Applications of Algebraic Fractions

Algebraic fractions are not only abstract algebraic expressions. They appear whenever one quantity is divided by another quantity that can change.

Many familiar formulas are algebraic fractions:

v = d/t

ρ = m/V

p = F/A

R = V/I

These describe speed, density, pressure, and electrical resistance.

Algebraic fractions are also used in rates, work problems, geometry, finance, engineering, and many other practical situations.

The key skills are to recognize the fraction, identify restrictions, simplify or solve the mathematics, and then interpret the result in context.

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Recognizing Algebraic Fractions

An algebraic fraction contains a variable in its numerator, denominator, or both.

Examples:

x/5

3/x

(x + 4)/(x − 2)

2x/(x² − 9)

Science formulas such as:

v = d/t

are also algebraic fractions because variables appear in a fraction.

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Why Algebraic Fractions Are Useful

Fractions naturally describe relationships involving:

• rates
• ratios
• averages
• quantities per unit
• sharing
• concentration
• density
• pressure
• speed
• electrical resistance
• efficiency

For example:

average speed = distance/time

The fraction compares distance traveled with the time taken.

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Restrictions Still Matter

Whenever a variable appears in a denominator, values that make the denominator zero must be excluded.

For:

5/x

restriction:

x ≠ 0

For:

3/(x − 4)

restriction:

x ≠ 4

For:

2/[x(x + 3)]

restrictions:

x ≠ 0, −3

These restrictions can have practical meanings.

For example, in:

v = d/t

we require:

t ≠ 0

because calculating average speed over zero elapsed time would involve division by zero.

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Solving Equations Containing Fractions

Consider:

x/5 = 7

Multiply both sides by 5:

x = 35

This works because multiplication by 5 removes the denominator.

The same principle can be extended to more complicated equations.

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Solving with a Variable in the Denominator

Consider:

12/x = 3

Restriction:

x ≠ 0

Multiply both sides by x:

12 = 3x

Divide by 3:

x = 4

Check:

12/4 = 3

Correct.

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Another Example

Solve:

20/(x + 1) = 5

Restriction:

x ≠ −1

Multiply both sides by x + 1:

20 = 5(x + 1)

Divide by 5:

4 = x + 1

Therefore:

x = 3

Check:

20/(3 + 1) = 5

Correct.

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Clearing Denominators

When an equation contains several fractions, it is often easiest to multiply the entire equation by the least common denominator (LCD).

For example:

x/3 + x/6 = 9

The LCD is:

6

Multiply every term by 6:

6(x/3) + 6(x/6) = 6(9)

Simplify:

2x + x = 54

3x = 54

Therefore:

x = 18

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Multiply Every Term

When clearing denominators, the LCD must multiply every term in the equation.

For:

x/4 + 3 = 7

multiply everything by 4:

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

Therefore:

x + 12 = 28

x = 16

A common error is multiplying only the fraction by the LCD.

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Two Different Denominators

Solve:

2/x + 1/3 = 1

Restriction:

x ≠ 0

The LCD is:

3x

Multiply every term by 3x:

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

Simplify:

6 + x = 3x

Subtract x:

6 = 2x

Therefore:

x = 3

Check:

2/3 + 1/3 = 1

Correct.

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Variable Expressions in Denominators

Solve:

3/(x − 2) = 6

Restriction:

x ≠ 2

Multiply by x − 2:

3 = 6(x − 2)

Divide by 6:

1/2 = x − 2

Therefore:

x = 2.5

This does not violate the restriction.

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Fractions on Both Sides

Solve:

5/x = 2/(x − 3)

Restrictions:

x ≠ 0, 3

The LCD is:

x(x − 3)

Multiply both sides:

5(x − 3) = 2x

Expand:

5x − 15 = 2x

Subtract 2x:

3x = 15

Therefore:

x = 5

Check:

Left:

5/5 = 1

Right:

2/(5 − 3) = 1

Correct.

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Cross Multiplication

For an equation of the form:

a/b = c/d

we can write:

ad = bc

provided the denominators are nonzero.

For example:

3/x = 6/10

Cross multiply:

3(10) = 6x

30 = 6x

Therefore:

x = 5

Cross multiplication is useful when there is one fraction on each side of an equation.

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Do Not Overuse Cross Multiplication

Cross multiplication is not appropriate for:

1/x + 2/3 = 5

because one side contains a sum.

Instead, find the LCD and clear the denominators.

Cross multiplication works directly for proportions such as:

a/b = c/d

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Equations Requiring Factoring

Solve:

1/(x − 2) = 3/(x² − 4)

Factor:

x² − 4 = (x − 2)(x + 2)

Restrictions:

x ≠ 2, −2

The LCD is:

(x − 2)(x + 2)

Multiply through:

x + 2 = 3

Therefore:

x = 1

This value is allowed.

Check:

1/(1 − 2) = −1

and:

3/(1 − 4) = −1

Correct.

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Extraneous Solutions

An extraneous solution is a value produced during algebraic manipulation that does not satisfy the original equation.

This is especially important with rational equations.

Suppose a calculation produces:

x = 4

but the original denominator contains:

x − 4

Then x = 4 is not allowed.

It must be rejected.

Therefore:

Always check solutions in the original equation.

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A Reliable Equation-Solving Strategy

Use:

Restrict → Factor → LCD → Clear → Solve → Check → Interpret

Restrict

Find values that make original denominators zero.

Factor

Factor denominators where necessary.

LCD

Identify the least common denominator.

Clear

Multiply every term by the LCD.

Solve

Solve the resulting equation.

Check

Substitute into the original equation.

Interpret

Explain what the answer means in context.

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Simplifying Formulas Containing Fractions

Formulas often contain several fractions that can be combined.

Consider:

1/R = 1/R₁ + 1/R₂

This formula describes two resistors connected in parallel.

The LCD on the right is:

R₁R₂

So:

1/R = R₂/(R₁R₂) + R₁/(R₁R₂)

Combine:

1/R = (R₁ + R₂)/(R₁R₂)

Take reciprocals:

R = R₁R₂/(R₁ + R₂)

The second form is often easier to use for calculations.

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Parallel Resistance Example

Suppose:

R₁ = 6 Ω

R₂ = 3 Ω

Use:

R = R₁R₂/(R₁ + R₂)

Substitute:

R = (6)(3)/(6 + 3)

R = 18/9

Therefore:

R = 2 Ω

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

That makes physical sense for resistors connected in parallel.

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Speed as an Algebraic Fraction

Average speed is:

v = d/t

where:

• v = average speed
• d = distance
• t = elapsed time

If:

d = 240 km

t = 4 h

then:

v = 240/4

Therefore:

v = 60 km/h

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Rearranging the Speed Formula

Start:

v = d/t

Multiply by t:

vt = d

Therefore:

d = vt

To find time:

v = d/t

Multiply by t:

vt = d

Divide by v:

t = d/v

So the three useful forms are:

v = d/t

d = vt

t = d/v

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Interpreting a Speed Result

Suppose:

v = 72 km/h

This does not simply mean "72."

It means the object travels an average of:

72 kilometres for every hour

The units are essential to the interpretation.

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Density

Density is:

ρ = m/V

where:

• ρ = density
• m = mass
• V = volume

If:

m = 540 g

V = 200 cm³

then:

ρ = 540/200

Therefore:

ρ = 2.7 g/cm³

This means every cubic centimetre of the material has an average mass of 2.7 g.

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Rearranging Density

Start:

ρ = m/V

Multiply by V:

ρV = m

Therefore:

m = ρV

To find volume:

V = m/ρ

These forms allow the same algebraic relationship to answer different questions.

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Pressure

Pressure is:

p = F/A

where:

• p = pressure
• F = force
• A = area

Suppose:

F = 800 N

A = 0.20 m²

Then:

p = 800/0.20

Therefore:

p = 4000 Pa

This means the force is distributed over the surface at 4000 newtons per square metre.

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Why Smaller Area Produces Greater Pressure

From:

p = F/A

if F remains constant while A decreases, p increases.

For example:

F = 600 N

If:

A = 0.30 m²

then:

p = 2000 Pa

But if:

A = 0.10 m²

then:

p = 6000 Pa

This is why sharp objects can produce high pressures with relatively modest forces.

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Electrical Resistance

Ohm's Law can be written:

R = V/I

where:

• R = resistance
• V = voltage
• I = current

Suppose:

V = 12 V

I = 3 A

Then:

R = 12/3

Therefore:

R = 4 Ω

Restriction:

I ≠ 0 when using this form as an algebraic division.

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Power

Power can be calculated using:

P = E/t

where:

• P = power
• E = energy transferred
• t = time

If:

E = 6000 J

t = 30 s

then:

P = 6000/30

Therefore:

P = 200 W

This means energy is transferred at a rate of 200 joules per second.

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Rates in Everyday Life

Algebraic fractions commonly describe rates.

Examples include:

speed = distance/time

unit price = cost/quantity

fuel economy = distance/fuel

hourly wage = earnings/time

flow rate = volume/time

population density = population/area

These are all quantities divided by other quantities.

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Unit Price

Suppose 8 notebooks cost $20.

Unit price:

p = C/n

where:

• C = total cost
• n = number of items

Substitute:

p = 20/8

Therefore:

p = $2.50 per notebook

This is an algebraic fraction used in everyday decision-making.

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Comparing Unit Prices

Shop A sells 6 bottles for $15.

15/6 = $2.50 per bottle

Shop B sells 10 bottles for $23.

23/10 = $2.30 per bottle

Therefore Shop B has the lower unit price.

The algebraic fraction allows a fair comparison even though the package sizes are different.

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Work-Rate Problems

Algebraic fractions are useful when modeling how quickly people or machines complete tasks.

Suppose a machine completes one job in t hours.

Its work rate is:

1/t

jobs per hour.

If another machine takes s hours, its rate is:

1/s

Together:

combined rate = 1/t + 1/s

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Two Workers Example

Worker A can complete a task in 6 hours.

Rate:

1/6

task per hour.

Worker B can complete the same task in 3 hours.

Rate:

1/3

Together:

1/6 + 1/3

Common denominator 6:

1/6 + 2/6 = 3/6

= 1/2

Together they complete:

1/2 of the task per hour

Therefore the complete task takes:

2 hours

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Solving a Work-Rate Equation

Suppose one machine can complete a job in 4 hours. A second machine takes x hours.

Together they complete the job in 3 hours.

Their rates give:

1/4 + 1/x = 1/3

Restriction:

x ≠ 0

LCD:

12x

Multiply every term:

3x + 12 = 4x

Therefore:

x = 12

Interpretation:

The second machine working alone would take 12 hours.

Check:

1/4 + 1/12 = 3/12 + 1/12 = 4/12 = 1/3

Correct.

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Distance and Travel Problems

Suppose two vehicles travel the same distance but at different speeds.

Since:

t = d/v

their travel times may be represented by algebraic fractions.

For example, a journey of d kilometres at 60 km/h takes:

d/60

hours.

At 80 km/h, the same journey takes:

d/80

hours.

These expressions can be used to compare travel times.

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A Travel-Time Equation

A journey takes 1 hour longer at 60 km/h than at 80 km/h.

Let d be the distance.

Then:

d/60 − d/80 = 1

LCD:

240

Multiply through:

4d − 3d = 240

Therefore:

d = 240 km

Check:

At 60 km/h:

240/60 = 4 h

At 80 km/h:

240/80 = 3 h

Difference:

1 h

Correct.

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Average Speed Requires Care

Suppose a car travels:

• 100 km at 50 km/h
• 100 km at 100 km/h

It might be tempting to average the speeds:

(50 + 100)/2 = 75 km/h

But this is incorrect because the car spends different amounts of time at each speed.

First find the times:

t₁ = 100/50 = 2 h

t₂ = 100/100 = 1 h

Total distance:

200 km

Total time:

3 h

Average speed:

v_avg = 200/3

Therefore:

v_avg ≈ 66.7 km/h

Algebraic fractions help model the situation correctly.

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Geometry Applications

Algebraic fractions can also arise when dimensions are calculated from formulas.

Rectangle area:

A = lw

Solve for width:

w = A/l

Suppose:

A = 84 cm²

l = 12 cm

Then:

w = 84/12

Therefore:

w = 7 cm

The rearranged formula is an algebraic fraction.

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Triangle Height

Triangle area:

A = 1/2 bh

Solve for h:

2A = bh

Therefore:

h = 2A/b

If:

A = 60 cm²

b = 15 cm

then:

h = 2(60)/15

h = 8 cm

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Circle Radius

Circle area:

A = πr²

Rearrange:

r² = A/π

Therefore:

r = √(A/π)

The fraction A/π is part of the rearranged formula.

If:

A = 314 cm²

using π ≈ 3.14:

r = √(314/3.14)

r = √100

Therefore:

r = 10 cm

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Concentration

Concentration can be represented by:

c = m/V

where:

• c = concentration
• m = mass of solute
• V = volume of solution

Suppose:

m = 15 g

V = 0.50 L

Then:

c = 15/0.50

Therefore:

c = 30 g/L

This means there are 30 grams of solute per litre of solution.

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Flow Rate

Flow rate can be represented as:

Q = V/t

where:

• Q = volume flow rate
• V = volume
• t = time

Suppose 120 L of water passes through a pipe in 4 minutes.

Q = 120/4

Therefore:

Q = 30 L/min

Again, the fraction describes a rate.

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Population Density

Population density can be modeled as:

D = P/A

where:

• D = population density
• P = population
• A = land area

Suppose a region has:

P = 600,000 people

A = 300 km²

Then:

D = 600,000/300

Therefore:

D = 2000 people/km²

The units help explain the meaning of the result.

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Modeling with Algebraic Fractions

Suppose x students share a cost of $240 equally.

The cost per student is:

C = 240/x

If there are 20 students:

C = 240/20 = $12

If there are 30 students:

C = 240/30 = $8

As the number of students increases, the cost per student decreases.

This is an example of an inverse relationship.

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Inverse Relationships

Many algebraic fractions have the form:

y = k/x

where k is constant.

If x increases, y decreases.

For example:

C = 240/x

If x doubles:

C = 240/(2x)

which is half the original value.

Examples of inverse relationships include:

• cost per person when sharing a fixed bill
• time for a fixed journey as speed increases
• pressure changes when force is constant and area changes
• some work-rate relationships

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Interpreting Restrictions in Context

Consider:

C = 240/x

Mathematically:

x ≠ 0

But context gives additional information.

If x represents the number of students:

• x cannot be negative
• x usually must be a whole number
• x must be greater than zero

Therefore the practical domain is more restrictive than the purely algebraic domain.

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Mathematical Versus Practical Solutions

Suppose an equation gives:

x = 4.5

If x represents a length, 4.5 may be perfectly reasonable.

If x represents the number of buses required, 4.5 buses is not a practical answer.

The context might require rounding up to:

5 buses

Therefore solving the equation is not always the final step.

The result must be interpreted.

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Units Are Part of the Interpretation

Consider:

ρ = 2.7

This is incomplete.

A proper answer might be:

ρ = 2.7 g/cm³

Similarly:

v = 60 km/h

p = 4000 Pa

Q = 30 L/min

Units tell us what the calculated number means.

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Checking Reasonableness

After solving a practical problem, ask:

• Is the value allowed?
• Does it violate a denominator restriction?
• Are the units correct?
• Is the sign reasonable?
• Is the size of the answer realistic?
• Does it satisfy the original equation?
• Does it make sense in the real situation?

This combines algebraic verification with practical reasoning.

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A Full Modeling Example

A group rents a bus for $900. The cost is shared equally among x students.

Cost per student:

C = 900/x

Suppose each student should pay $30.

Set:

900/x = 30

Restriction:

x ≠ 0

Multiply by x:

900 = 30x

Divide by 30:

x = 30

Interpretation:

30 students are needed for the cost to be $30 per student.

Check:

900/30 = 30

Correct.

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A More Complex Cost Problem

A group pays a fixed $120 booking fee plus $15 per person.

For x people, total cost is:

120 + 15x

Cost per person is:

C = (120 + 15x)/x

Separate the fraction:

C = 120/x + 15

This form is useful for interpretation.

It shows that each person pays:

• $15 individual cost
• plus a share of the $120 fixed cost

As x increases, the shared portion:

120/x

decreases.

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Solving the Cost Model

Suppose the cost per person should be $20.

Use:

120/x + 15 = 20

Subtract 15:

120/x = 5

Multiply by x:

120 = 5x

Therefore:

x = 24

Interpretation:

24 people are needed to reduce the cost to $20 per person.

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Combining Formulas

Algebraic fractions become especially useful when formulas are substituted into one another.

Suppose:

v = d/t

and:

d = Pt/F

Substitute d into the speed formula:

v = (Pt/F)/t

Rewrite:

v = Pt/(Ft)

Cancel t:

v = P/F

provided the relevant quantities are nonzero.

This shows how complicated formulas can sometimes simplify dramatically.

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Formula Simplification

Consider:

A = (xy/xz)

Restrictions:

x ≠ 0

z ≠ 0

Cancel x:

A = y/z

The simplified formula may be easier to interpret and calculate with.

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Comparing Two Rates

Suppose Machine A produces x items in 4 hours.

Rate:

x/4

Machine B produces x items in 6 hours.

Rate:

x/6

Combined rate:

x/4 + x/6

LCD:

12

Rewrite:

3x/12 + 2x/12

Therefore:

5x/12

items per hour.

This demonstrates how adding algebraic fractions can model combined rates.

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A Multi-Step Equation

Solve:

2/(x − 1) + 1/(x + 1) = 1

Restrictions:

x ≠ 1, −1

LCD:

(x − 1)(x + 1)

Multiply through:

2(x + 1) + (x − 1) = (x − 1)(x + 1)

Simplify the left:

2x + 2 + x − 1 = x² − 1

3x + 1 = x² − 1

Move everything to one side:

x² − 3x − 2 = 0

This quadratic does not factor neatly using integers.

Use the quadratic formula:

x = [3 ± √17]/2

Neither value equals ±1, so both are permitted.

This shows that rational equations can lead to other types of equations.

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Graphical Meaning

An equation such as:

1/x = 2/(x + 3)

can also be interpreted as asking:

At which x-value do the graphs y = 1/x and y = 2/(x + 3) have the same y-value?

Solve algebraically:

x + 3 = 2x

Therefore:

x = 3

At x = 3, the two expressions are equal.

This connects rational equations with graph intersections.

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When There Is No Valid Solution

Suppose algebraic manipulation produces a value that makes an original denominator zero.

That value must be rejected.

For example, if the original expression contains:

1/(x − 5)

then:

x = 5

can never be a valid solution.

A rational equation can therefore produce:

• one valid solution
• several valid solutions
• no valid solutions

depending on the equation.

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Common Mistakes

Mistake: Ignoring restrictions.

Always identify values that make original denominators zero.

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Mistake: Multiplying only some terms by the LCD.

When clearing denominators, multiply every term.

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Mistake: Using cross multiplication when the equation is not a simple proportion.

Cross multiplication works directly for:

a/b = c/d

not for arbitrary sums of fractions.

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Mistake: Failing to factor denominators first.

Factoring can reveal the correct LCD.

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Mistake: Accepting an extraneous solution.

Check solutions in the original equation.

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Mistake: Giving only a number in a practical problem.

Include units and explain what the number represents.

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Mistake: Ignoring practical restrictions.

A negative time or fractional number of people may not make sense even if the algebra permits it.

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Mistake: Averaging rates incorrectly.

Average speed should generally be:

total distance/total time

not simply the arithmetic mean of different speeds.

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Mistake: Treating algebraic fractions as unrelated to formulas.

Many common science formulas are algebraic fractions.

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Mistake: Stopping before simplifying a formula.

Factoring and cancelling common factors may reveal a much simpler relationship.

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Key Terms

Algebraic fraction — fraction containing variables or algebraic expressions.

Rational equation — equation containing one or more algebraic fractions.

Restriction — value excluded because it makes an expression undefined.

Least common denominator (LCD) — smallest expression containing all required denominator factors.

Clear the denominator — multiply an equation by an appropriate common denominator to remove fractions.

Cross multiplication — method for solving a proportion by multiplying opposite numerator-denominator pairs.

Extraneous solution — result obtained during solving that does not satisfy the original equation.

Rate — comparison of two quantities measured in different units.

Ratio — comparison of two quantities by division.

Model — mathematical representation of a real situation.

Average speed — total distance divided by total elapsed time.

Density — mass per unit volume.

Pressure — force per unit area.

Resistance — ratio of voltage to current.

Flow rate — volume passing a point per unit time.

Unit rate — rate expressed for one unit of another quantity.

Inverse relationship — relationship in which one quantity decreases as another increases according to a reciprocal-type relationship.

Practical domain — values that make sense within the real-world situation being modeled.

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Key Takeaways

• Algebraic fractions occur naturally whenever one variable quantity is divided by another.
• Many familiar science formulas are algebraic fractions.
• Examples include v = d/t, ρ = m/V, p = F/A, and R = V/I.
• Restrictions must be identified before solving rational equations.
• Denominators cannot equal zero.
• Rational equations can often be solved by multiplying every term by the LCD.
• Factor denominators before determining the LCD.
• Cross multiplication is useful for equations consisting of one fraction equal to another fraction.
• Cross multiplication should not be used blindly with sums or differences.
• Solutions should always be checked in the original equation.
• Values that make an original denominator zero must be rejected.
• Algebraic fractions can model speed, density, pressure, resistance, work rates, unit prices, travel times, concentration, and many other quantities.
• Work-rate problems often use fractions such as 1/t.
• Travel-time problems frequently use t = d/v.
• Fixed costs shared among people often produce models such as C = k/x.
• Algebraic fractions commonly represent inverse relationships.
• Formulas containing fractions can often be simplified by factoring and cancelling.
• Combining formulas may create algebraic fractions that can then be simplified.
• Units are an essential part of interpreting practical answers.
• The mathematical domain and practical domain may be different.
• A mathematically possible solution may still be unrealistic in context.
• A complete solution should include the algebra, restrictions, verification, units, and interpretation.

Check Your Understanding

1. Solve x/5 = 8.
2. Solve 18/x = 6 and state the restriction.
3. Solve 20/(x + 2) = 4 and check your answer.
4. Solve x/3 + x/6 = 12.
5. Solve 2/x + 1/4 = 1.
6. Solve 3/x = 6/(x + 4) and state all restrictions.
7. Explain when cross multiplication can be used.
8. Explain why solutions to rational equations should be checked in the original equation.
9. Rearrange v = d/t to make t the subject.
10. A car travels 360 km in 4.5 hours. Calculate its average speed and interpret your result.
11. A substance has mass 675 g and volume 250 cm³. Calculate its density.
12. A force of 900 N acts over an area of 0.30 m². Calculate the pressure.
13. A device has a voltage of 24 V and current of 3 A. Calculate its resistance.
14. Two resistors have resistances of 4 Ω and 12 Ω. Use R = R₁R₂/(R₁ + R₂) to calculate their equivalent parallel resistance.
15. A machine completes a job in 5 hours. Write an algebraic fraction representing its work rate.
16. One worker completes a task in 4 hours and another in 12 hours. Determine how long they take working together.
17. A group shares a $600 cost equally. Write a formula for the cost per person when x people participate. How many people are needed for the cost to be $25 each?
18. A 180 km journey is made at speed v. Write an algebraic expression for the travel time. State the restriction and explain its practical meaning.
19. A journey takes 2 hours longer at 40 km/h than at 60 km/h. Create and solve an algebraic-fraction equation to determine the distance.
20. Describe the complete Restrict → Factor → LCD → Clear → Solve → Check → Interpret process for solving and interpreting a practical problem involving algebraic fractions.