Rearranging Equations and Formulas

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