

Rearranging to Find Final Velocity

The same formula can be rearranged:

a = (v − u)/t

Multiply by t:

at = v − u

Add u:

v = u + at

Verification

v = 0 + 3(8)

v = 24 m/s

The result agrees with the original information.

Important Observation

Rearranging a formula allows different quantities to be calculated from the same scientific relationship.

Application: Calculating Acceleration from Force

The electric vehicle has a mass of 12 kg.

The resultant force acting on it is 36 N.

Calculate its acceleration.

Formula

F = ma

Make a the Subject

Divide both sides by m:

a = F/m

Substitute

a = 36/12

a = 3 m/s²

Interpretation

The acceleration calculated using Newton's Second Law agrees with the acceleration calculated from the velocity measurements.

This agreement supports the consistency of the two calculations.

Important Observation

Scientists often use different equations to calculate the same quantity and compare the results.

Application: Calculating Braking Distance

A vehicle travelling at 20 m/s brakes uniformly until it stops.

Its acceleration is −5 m/s².

Calculate the braking distance.

We can use:

v² = u² + 2as

where:

• v = final velocity

• u = initial velocity

• a = acceleration

• s = displacement

Make s the Subject

Start:

v² = u² + 2as

Subtract u²:

v² − u² = 2as

Divide by 2a:

s = (v² − u²)/(2a)

This requires a ≠ 0.

Substitute

v = 0 m/s

u = 20 m/s

a = −5 m/s²

s = (0² − 20²)/[2(−5)]

= (0 − 400)/(−10)

s = 40 m

Verification

v² = u² + 2as

0² = 20² + 2(−5)(40)

0 = 400 − 400

0 = 0

The result is correct.

Important Observation

Negative acceleration does not necessarily mean negative displacement. The signs depend on the chosen positive direction.

Application: Designing a Cylindrical Water Tank

An engineer needs to design a cylindrical water tank.

The tank must hold 5000 litres of water.

Its internal radius will be 1 m.

Calculate the required internal height.

Step 1: Identify the Formula

The volume of a cylinder is:

V = πr²h

Step 2: Make h the Subject

Divide both sides by πr²:

h = V/(πr²)

Step 3: Convert Units

1 litre = 0.001 m³

Therefore:

5000 litres = 5 m³

Step 4: Substitute

h = 5/(π × 1²)

h = 5/π

h ≈ 1.59 m

Step 5: Verify

V = πr²h

Using the unrounded height:

V = π(1²)(5/π)

V = 5 m³

Interpretation

The tank needs an internal height of approximately 1.59 m to hold 5000 litres when completely full.

In practice, an engineer might provide additional space above the maximum operating water level.

Important Observation

Correct unit conversion is essential when applying geometric formulae to real-world engineering problems.

Application: Calculating the Radius of a Water Tank

Suppose the same tank must have a volume of 5 m³, but its height is restricted to 2 m.

What radius is required?

Formula

V = πr²h

Make r the Subject

Divide by πh:

V/(πh) = r²

Take the nonnegative square root:

r = √[V/(πh)]

Substitute

r = √[5/(π × 2)]

r = √(5/6.2832)

r ≈ √0.7958

r ≈ 0.892 m

Verification

Using the unrounded radius:

V = πr²h

V = π5/(2π)

V = 5 m³

Important Observation

The same formula can be rearranged in different ways to accommodate engineering design constraints.

Application: Investigating Density

A student measures the mass and volume of several metal samples.

The density formula is:

ρ = m/V

Suppose the student wants to predict the mass of a metal sample from its volume.

Rearranging the Formula

ρ = m/V

Multiply by V:

m = ρV

Worked Example

A metal has a density of 8.0 g/cm³.

Calculate the mass of a sample with a volume of 35 cm³.

m = ρV

m = 8.0 × 35

m = 280 g

Verification

ρ = m/V

ρ = 280/35

ρ = 8.0 g/cm³

Connecting the Formula to a Graph

If density is constant:

m = ρV

This has the same mathematical structure as:

y = mx

For a graph of mass against volume:

• Horizontal axis: volume

• Vertical axis: mass

• Gradient: density

3. Working with Formulas Containing Fractions

Introduction

Many mathematical and scientific formulas contain fractions. These fractions often represent important relationships between quantities, such as speed, density, pressure, and electrical resistance.

For example:

• Speed = distance ÷ time

• Density = mass ÷ volume

• Pressure = force ÷ area

• Average speed = total distance ÷ total time

When rearranging formulas, fractions can make it difficult to isolate a particular variable. Fortunately, we can often eliminate fractions by multiplying both sides of an equation by an appropriate denominator.

This technique is especially useful when working with algebraic equations, geometry formulas, and scientific relationships.

By understanding how to eliminate fractions correctly, we can simplify complicated formulas and solve problems more efficiently.

Understanding Fractions in Algebraic Formulas

A fraction represents division.

For example:

a/b means a divided by b, where b ≠ 0.

The upper part of a fraction is called the numerator, while the lower part is called the denominator.

Consider the formula:

v = d/t

where:

• v = speed

• d = distance

• t = time

This formula tells us that speed is calculated by dividing distance by time.

However, suppose we want to calculate distance rather than speed.

We must rearrange the formula so that d is the subject.

Starting with:

v = d/t

Multiply both sides by t:

vt = (d/t) × t

The t in the denominator cancels with the t in the numerator:

vt = d

Therefore:

d = vt

We have successfully eliminated the fraction and isolated distance.

Why Multiplying by the Denominator Works

An equation represents a balance between two expressions.

Whatever operation we perform on one side must also be performed on the other side to maintain equality.

Consider:

x/5 = 7

The variable x is being divided by 5.

To reverse division by 5, we multiply by 5.

(x/5) × 5 = 7 × 5

x = 35

We can check the result:

35/5 = 7

Since both sides are equal, the solution is correct.

This illustrates an important algebraic principle:

Multiplication and division are inverse operations.

Multiplying by a denominator reverses division by that denominator, provided the denominator is not zero.

Eliminating a Single Fraction

The simplest equations containing fractions have only one fractional expression.

Worked Example: Solving for a Numerator

Solve for x:

x/8 = 12

Step 1: Identify the denominator.

The denominator is 8.

Step 2: Multiply both sides by 8.

8 × (x/8) = 12 × 8

Step 3: Simplify.

x = 96

Step 4: Check.

96/8 = 12

The solution is correct.

Worked Example: Rearranging a Density Formula

Density is calculated using:

ρ = m/V

where:

• ρ = density

• m = mass

• V = volume

Suppose we want to make mass the subject.

Step 1: Start with the original formula.

ρ = m/V

Step 2: Multiply both sides by V.

ρV = (m/V) × V

Step 3: Simplify.

ρV = m

Therefore:

m = ρV

Real-World Application: Calculating the Mass of a Metal Block

A metal block has:

• Density = 7.8 g/cm³

• Volume = 25 cm³

Calculate its mass.

Using:

m = ρV

Substitute:

m = 7.8 × 25

m = 195 g

The mass of the block is 195 g.

This rearrangement is useful in materials science, manufacturing, and laboratory investigations.

Eliminating Fractions with Numerical Coefficients

Sometimes a formula contains both a fraction and a numerical coefficient.

Consider:

3x/4 = 15

We need to eliminate the denominator before isolating x.

Worked Example

Step 1: Multiply both sides by 4.

4 × (3x/4) = 15 × 4

Step 2: Simplify.

3x = 60

Step 3: Divide both sides by 3.

x = 60/3

Step 4: Calculate.

x = 20

Check:

3(20)/4 = 60/4 = 15

The answer is correct.

A More Efficient Method

We can also rearrange the equation in a single combined step:

3x/4 = 15

x = (15 × 4)/3

x = 20

Both methods produce the same answer.

The first method is generally easier for beginners because each operation is shown separately.

The second method becomes useful when students are confident with inverse operations.

Eliminating Fractions on Both Sides

Some equations contain fractions on both sides.

For example:

x/3 = 8/5

We could multiply both sides by 3 and then simplify, but another method is often more efficient.

Cross-Multiplication

For an equation of the form:

a/b = c/d

where b ≠ 0 and d ≠ 0, we can multiply both sides by bd to obtain:

ad = bc

This is commonly called cross-multiplication.

Worked Example

Solve:

x/3 = 8/5

Cross-multiply:

5x = 3 × 8

5x = 24

Divide by 5:

x = 24/5

x = 4.8

Check:

4.8/3 = 1.6

8/5 = 1.6

Both sides are equal.

Important Limitation

Cross-multiplication is appropriate when two complete fractions are equal.

For example:

(x + 2)/5 = 7/3

But it should not be applied directly to an equation containing additional terms outside the fractions, such as:

x/3 + 2 = 5/4

In that situation, we must first rearrange the equation or multiply every term by a suitable common denominator.

Introducing the Lowest Common Denominator

When an equation contains several fractions with different denominators, we need a systematic method to eliminate them.

The lowest common denominator (LCD) is the least common multiple of all the denominators.

Consider:

x/2 + x/3 = 10

The denominators are 2 and 3.

Their lowest common multiple is 6.

Therefore, the LCD is 6.

Worked Example: Eliminating Two Fractions

Solve:

x/2 + x/3 = 10

Step 1: Identify the LCD.

LCD = 6

Step 2: Multiply every term by 6.

6(x/2) + 6(x/3) = 6(10)

Step 3: Simplify each term.

3x + 2x = 60

Step 4: Collect like terms.

5x = 60

Step 5: Divide by 5.

x = 12

Check:

12/2 + 12/3 = 10

6 + 4 = 10

The solution is correct.

Why Every Term Must Be Multiplied

A frequent mistake is to multiply only the fractions and forget terms that are already whole numbers.

Consider:

x/4 + 3 = 7

Multiplying through by 4 means multiplying every term:

4(x/4) + 4(3) = 4(7)

Therefore:

x + 12 = 28

x = 16

It would be incorrect to write:

x + 3 = 28

because the constant 3 must also be multiplied by 4.

Choosing an Efficient Common Denominator

We can multiply by any nonzero common multiple of the denominators, but choosing the lowest common denominator usually makes the calculations easier.

Consider:

x/4 + x/6 = 5

Possible common denominators include 12, 24, 36, and 48.

The LCD is 12.

Multiply through by 12:

12(x/4) + 12(x/6) = 12(5)

3x + 2x = 60

5x = 60

x = 12

Using 24 would also work, but it would produce larger numbers:

6x + 4x = 120

10x = 120

x = 12

The answer is identical, but the LCD produces simpler intermediate calculations.

Practice Questions — Part 1

Try these questions before moving on to more complicated formulas.

1. Solve: x/7 = 9

2. Solve: 5x/6 = 20

3. Rearrange P = F/A to make F the subject.

4. Rearrange v = d/t to make d the subject.

5. Solve: x/4 = 3/8

6. Solve: x/3 + x/5 = 16

7. Solve: x/2 + 4 = 11

8. Explain why every term in an equation must be multiplied when eliminating fractions.

   Rearranging Formulas with Several Fractions

   In Part 1, we learned how to eliminate fractions by multiplying both sides of an equation by a denominator or the lowest common denominator (LCD).

   However, many formulas contain several fractions, different variables, or combinations of addition, subtraction, multiplication, and division.

   To rearrange these formulas successfully, we must apply algebraic operations in the correct order.

   The general strategy is:

   1. Identify the variable we want to isolate.

   2. Identify the denominators in the equation.

   3. Find a suitable common denominator.

   4. Multiply every term by that denominator.

   5. Expand brackets and collect like terms.

   6. Factor the required variable if necessary.

   7. Divide to isolate the variable.

   8. Check the result by substitution or algebraic verification.

   These steps work for many formulas used in mathematics, physics, chemistry, and engineering.

   Working with Variables in Denominators

   A formula becomes more challenging when the denominator contains a variable.

   Consider:

   A = b/h

   Suppose we want to make h the subject.

   Step 1: Multiply both sides by h.

   Ah = b

   Step 2: Divide both sides by A.

   h = b/A

   Therefore:

   h = b/A

   This rearrangement assumes that h and A are nonzero.

   Worked Example: Rearranging an Electrical Formula

   Electrical resistance is related to voltage and current by:

   R = V/I

   where:

   • R = resistance, measured in ohms (Ω)

   • V = voltage, measured in volts (V)

   • I = current, measured in amperes (A)

   Suppose we want to make current the subject.

   Starting with:

   R = V/I

   Multiply both sides by I:

   RI = V

   Divide both sides by R:

   I = V/R

   

    

    

    

   Real-World Application: Finding Electric Current

   A resistor has a resistance of 8 Ω and a potential difference of 24 V across it.

   Calculate the current.

   Using:

   I = V/R

   I = 24/8

   I = 3 A

   The current through the resistor is 3 amperes.

   This type of rearrangement is essential when analyzing electrical circuits.

   Rearranging Formulas with Addition and Fractions

   Some formulas contain a fraction together with an additional term.

   Consider:

   y = x/4 + 7

   We want to make x the subject.

   Worked Example

   Step 1: Subtract 7 from both sides.

   y − 7 = x/4

   Step 2: Multiply both sides by 4.

   4(y − 7) = x

   Therefore:

   x = 4(y − 7)

   We could also expand the brackets:

   x = 4y − 28

   Both forms are correct.

   Checking the Rearrangement

   Suppose y = 10.

   Using the rearranged formula:

   x = 4(10 − 7)

   x = 12

   Check in the original equation:

   10 = 12/4 + 7

   10 = 3 + 7

   10 = 10

   The rearrangement is correct.

   Why the Order Matters

   It is important to recognize that the original equation contains two operations:

   • Division of x by 4.

   • Addition of 7.

   We can reverse these operations by subtracting 7 first, then multiplying by 4.

   Alternatively, we could multiply the entire equation by 4 first:

   4y = x + 28

   x = 4y − 28

   Both approaches are valid because every algebraic operation is applied consistently to both sides.

   Eliminating Multiple Fractions with Different Denominators

   Consider the equation:

   x/3 + x/4 = 14

   The denominators are 3 and 4.

   The LCD is 12.

   Worked Example

   Step 1: Multiply every term by 12.

   12(x/3) + 12(x/4) = 12(14)

   Step 2: Simplify.

   4x + 3x = 168

   Step 3: Collect like terms.

   7x = 168

   Step 4: Divide by 7.

   x = 24

   Check:

   24/3 + 24/4 = 14

   8 + 6 = 14

   The solution is correct.

   A More Challenging Example

   Solve:

   (x + 2)/3 + (x − 1)/4 = 5

   Step 1: Identify the LCD.

   LCD = 12

   Step 2: Multiply every term by 12.

   12(x + 2)/3 + 12(x − 1)/4 = 12(5)

   Step 3: Simplify the fractions.

   4(x + 2) + 3(x − 1) = 60

   Step 4: Expand the brackets.

   4x + 8 + 3x − 3 = 60

   Step 5: Collect like terms.

   7x + 5 = 60

   7x = 55

   Step 6: Solve.

   x = 55/7

   Approximately:

   x ≈ 7.86

   Notice that we multiplied the entire numerator by the appropriate factor.

   For example:

   12(x + 2)/3 = 4(x + 2)

   We must not multiply only x and forget the constant 2.

   Rearranging Formulas with Two Fractional Terms

   Sometimes a formula contains two fractions that involve different variables.

   Consider:

   P = a/b + c/d

   Suppose we want to make a the subject.

   Worked Example

   Step 1: Subtract c/d from both sides.

   P − c/d = a/b

   Step 2: Multiply both sides by b.

   b(P − c/d) = a

   Therefore:

   a = b(P − c/d)

   Expanding gives:

   a = bP − bc/d

   Both forms are equivalent.

   Alternative Method: Multiply by a Common Denominator

   Starting with:

   P = a/b + c/d

   Multiply every term by bd:

   bdP = ad + bc

   Subtract bc:

   ad = bdP − bc

   Divide by d:

   a = bP − bc/d

   This is the same result.

   The first approach is usually shorter, but the second demonstrates how common denominators can eliminate multiple fractions simultaneously.

   Rearranging Formulas Where the Variable Appears More Than Once

   A particularly important algebraic skill is rearranging formulas in which the required variable appears in several terms.

   Consider:

   P = x/a + x/b

   We want to make x the subject.

   Because x appears twice, we cannot isolate it by simply multiplying one fraction by its denominator.

   Instead, we must collect and factor the terms containing x.

   Worked Example

   Step 1: Identify the common denominator.

   The denominators are a and b.

   A suitable common denominator is ab.

   Step 2: Multiply every term by ab.

   abP = bx + ax

   Step 3: Factor out x.

   abP = x(a + b)

   Step 4: Divide by (a + b).

   x = abP/(a + b)

   This result assumes a ≠ 0, b ≠ 0, and a + b ≠ 0.

   Numerical Application

   Suppose:

   P = 10

   a = 2

   b = 3

   Calculate x.

   Using:

   x = abP/(a + b)

   x = (2 × 3 × 10)/(2 + 3)

   x = 60/5

   x = 12

   Check:

   P = x/a + x/b

   P = 12/2 + 12/3

   P = 6 + 4

   P = 10

   The result is correct.

   Application: The Parallel Resistance Formula

   One important application of fractions occurs in electrical circuits containing resistors connected in parallel.

   For two resistors:

   1/R = 1/R₁ + 1/R₂

   where:

   • R = equivalent resistance

   • R₁ = resistance of the first resistor

   • R₂ = resistance of the second resistor

   

    

    

    

   Rearranging to Find Equivalent Resistance

   Starting with:

   1/R = 1/R₁ + 1/R₂

   Step 1: Combine the fractions on the right.

   1/R = (R₁ + R₂)/(R₁R₂)

   Step 2: Cross-multiply.

   R₁R₂ = R(R₁ + R₂)

   Step 3: Divide by (R₁ + R₂).

   R = R₁R₂/(R₁ + R₂)

   This is the familiar formula for the equivalent resistance of two resistors connected in parallel.

   For positive resistances, all the denominators in this derivation are nonzero.

   Worked Example

   Two resistors have resistances of:

   R₁ = 6 Ω

   R₂ = 3 Ω

   Calculate their equivalent resistance.

   R = (6 × 3)/(6 + 3)

   R = 18/9

   R = 2 Ω

   Notice that the equivalent resistance is smaller than either individual resistance.

   This is an important property of parallel circuits.

   Application: The Lens Formula

   In optics, the relationship between object distance, image distance, and focal length is often expressed as:

   1/f = 1/u + 1/v

   where:

   • f = focal length

   • u = object distance

   • v = image distance

   Here we use the positive-distance convention for a real object and a real image formed by a converging thin lens.

   

    

    

    

   Suppose we want to make v the subject.

   Worked Example: Rearranging the Lens Formula

   Step 1: Subtract 1/u from both sides.

   1/v = 1/f − 1/u

   Step 2: Combine the fractions.

   1/v = (u − f)/(fu)

   Step 3: Cross-multiply.

   fu = v(u − f)

   Step 4: Divide by (u − f).

   v = fu/(u − f)

   This formula is valid for nonzero f and u, with u ≠ f.

   If u = f, the rays emerge parallel and there is no finite real image distance.

   Numerical Example

   A converging lens has:

   • Focal length = 10 cm

   • Object distance = 30 cm

   Calculate the image distance.

   v = fu/(u − f)

   v = (10 × 30)/(30 − 10)

   v = 300/20

   v = 15 cm

   The real image forms 15 cm from the lens on the opposite side from the object.

   Rearranging Formulas with Algebraic Denominators

   A denominator may contain more than one term.

   For example:

   y = a/(x + b)

   Suppose we want to make x the subject.

   Worked Example

   Step 1: Multiply both sides by (x + b).

   y(x + b) = a

   Step 2: Divide both sides by y.

   x + b = a/y

   Step 3: Subtract b.

   x = a/y − b

   This rearrangement requires y ≠ 0 and the original denominator x + b ≠ 0.

   Numerical Example

   Given:

   y = 4

   a = 24

   b = 2

   Calculate x.

   x = 24/4 − 2

   x = 6 − 2

   x = 4

   Check:

   y = 24/(4 + 2)

   y = 24/6

   y = 4

   The solution is correct.

   Application: Average Speed for a Two-Part Journey

   Suppose a vehicle travels two distances at different speeds.

   The total time is:

   T = d₁/v₁ + d₂/v₂

   where:

   • T = total travel time

   • d₁ = distance travelled during the first part

   • d₂ = distance travelled during the second part

   • v₁ = speed during the first part

   • v₂ = speed during the second part

    

    

    

   Suppose we want to find d₂.

   Worked Example

   Starting with:

   T = d₁/v₁ + d₂/v₂

   Step 1: Subtract d₁/v₁ from both sides.

   T − d₁/v₁ = d₂/v₂

   Step 2: Multiply both sides by v₂.

   v₂(T − d₁/v₁) = d₂

   Therefore:

   d₂ = v₂(T − d₁/v₁)

   Numerical Application

   A vehicle travels:

   • 120 km at 60 km/h

   • A second distance at 80 km/h

   • Total journey time = 5 hours

   Calculate the second distance.

   d₂ = 80(5 − 120/60)

   d₂ = 80(5 − 2)

   d₂ = 80 × 3

   d₂ = 240 km

   The total distance travelled is:

   120 + 240 = 360 km

   The average speed for the entire journey is:

   Average speed = total distance/total time

   Average speed = 360/5

   Average speed = 72 km/h

   Notice that the average speed is not simply the arithmetic mean of 60 km/h and 80 km/h.

   The distances and times travelled at each speed must be considered.

   When the Variable Appears in Both Numerators and Denominators

   Some formulas are more complicated because the variable we want to isolate appears in several locations.

   Consider:

   P = x/(x + 2) + 1/3

   We want to solve for x.

   Worked Example

   Step 1: Subtract 1/3 from both sides.

   P − 1/3 = x/(x + 2)

   Step 2: Combine the left-hand expression.

   (3P − 1)/3 = x/(x + 2)

   Step 3: Cross-multiply.

   (3P − 1)(x + 2) = 3x

   Step 4: Expand.

   (3P − 1)x + 2(3P − 1) = 3x

   Step 5: Collect the terms containing x.

   2(3P − 1) = 3x − (3P − 1)x

   Step 6: Factor out x.

   2(3P − 1) = x[3 − (3P − 1)]

   Simplify:

   2(3P − 1) = x(4 − 3P)

   Step 7: Divide to isolate x.

   x = 2(3P − 1)/(4 − 3P)

   The original equation excludes x = −2. The rearranged expression requires P ≠ 4/3; that value cannot occur for any permitted x.

   Checking with a Numerical Value

   Suppose P = 1.

   x = 2(3 × 1 − 1)/(4 − 3 × 1)

   x = 2(2)/1

   x = 4

   Check:

   P = 4/(4 + 2) + 1/3

   P = 4/6 + 1/3

   P = 2/3 + 1/3

   P = 1

   The result is correct.

   Avoiding Undefined Expressions

   When rearranging formulas containing fractions, we must remember that division by zero is undefined.

   For example:

   y = 5/(x − 3)

   The denominator cannot equal zero.

   Therefore:

   x − 3 ≠ 0

   x ≠ 3

   This is a restriction on the possible values of x.

   Why Restrictions Matter

   Consider:

   x/x = 1

   For every nonzero value of x, this statement is true.

   However, when x = 0:

   0/0

   is undefined.

   Therefore, cancelling x from the numerator and denominator is only valid when x ≠ 0.

   Similarly, when dividing by an algebraic expression such as (a + b), we must ensure that:

   a + b ≠ 0

   Otherwise, we may divide by zero or accidentally remove possible solutions.

   Practice Questions — Part 2

   Rearrange each formula to make the indicated variable the subject.

   1. A = b/c + d. Make b the subject.

   2. P = x/3 + x/5. Make x the subject.

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

   4. T = d₁/v₁ + d₂/v₂. Make v₂ the subject.

   5. 1/R = 1/R₁ + 1/R₂. Make R₂ the subject.

   6. Solve: (x + 3)/4 + (x − 2)/6 = 5.

   7. Solve: 3x/5 − x/2 = 4.

   8. Rearrange: P = a/x + b/x. Make x the subject.

   9. Rearrange: y = (2x + 3)/(x − 1). Make x the subject.

   10. Explain why dividing both sides of an equation by an expression containing a variable may require checking for excluded values.

   Advanced Techniques for Rearranging Fractional Formulas

   In Parts 1 and 2, we explored how to eliminate fractions, identify common denominators, and rearrange formulas containing variables in numerators and denominators.

   We can now apply these skills to more complicated equations involving multiple fractions, brackets, and variables appearing more than once.

   The most important principle remains the same:

   Whatever operation we perform on one side of an equation, we must also perform on the other side.

   When rearranging complicated formulas, the goal is to simplify the equation systematically while preserving equality.

   Working with Nested Fractions

   A nested fraction, sometimes called a complex fraction, contains another fraction in its numerator or denominator.

   For example:

   y = (a/b)/c

   This can be rewritten as:

   y = a/(bc)

   because dividing by c is equivalent to multiplying by 1/c.

   Worked Example: Simplifying a Nested Fraction

   Simplify:

   x = (6/5)/(3/4)

   Step 1: Rewrite division as multiplication by the reciprocal.

   x = (6/5) × (4/3)

   Step 2: Multiply the numerators and denominators.

   x = 24/15

   Step 3: Simplify.

   x = 8/5 = 1.6

   This technique is useful when rearranging formulas that contain ratios of ratios.

   Worked Example: Rearranging a Nested Fraction

   Consider:

   P = (a/b)/(c/d)

   We want to make a the subject.

   Step 1: Simplify the nested fraction.

   P = (a/b) × (d/c)

   P = ad/(bc)

   Step 2: Multiply both sides by bc.

   Pbc = ad

   Step 3: Divide by d.

   a = Pbc/d

   This assumes b, c, and d are nonzero.

   Rearranging Formulas with Several Variables

   Sometimes the variable we want to isolate appears in several terms, making direct rearrangement difficult.

   Consider:

   A = x/a + x/b + c

   We want to make x the subject.

   Worked Example

   Step 1: Subtract c from both sides.

   A − c = x/a + x/b

   Step 2: Multiply by the common denominator ab.

   ab(A − c) = bx + ax

   Step 3: Factor out x.

   ab(A − c) = x(a + b)

   Step 4: Divide by (a + b).

   x = ab(A − c)/(a + b)

   The formula is valid when a ≠ 0, b ≠ 0, and a + b ≠ 0.

   This method illustrates an important technique:

   When the required variable appears more than once, collect the terms containing that variable and factor it out.

   Application: The Trapezium Area Formula

   The area of a trapezium is:

   A = (a + b)h/2

   where:

   • A = area

   • a = length of one parallel side

   • b = length of the other parallel side

   • h = perpendicular height

   

    

    

    

   Suppose we want to make b the subject.

   Worked Example

   Step 1: Multiply both sides by 2.

   2A = (a + b)h

   Step 2: Divide both sides by h.

   2A/h = a + b

   Step 3: Subtract a.

   b = 2A/h − a

   This rearrangement assumes h ≠ 0.

   Numerical Application

   A trapezium has:

   • Area = 96 cm²

   • One parallel side = 10 cm

   • Height = 8 cm

   Find the length of the other parallel side.

   b = 2A/h − a

   b = (2 × 96)/8 − 10

   b = 192/8 − 10

   b = 24 − 10

   b = 14 cm

   Check:

   A = (10 + 14) × 8/2

   A = 24 × 4

   A = 96 cm²

   The answer is correct.

   Application: The Kinetic Energy Formula

   Kinetic energy is the energy an object possesses because of its motion.

   The formula is:

   KE = mv²/2

   where:

   • KE = kinetic energy, measured in joules (J)

   • m = mass, measured in kilograms (kg)

   • v = speed, measured in metres per second (m/s)

   

    

    

    

   Suppose we want to make mass the subject.

   Worked Example: Finding Mass

   Step 1: Start with the formula.

   KE = mv²/2

   Step 2: Multiply both sides by 2.

   2KE = mv²

   Step 3: Divide both sides by v².

   m = 2KE/v²

   This requires v ≠ 0.

   Numerical Example

   A moving object has:

   • Kinetic energy = 400 J

   • Speed = 10 m/s

   Calculate its mass.

   m = 2KE/v²

   m = (2 × 400)/10²

   m = 800/100

   m = 8 kg

   Rearranging to Find Speed

   Starting with:

   KE = mv²/2

   Multiply by 2:

   2KE = mv²

   Divide by m:

   v² = 2KE/m

   Take the square root:

   v = √(2KE/m)

   Because v represents speed, we take the nonnegative square root.

   If the variable represented signed velocity in one dimension, the algebraic equation would permit both positive and negative roots.

   For the previous example:

   v = √(2 × 400/8)

   v = √100

   v = 10 m/s

   This example demonstrates that rearranging a fractional formula may require more than multiplication and division.

   Sometimes we must also apply powers or roots.

   Application: Gravitational Field Strength

   Gravitational field strength is related to gravitational force and mass by:

   g = F/m

   where:

   • g = gravitational field strength, measured in N/kg

   • F = gravitational force, measured in newtons (N)

   • m = mass, measured in kilograms (kg)

   

    

    

    

   Rearranging to Find Gravitational Force

   Starting with:

   g = F/m

   Multiply both sides by m:

   gm = F

   Therefore:

   F = mg

   This is the familiar equation for calculating weight.

   Numerical Application

   An astronaut has a mass of 75 kg.

   The gravitational field strength on the Moon is approximately 1.62 N/kg.

   Calculate the astronaut's weight on the Moon.

   F = mg

   F = 75 × 1.62

   F = 121.5 N

   The astronaut's mass remains 75 kg, but their weight is much smaller than on Earth because the Moon has a weaker gravitational field.

   Application: The Ideal Gas Equation

   The ideal gas equation describes the relationship between pressure, volume, temperature, and the amount of gas.

   PV = nRT

   where:

   • P = pressure, measured in pascals (Pa)

   • V = volume, measured in cubic metres (m³)

   • n = amount of gas, measured in moles (mol)

   • R = ideal gas constant, approximately 8.314 J/(mol·K)

   • T = absolute temperature, measured in kelvin (K)

   

    

    

    

   Rearranging to Find Pressure

   Starting with:

   PV = nRT

   Divide both sides by V:

   P = nRT/V

   Rearranging to Find Temperature

   Starting with:

   PV = nRT

   Divide both sides by nR:

   T = PV/(nR)

   These rearrangements assume the denominators are nonzero.

   Numerical Application

   A container contains:

   • n = 2.0 mol

   • T = 300 K

   • V = 0.050 m³

   Calculate the pressure.

   P = nRT/V

   P = (2.0 × 8.314 × 300)/0.050

   P = 4988.4/0.050

   P = 99,768 Pa

   To an appropriate number of significant figures:

   P ≈ 1.0 × 10⁵ Pa

   This is approximately atmospheric pressure at sea level.

   Application: Rearranging the Combined Gas Law

   The combined gas law relates the pressure, volume, and temperature of a fixed amount of ideal gas between two states.

   P₁V₁/T₁ = P₂V₂/T₂

   Suppose we want to make V₂ the subject.

   Worked Example

   Step 1: Multiply both sides by T₂.

   P₁V₁T₂/T₁ = P₂V₂

   Step 2: Divide both sides by P₂.

   V₂ = P₁V₁T₂/(P₂T₁)

   This formula is useful when predicting how the volume of a gas changes under different conditions.

   Numerical Example

   A gas initially has:

   • P₁ = 100 kPa

   • V₁ = 2.0 L

   • T₁ = 300 K

   Its final conditions are:

   • P₂ = 150 kPa

   • T₂ = 330 K

   Calculate V₂.

   V₂ = (100 × 2.0 × 330)/(150 × 300)

   V₂ = 66,000/45,000

   V₂ ≈ 1.47 L

   The final volume is approximately 1.47 litres.

   Although the temperature increases, the larger increase in pressure causes the volume to decrease.

   Solving Equations with Fractions and Brackets

   Now consider a more challenging algebraic equation:

   (2x + 3)/4 − (x − 1)/6 = 5

   Worked Example

   Step 1: Find the lowest common denominator.

   The denominators are 4 and 6.

   LCD = 12

   Step 2: Multiply every term by 12.

   3(2x + 3) − 2(x − 1) = 60

   Step 3: Expand the brackets.

   6x + 9 − 2x + 2 = 60

   Step 4: Collect like terms.

   4x + 11 = 60

   Step 5: Subtract 11.

   4x = 49

   Step 6: Divide by 4.

   x = 49/4 = 12.25

   Checking the Answer

   Substitute x = 12.25:

   Left side = (2 × 12.25 + 3)/4 − (12.25 − 1)/6

   = 27.5/4 − 11.25/6

   = 6.875 − 1.875

   = 5

   The answer is correct.

   Solving Equations with Variables in Both Denominators

   Consider:

   3/(x − 1) = 2/(x + 2)

   We want to solve for x.

   Worked Example

   Step 1: Identify excluded values.

   x − 1 ≠ 0, so x ≠ 1.

   x + 2 ≠ 0, so x ≠ −2.

   Step 2: Cross-multiply.

   3(x + 2) = 2(x − 1)

   Step 3: Expand.

   3x + 6 = 2x − 2

   Step 4: Collect the variables.

   x = −8

   Step 5: Check the restrictions.

   −8 is neither 1 nor −2, so the solution is permitted.

   Step 6: Verify by substitution.

   Left side = 3/(−8 − 1) = −1/3

   Right side = 2/(−8 + 2) = −1/3

   Both sides are equal.

   Therefore:

   x = −8

   This example shows why identifying excluded values is an important part of solving rational equations.

   Common Mistakes

   Multiplying Only One Term

   Incorrect:

   x/3 + 5 = 9

   Multiplying by 3 and writing:

   x + 5 = 27

   Correct:

   3(x/3) + 3(5) = 3(9)

   x + 15 = 27

   x = 12

   Every term must be multiplied by the common denominator.

   Forgetting to Expand Brackets

   Incorrect:

   4(x + 2) = 4x + 2

   Correct:

   4(x + 2) = 4x + 8

   Multiplication must apply to every term inside the brackets.

   Cancelling Terms Across Addition

   Incorrect:

   (a + b)/a = b

   Correct:

   (a + b)/a = a/a + b/a

   = 1 + b/a

   Cancellation is valid for common factors, not arbitrary terms in a sum.

   Cross-Multiplying Incorrectly

   Incorrect:

   x/3 + 2 = 5/4

   Writing:

   4x = 15

   ignores the additional term 2.

   Correct:

   x/3 = 5/4 − 2

   x/3 = −3/4

   4x = −9

   x = −9/4

   Dividing by an Expression That Could Be Zero

   Consider:

   x(x − 2) = 0

   If we divide both sides by x, we obtain:

   x − 2 = 0

   x = 2

   But this loses the solution x = 0.

   The correct approach is to use the zero-product property:

   x = 0 or x − 2 = 0

   Therefore:

   x = 0 or x = 2

   Never divide by an expression containing a variable without considering whether it could equal zero.

   Forgetting Units

   In scientific applications, a correct numerical calculation may still be incomplete without appropriate units.

   For example:

   m = ρV

   If density is measured in kg/m³ and volume in m³, then:

   m = (kg/m³) × m³

   m = kg

   Units can help us check whether a rearranged formula makes physical sense.

   Key Terms

   • Fraction: A mathematical expression representing division.

   • Numerator: The upper part of a fraction.

   • Denominator: The lower part of a fraction.

   • Common denominator: A denominator shared by two or more equivalent fractional expressions.

   • Lowest common denominator (LCD): The smallest positive common multiple of the denominators.

   • Equivalent equations: Equations that have the same solution set, subject to any stated restrictions.

   • Inverse operations: Operations that reverse each other, such as multiplication and division.

   • Subject of a formula: The variable isolated on one side of an equation.

   • Rearranging a formula: Using algebraic operations to make a different variable the subject.

   • Cross-multiplication: A technique for eliminating denominators when two fractions are equal.

   • Complex fraction: A fraction containing another fraction in its numerator or denominator.

   • Factorisation: Rewriting an expression as a product of factors.

   • Excluded value: A value that makes an original expression undefined.

   • Rational expression: A fraction whose numerator and denominator are polynomials.

   • Verification: Checking a solution by substituting it into the original equation.

   Key Takeaways

   • Fractions can often be eliminated by multiplying both sides of an equation by a suitable denominator.

   • When several denominators are present, using the lowest common denominator usually makes calculations simpler.

   • Every term in an equation must be multiplied when clearing denominators.

   • Brackets must be expanded carefully, especially when negative signs are involved.

   • When the required variable appears in several terms, collect those terms and factor out the variable.

   • Variables in denominators require special attention because division by zero is undefined.

   • Cross-multiplication is useful when two complete fractions are equal.

   • Scientific formulas can often be rearranged using the same algebraic techniques as ordinary equations.

   • Always check rearranged formulas and numerical solutions where possible.

   • Correct units and reasonable numerical results provide additional ways to verify scientific calculations.

   Check Your Understanding

   Recall and Understanding

   1. What is the purpose of multiplying an equation by a common denominator?

   2. Explain why every term must be multiplied when eliminating fractions.

   3. Find the lowest common denominator of 6, 8, and 12.

   4. Explain why division by zero is not allowed.

   5. State when cross-multiplication can be used.

   Algebraic Skills

   6. Solve:

   x/7 = 12

   7. Solve:

   3x/5 = 18

   8. Solve:

   x/4 + x/6 = 15

   9. Solve:

   (x + 2)/3 = (x − 1)/2

   10. Solve:

   (2x − 1)/5 + (x + 3)/4 = 7

   Rearranging Formulas

   11. Rearrange the formula:

   P = F/A

   Make A the subject.

   12. Rearrange:

   y = a/x + b

   Make x the subject.

   13. Rearrange:

   T = x/a + x/b

   Make x the subject.

   14. Rearrange:

   1/f = 1/u + 1/v

   Make u the subject.

   15. Rearrange:

   E = mv²/2

   Make m the subject.

   Application and Analysis

   16. A rectangular prism has a volume of 360 cm³, a length of 12 cm, and a width of 5 cm. Using V = lwh, calculate its height.

   17. Two resistors of 12 Ω and 6 Ω are connected in parallel. Calculate their equivalent resistance.

   18. A vehicle travels 150 km at 75 km/h, followed by another journey at 60 km/h. The total travel time is 5 hours. Calculate the distance travelled during the second part.

   19. A gas has a pressure of 120 kPa, volume of 3.0 L, and temperature of 300 K. Its pressure increases to 180 kPa while its temperature remains constant. Calculate its final volume.

   20. Explain why the equation x/x = 1 does not mean that x can be any real number.

   Answer key — 20 questions
   Hide answers
   1.
   Multiplying by a common denominator eliminates fractions and simplifies the equation.

   2.
   Every term must be multiplied to preserve equality.

   3.
   LCD = 24.

   4.
   Division by zero is undefined.

   5.
   When two complete fractions are equal and both denominators are nonzero.

   6.
   x = 84.

   7.
   x = 30.

   8.
   x = 36.

   9.
   x = 7.

   10.
   x = 129/13 ≈ 9.92.

   11.
   A = F/P, assuming P ≠ 0.

   12.
   x = a/(y − b), assuming y ≠ b and the original denominator is nonzero.

   13.
   x = abT/(a + b), assuming a, b, and a + b are nonzero.

   14.
   u = fv/(v − f), for the permitted nonzero denominators.

   15.
   m = 2E/v², for v ≠ 0.

   16.
   h = V/(lw) = 360/(12 × 5) = 6 cm.

   17.
   R = (12 × 6)/(12 + 6) = 4 Ω.

   18.
   First time = 150/75 = 2 h. Remaining time = 3 h. Distance = 60 × 3 = 180 km.

   19.
   V₂ = P₁V₁/P₂ = (120 × 3.0)/180 = 2.0 L.

   20.
   x = 0 makes the denominator zero, so the original expression is undefined. Therefore x ≠ 0.