Rearranging Equations and Formulas

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.