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Challenge

Egyptian Fractions

Ramsha is studying Egyptian history for class.She learned that ancient Egyptians had an interesting way to represent fractions. They used unit fractions, which are fractions of the form $\frac{1}{N},$ to represent all other fractions. The examples of the Egyptian fractions was found in the image of the Eye of Horus. Each part of the eye represents a different fraction.

$Eye of Horus and Egyptian fractions$

External credits: Benoît Stella

The Egyptians were able to write any fraction as a sum of different unit fractions.

\frac{3}{4} = \frac{1}{2} + \frac{1}{4}

However, the Egyptians did not use a given unit fraction more than once, so they would not have written

\frac{3}{4} = \frac{1}{4} + \frac{1}{4} + \frac{1}{4} .

a How might the Egyptians have expressed

\frac{2}{7} ?

b Find the sum of the rational expressions for any

x > 0 .

\frac{1}{x + 1} + \frac{1}{x ( x + 1 )}

c Use the result from Part B to show that each unit fraction

\frac{1}{N},

with

N \geq 2,

can be written as a sum of two or more different unit fractions.

d Describe a procedure for writing any positive rational number,

\frac{p}{q}

with

p > 0

and

q > 2,

as an Egyptian fraction.

Discussion

Least Common Multiple of Polynomials and How to Find It

Concept

Least Common Multiple

The least common multiple (LCM) of two whole numbers $a$ and $b$ is the smallest whole number that is a multiple of both $a$ and $b .$ It is denoted as $LCM (a, b) .$ The least common multiple of $a$ and $b$ is the smallest whole number that is divisible by both $a$ and $b .$ Some examples can be seen in the table below.

Numbers	Multiples of Numbers	Common Multiples	Least Common Multiple
$2$ and $3$	$Multiples of 2 : Multiples of 3 : 2, 4, 6, 8, 10, 12, \dots 3, 6, 9, 12, 15, \dots$	$\underline{6}, 12, 18, 24, \dots$	$LCM (2, 3) = 6$
$8$ and $12$	$Multiples of 8 : Multiples of 12 : 8, 16, 24, 32, 40, 48, \dots 12, 24, 36, 48, \dots$	$\underline{24}, 48, 72, 96, \dots$	$LCM (8, 12) = 24$

A special procedure exists for finding the $LCM (a, b)$ of a pair of numeric expressions. The LCM of two relatively prime numbers is always equal to their product.

Coprimes	LCM
$3$ and $5$	$15$
$5$ and $4$	$20$
$4$ and $9$	$36$

Polynomials can also have a least common multiple. The LCM of two or more polynomials is the smallest multiple of both polynomials. In other words, the LCM is the smallest expression that can be evenly divided by each of the given polynomials.

Polynomials	Factor	LCM	Explanation
$4 x^{3} and 6 x y^{2}$	$2^{2} \cdot x^{3} and 2 \cdot 3 \cdot x \cdot y^{2}$	$12 x^{3} y^{2}$	$12 x^{3} y^{2}$ is the smallest expression that is divisible by both $4 x^{3}$ and $6 x y^{2} .$
$3 x^{3} + 3 x^{2} and x^{2} + 5 x + 4$	$3 \cdot x^{2} \cdot (x + 1) and (x + 1) (x + 4)$	$3 x^{2} (x + 1) (x + 4)$	$3 x^{2} (x + 1) (x + 4)$ is the smallest expression that is divisible by both $3 x^{3} + 3 x^{2}$ and $x^{2} + 5 x + 4 .$

Finding the LCM of polynomials requires identifying the factors with the highest power that appear in each polynomial.

Method

Finding the Least Common Multiple of Polynomials

To find the least common multiple (LCM) of two or more polynomials, the polynomials must be factored completely. The LCM is the product of the factors with the highest power that appear in any of the polynomials. To show an example, the LCM of the following polynomials will be found.

Polynomial I : Polynomial II : 12 x^{2} y + 48 x y + 48 y 3 x^{3} y - 18 x^{2} y - 48 x y

The procedure of finding the LCM of the polynomials involves three steps.

Factor Each Polynomial and Write Numerical Factors as Product of Primes

Each polynomial will be factored. Start by factoring the first polynomial. Note that

12 y

is used in all of its terms. Therefore, it will be factored out.

12 x^{2} y + 48 x y + 48 y

SplitIntoFactors

Split into factors

12 y \cdot x^{2} + 12 y \cdot 4 x + 12 y \cdot 4

FactorOut

Factor out $12 y$

12 y (x^{2} + 4 x + 4)

FacPosPerfectSquare

$a^{2} + 2 a b + b^{2} = (a + b)^{2}$

12 y (x + 2)^{2}

Then the numerical factor can be written as a product of

2^{2}

and

3 .

Polynomial I 12 x^{2} y + 48 x y + 48 y ⇕ 2^{2} \cdot 3 \cdot y (x + 2)^{2}

Now the other polynomial will be factored. Since the factor

3 x y

can be seen in each of its terms, start by factoring it out.

3 x^{3} y - 18 x^{2} y - 48 x y

SplitIntoFactors

Split into factors

3 x y \cdot x^{2} - 3 x y \cdot 6 x - 3 x y \cdot 16

FactorOut

Factor out $3 x y$

3 x y (x^{2} - 6 x - 16)

Factor

(x^{2} - 6 x - 16)

WriteSum

Write as a sum

3 x y (x^{2} - 8 x + 2 x - 16)

FactorOut

Factor out $x$

3 x y (x (x - 8) + 2 x - 16)

FactorOut

Factor out $2$

3 x y (x (x - 8) + 2 (x - 8))

FactorOut

Factor out $(x - 8)$

3 x y (x + 2) (x - 8)

Identify the Factors With the Highest Power

Both polynomials are now written in factored form. Now, write all the missing related factors to identify the factor with the highest power.

	Standard Form	Factored Form	All Related Factors
Polynomail I	$12 x^{2} y + 48 x y + 48 y$	$2^{2} \cdot 3 \cdot y \cdot (x + 2)^{2}$	$2^{2} \cdot 3^{1} \cdot x^{0} \cdot y^{1} \cdot (x + 2)^{2} \cdot (x - 8)^{1}$
Polynomail II	$3 x^{3} y - 18 x^{2} y - 48 x y$	$3 \cdot x \cdot y \cdot (x + 2) \cdot (x - 8)$	$2^{0} \cdot 3^{1} \cdot x^{1} \cdot y^{1} \cdot (x + 2)^{1} \cdot (x - 8)^{1}$

The highest power of each prime factor can be listed as follows.

2^{2}, 3^{1}, x^{1}, y^{1}, (x + 2)^{2}, and (x - 8)^{1}

Multiply the Highest Power of the Factors

The LCM of the polynomials is the product of the factors listed in the previous step.

2^{2} \cdot 3^{1} \cdot x^{1} \cdot y^{1} \cdot (x + 2)^{2} \cdot (x - 8)^{1}

CalcPow

Calculate power

4 \cdot 3 \cdot x \cdot y \cdot (x + 2)^{2} \cdot (x - 8)

Multiply

12 x y (x + 2)^{2} (x - 8)

As shown, the least common multiple of the given polynomials is another polynomial.

Example

Writing an Expression for the Meeting Time

After reading some interesting facts about the Egyptians, Ramsha dreams of cycling around the Great Pyramid of Giza with her friend Heichi.

Suppose that the time it takes for each of them to complete a lap are represented by the following polynomials.

Heichi : Ramsha : 6 x^{2} y - 6 x 3 x^{3} y^{2} - 6 x^{2} y + 3 x

If they start at the same place and at the same time and go in the same direction, write an expression for the time when they will meet again at the starting point.

Hint

What is the least common multiple of the given polynomials?

Solution

Imagine that a pair of numbers was given instead of a pair of polynomials. For example, suppose Heichi completes a lap in

2

minutes and Ramsha completes a lap in

3

minutes. Then, they would meet at the starting point after

6

minutes, which is the least common multiple of

2

and

3 .

LCM (2, 3) = 6

Following this reasoning, the least common multiple of the given pair of polynomials needs to be determined.

Heichi : Ramsha : 6 x^{2} y - 6 x 3 x^{3} y^{2} - 6 x^{2} y + 3 x

To do so, each expression will be factored completely. Start by factoring the first polynomial. Note that all its terms contain a common factor,

6 x .

6 x^{2} y - 6 x

SplitIntoFactors

Split into factors

6 x \cdot x y + 6 x \cdot (- 1)

FactorOut

Factor out $6 x$

6 x (x y - 1)

SplitIntoPrimeFactors

Split into prime factors

2 \cdot 3 \cdot x (x y - 1)

Now the other polynomial will be factored. The factor

3 x

can be seen in each of its terms, so start by factoring it out.

3 x^{3} y^{2} - 6 x^{2} y + 3 x

SplitIntoFactors

Split into factors

3 x \cdot x^{2} y^{2} + 3 x \cdot (- 2 x y) + 3 x \cdot 1

FactorOut

Factor out $3 x$

3 x (x^{2} y^{2} - 2 x y + 1)

Factor

(x^{2} y^{2} - 2 x y + 1)

ProdPow

$a^{m} \cdot b^{m} = (a \cdot b)^{m}$

3 x ((x y)^{2} - 2 x y + 1)

SplitIntoFactors

Split into factors

3 x ((x y)^{2} - 2 (x y) 1 + 1)

WritePow

Write as a power

3 x ((x y)^{2} - 2 (x y) 1 + 1^{2})

FacNegPerfectSquare

$a^{2} - 2 a b + b^{2} = (a - b)^{2}$

3 x (x y - 1)^{2}

Both polynomials are now factored and the numerical factors are written as a product of prime factors.

	Given Form	Factored Form	All Related Factors
Heichi	$6 x^{2} y - 6 x$	$2 \cdot 3 \cdot x \cdot (x y - 1)$	$2^{1} \cdot 3^{1} \cdot x^{1} \cdot (x y - 1)^{1}$
Ramsha	$3 x^{3} y^{2} - 6 x^{2} y + 3 x$	$3 \cdot x \cdot (x y - 1)^{2}$	$2^{0} \cdot 3^{1} \cdot x^{1} \cdot (x y - 1)^{2}$

The LCM of these polynomials is the product of the highest power of each factor.

2^{1} \cdot 3^{1} \cdot x^{1} \cdot (x y - 1)^{2}

Multiply

6 x (x y - 1)^{2}

This expression represents the time when Heichi and Ramsha meet again at the starting point.

Discussion

Using the Least Common Denominator to Add or Subtract Rational Expressions

Method

Adding and Subtracting Rational Expressions

When adding and subtracting rational expressions, the same rules apply as when adding and subtracting fractions.

Adding and Subtracting With Like Denominators

If the rational expressions have a common denominator, the numerators can be added or subtracted directly.

$\frac{P ( x )}{Q ( x )} \pm \frac{R ( x )}{Q ( x )} = \frac{P ( x ) \pm R ( x )}{Q ( x )}$

Here, $P (x),$ $Q (x),$ and $R (x)$ are polynomials and $Q (x) \neq = 0 .$

Adding and Subtracting With Unlike Denominators

If the denominators are different, the expressions have to be manipulated to find a common denominator before they can be added or subtracted. One way of doing this is to multiply both numerator and denominator of one of the rational expressions with the denominator of the other, and vice versa.

$\frac{P ( x )}{Q ( x )} \pm \frac{R ( x )}{S ( x )} = \frac{P ( x ) S ( x ) \pm R ( x ) Q ( x )}{Q ( x ) S ( x )}$

Here,

P (x),

Q (x),

R (x),

and

S (x)

are polynomials,

Q (x) \neq = 0,

and

S (x) \neq = 0 .

Another way is to find the least common multiple of the denominators, or the least common denominator. Consider adding the following rational expressions to put theory into practice.

\frac{x + 2}{2 x - 2} + \frac{2 x + 1}{x ^{2} - 4 x + 3}

The result can be found in four steps.

Find the Least Common Denominator

Recall that the least common denominator is the least common multiple of the expressions in the denominators.

\frac{x + 2}{2 x - 2} + \frac{2 x + 1}{x ^{2} - 4 x + 3}

To find the LCM of the polynomials, they need to be factored.

Denominator	Factored Form
$2 x - 2$	$2 (x - 1)$
$x^{2} - 4 x + 3$	$(x - 1) (x - 3)$

The least common denominator is the product of the highest power of each prime factor.

LCD : 2 (x - 1) (x - 3)

Rewrite Each Expression with the LCD

Now the rational expressions will be multiplied by the appropriate factors to obtain the LCD. To do so, expand the first fraction by

(x - 3)

and the second fraction by

2 .

\frac{x + 2}{2 ( x - 1 )} + \frac{2 x + 1}{( x - 1 ) ( x - 3 )}

Expand by

(x - 3)

ExpandFrac

$\frac{a}{b} = \frac{a \cdot ( x - 3 )}{b \cdot ( x - 3 )}$

\frac{( x + 2 ) \cdot ( x - 3 )}{2 ( x - 1 ) \cdot ( x - 3 )} + \frac{2 x + 1}{( x - 1 ) ( x - 3 )}

MultPar

Multiply parentheses

\frac{x ^{2} - 3 x + 2 x - 6}{2 ( x - 1 ) \cdot ( x - 3 )} + \frac{2 x + 1}{( x - 1 ) ( x - 3 )}

AddTerms

Add terms

\frac{x ^{2} - x - 6}{2 ( x - 1 ) \cdot ( x - 3 )} + \frac{2 x + 1}{( x - 1 ) ( x - 3 )}

Expand by

2

ExpandFrac

$\frac{a}{b} = \frac{a \cdot 2}{b \cdot 2}$

\frac{x ^{2} - x - 6}{2 ( x - 1 ) \cdot ( x - 3 )} + \frac{( 2 x + 1 ) \cdot 2}{( x - 1 ) ( x - 3 ) \cdot 2}

Distr

Distribute $2$

\frac{x ^{2} - x - 6}{2 ( x - 1 ) \cdot ( x - 3 )} + \frac{4 x + 2}{( x - 1 ) ( x - 3 ) \cdot 2}

Multiply

\frac{x ^{2} - x - 6}{2 ( x - 1 ) ( x - 3 )} + \frac{4 x + 2}{2 ( x - 1 ) ( x - 3 )}

Add the Numerators

The expressions in the numerators can now be added because both fractions have the same denominator.

\frac{x ^{2} - x - 6}{2 ( x - 1 ) ( x - 3 )} + \frac{4 x + 2}{2 ( x - 1 ) ( x - 3 )}

AddFrac

Add fractions

\frac{x ^{2} - x - 6 + 4 x + 2}{2 ( x - 1 ) ( x - 3 )}

AddTerms

Add terms

\frac{x ^{2} + 3 x - 4}{2 ( x - 1 ) ( x - 3 )}

Simplify the Resulting Expression

Check if the resulting rational expression can be simplified or not.

\frac{x ^{2} + 3 x - 4}{2 ( x - 1 ) ( x - 3 )}

Factor the numerator

WriteDiff

Write as a difference

\frac{x ^{2} + 4 x - x - 4}{2 ( x - 1 ) ( x - 3 )}

FactorOut

Factor out $x$

\frac{x ( x + 4 ) - x - 4}{2 ( x - 1 ) ( x - 3 )}

FactorOut

Factor out $- 1$

\frac{x ( x + 4 ) - 1 ( x + 4 )}{2 ( x - 1 ) ( x - 3 )}

FactorOut

Factor out $(x + 4)$

\frac{( x - 1 ) ( x + 4 )}{2 ( x - 1 ) ( x - 3 )}

Simplify

CancelCommonFac

Cancel out common factors

\frac{( x - 1 ) ( x + 4 )}{2 ( x - 1 ) ( x - 3 )}

SimpQuot

Simplify quotient

\frac{x + 4}{2 ( x - 3 )}

The result is in simplest form. Note that the excluded values for the sum are

1

and

3 .

\frac{x + 4}{2 ( x - 3 )}, x \neq = 1

The restrictions on the domain of a sum or difference of rational expressions consist of the restrictions to the domains of each expression.

Note that the process for subtracting rational expressions is similar.

Example

Acidity of the Mouth

Ramsha is very enthusiastic about learning more about Egypt. She finds an Egyptian pen pal, Izabella. Izabella mentions that her favorite dessert is basbousa.

The talk of dessert reminds Ramsha of something she learned in chemistry class. The pH, or acidity, of a person's mouth changes after eating a dessert. The pH level

L

of a mouth

t

minutes after eating a dessert is modeled by the following formula.

L = 6 - \frac{2 0 . 4 t}{t ^{2} + 3 6}

a Simplify the formula.

b What would the acidity of a person's mouth be after

15

minutes? Round the answer to the two decimal place.

Hint

a Rewrite

6

as a fraction and then expand it to have the same denominator as the other fraction.

b Substitute

15

for

t

in the formula from Part A.

Solution

a It is given that the acidity

L

of a person's mouth

t

minutes after eating a dessert can be determined by the following formula.

L = 6 - \frac{2 0 . 4 t}{t ^{2} + 3 6}

To simplify the formula,

6

will be rewritten as a fraction whose denominator is

1 .

L = \frac{6}{1} - \frac{2 0 . 4 t}{t ^{2} + 3 6}

In order to subtract fractions, they need to have a common denominator. The least common multiple of

1

and

t^{2} + 36

t^{2} + 36,

so expand the first fraction by this expression.

L = \frac{6 ( t ^{2} + 3 6 )}{t ^{2} + 3 6} - \frac{2 0 . 4 t}{t ^{2} + 3 6}

Now it is possible to subtract the fractions and simplify the expressions.

L = \frac{6 ( t ^{2} + 3 6 )}{t ^{2} + 3 6} - \frac{2 0 . 4 t}{t ^{2} + 3 6}

Distr

Distribute $6$

L = \frac{6 t ^{2} + 2 1 6}{t ^{2} + 3 6} - \frac{2 0 . 4 t}{t ^{2} + 3 6}

SubFrac

Subtract fractions

L = \frac{6 t ^{2} + 2 1 6 - 2 0 . 4 t}{t ^{2} + 3 6}

CommutativePropAdd

\CommutativePropAdd

L = \frac{6 t ^{2} - 2 0 . 4 t + 2 1 6}{t ^{2} + 3 6}

This is the simplification of the given formula.

b To find the acidity of a person's mouth

15

minutes after they eat a dessert, substitute

15

for

t

in the simplified formula from Part A.

L = \frac{6 t ^{2} - 2 0 . 4 t + 2 1 6}{t ^{2} + 3 6}

Substitute

$t = 15$

L = \frac{6 ( 1 5 ) ^{2} - 2 0 . 4 ( 1 5 ) + 2 1 6}{1 5 ^{2} + 3 6}

CalcPow

Calculate power

L = \frac{6 ( 2 2 5 ) - 2 0 . 4 ( 1 5 ) + 2 1 6}{2 2 5 + 3 6}

Multiply

L = \frac{1 3 5 0 - 3 0 6 + 2 1 6}{2 2 5 + 3 6}

AddSubTerms

Add and subtract terms

L = \frac{1 2 6 0}{2 6 1}

UseCalc

Use a calculator

L = 4.827586 \dots

RoundDec

Round to $2$ decimal place(s)

L \approx 4.83

The pH of the person's mouth

15

minutes after eating a dessert would be about

4.83 .

Example

Finding the Time It Takes for a Pipe to Fill a Pool

In one of her emails, Izabella tells Ramsha about her father's job. He is in charge of pool maintenance work at a local hotel.

Her father gives Isabella the following set of information.

Three pipes take $6$ hours to fill the pool in the hotel.
Pipe II takes $2$ times as long to fill the pool as Pipe I.
Pipe III takes $1.5$ times as long to fill the pool as Pipe II.

Izabella is asked to find how long it takes to fill the pool when only the fastest pipe is open. Izabella and Ramsha bounce ideas off each other but they are unable to come up with a reasonable answer. Help them to find the answer.

Hint

Let $t$ be the time it takes for the first pipe to fill the pool. Then, the fraction $\frac{1}{t}$ represents the part of the pool that the first pipe fills in one hour.

Solution

Let $t$ be the time it takes for Pipe I to fill the pool. Since Pipe II takes 2 times as long to fill the pool as Pipe I, the time needed for Pipe II to fill the pool will be $2$ times $t .$ Similarly, the time needed for Pipe III will be $1.5$ times $2 t$ because Pipe III takes $1.5$ times as long to fill the pool as Pipe II.

Pipe	Time Needed to Fill the Pool (hours)
Pipe I	$t$
Pipe II	$2 \cdot t = 2 t$
Pipe III	$1.5 \cdot 2 t = 3 t$
All of Them	$6$

Now, think about how much of the pool is filled by each pipe, alone or together, in one hour. For example, it takes all three pipes $6$ hours to fill the pool. Therefore, they would fill $\frac{1}{6}$ of the pool in one hour. The filling rates of the other pipes can be determined by this same logic.

Pipe	Time Needed to Fill the Pool (hours)	Filling Rate (per hour)
Pipe I	$t$	$\frac{1}{t}$
Pipe II	$2 t$	$\frac{1}{2 t}$
Pipe III	$3 t$	$\frac{1}{3 t}$
All of Them	$6$	$\frac{1}{6}$

The sum of the filling rates of all pipes is equal to the filling rate when all of the pipes are open.

\frac{1}{t} + \frac{1}{2 t} + \frac{1}{3 t} = \frac{1}{6}

To add the rational expressions, all the denominators must be factored.

Denominator	Factored Form
$t$	$t$
$2 t$	$2 \cdot t$
$3 t$	$3 \cdot t$
$6$	$2 \cdot 3$

The least common denominator (LCD) is

2 \cdot 3 \cdot t,

6 t .

Now, expand each fraction such that all the denominators are equal to the LCD.

\frac{1}{t} + \frac{1}{2 t} + \frac{1}{3 t} = \frac{1}{6}

Expand by

LCD

ExpandFrac

$\frac{a}{b} = \frac{a \cdot 6}{b \cdot 6}$

\frac{1}{t} \cdot \frac{6}{6} + \frac{1}{2 t} + \frac{1}{3 t} = \frac{1}{6}

ExpandFrac

$\frac{a}{b} = \frac{a \cdot 3}{b \cdot 3}$

\frac{1}{t} \cdot \frac{6}{6} + \frac{1}{2 t} \cdot \frac{3}{3} + \frac{1}{3 t} = \frac{1}{6}

ExpandFrac

$\frac{a}{b} = \frac{a \cdot 2}{b \cdot 2}$

\frac{1}{t} \cdot \frac{6}{6} + \frac{1}{2 t} \cdot \frac{3}{3} + \frac{1}{3 t} \cdot \frac{2}{2} = \frac{1}{6}

ExpandFrac

$\frac{a}{b} = \frac{a \cdot t}{b \cdot t}$

\frac{1}{t} \cdot \frac{6}{6} + \frac{1}{2 t} \cdot \frac{3}{3} + \frac{1}{3 t} \cdot \frac{2}{2} = \frac{1}{6} \cdot \frac{t}{t}

MultFrac

Multiply fractions

\frac{6}{6 t} + \frac{3}{6 t} + \frac{2}{6 t} = \frac{t}{6 t}

AddFrac

Add fractions

\frac{1 1}{6 t} = \frac{t}{6 t}

Since the denominators on the both sides are the same, the numerators can be set equal. Therefore,

t = 11 .

This means that it would take

11

hours for the fastest pipe to fill the pool.

Example

Finding the Total Time of Izabella's Trip

Izabella is planning a trip that involves a $90 -$ kilometer bus ride and a high-speed train ride. The entire trip is $300$ kilometers.

External credits: Eric Gaba and NordNordWest

The average speed of the high-speed train is $30$ kilometers per hour more than twice the average speed of the bus.

a In terms of the average speed of the bus, write an expression for the amount of time

t_{1}

Izabella is on the bus and an expression for the amount of time

t_{2}

that she rides the train.

b Write an expression for the total duration

T

of trip by using the expressions found in Part A.

Hint

a Distance traveled is equal to the product of speed and time,

d = r \cdot t .

Let

x

be the average speed of the bus. Write

t_{1}

in terms of

x .

b Find the sum of the expressions from Part A.

Solution

a Recall that distance traveled is equal to the product of speed and time,

d = r \cdot t .

Using this formula, expressions for the lengths of time

t_{1}

and

t_{2}

that Izabella spends on the bus and train, respectively, can be written. Start by solving the formula for

t .

d = r \cdot t \Rightarrow t = \frac{d}{r}

It is given that Izabella travels

90

kilometers by bus. If the average speed of the bus is

x,

then the ratio of

90

x

will give the time

t_{1}

she spent traveling by bus.

t_{1} = \frac{9 0}{x}

The remaining

300 - 90 = 210

kilometers of the trip is traveled by train. Since the average speed of the high-speed train is

30

kilometers per hours higher than twice the average speed of the bus, the speed of the train is

2 x + 30 .

With this information, an expression for the length of time that Izabella travels on the train

t_{2}

can be written.

t_{2} = \frac{2 1 0}{2 x + 3 0} \Leftrightarrow t_{2} = \frac{1 0 5}{x + 1 5}

b The total duration of the trip is the sum of the rational expressions written in Part A.

T = t_{1} + t_{2} \Rightarrow T = \frac{9 0}{x} + \frac{1 0 5}{x + 1 5}

It appears that the denominators have no common factors. Therefore, it is convenient to multiply both the numerator and denominator of one rational expression by the denominator of the other to obtain a common denominator.

T = \frac{9 0}{x} + \frac{1 0 5}{x + 1 5}

Expand by

(x + 15)

ExpandFrac

$\frac{a}{b} = \frac{a \cdot ( x + 1 5 )}{b \cdot ( x + 1 5 )}$

T = \frac{9 0}{x} \cdot \frac{( x + 1 5 )}{( x + 1 5 )} + \frac{1 0 5}{x + 1 5}

MultFrac

Multiply fractions

T = \frac{9 0 x + 1 3 5 0}{x ^{2} + 1 5 x} + \frac{1 0 5}{x + 1 5}

Expand by

x

ExpandFrac

$\frac{a}{b} = \frac{a \cdot x}{b \cdot x}$

T = \frac{9 0 x + 1 3 5 0}{x ^{2} + 1 5 x} + \frac{1 0 5}{x + 1 5} \cdot \frac{x}{x}

MultFrac

Multiply fractions

T = \frac{9 0 x + 1 3 5 0}{x ^{2} + 1 5 x} + \frac{1 0 5 x}{x ^{2} + 1 5 x}

AddFrac

Add fractions

T = \frac{1 9 5 x + 1 3 5 0}{x ^{2} + 1 5 x}

This expression represents the total length of time of the trip, written in terms of

x,

the speed of the bus.

Example

Finding Monthly Payment

Ramsha shares what she has learned about Egypt with her mother. She sees how enthusiastic Ramsha is about life in Egypt and considers taking out a loan for the family to take a trip to Egypt.

If she borrows

P

dollars and agrees to pay back it over

t

years at a monthly interest rate of

r,

her monthly payment is calculated by the following formula.

M = \frac{P r}{1 - ( \frac{1}{1 + r} ) ^{12 t}}

a Simplify the formula.

b If Ramsha's mother borrows

$ 1500

at a monthly interest rate of

0.5 %

and pays it back over

2

years, find her monthly payment. Round the answer to two decimal place.

Hint

a Start by using the Power of a Quotient Property,

(\frac{a}{b})^{n} = \frac{a ^{n}}{b ^{n}} .

b Substitute the values into the simplified formula.

Solution

a To simplify the formula, the Power of a Quotient Property will be used first.

M = \frac{P r}{1 - ( \frac{1}{1 + r} ) ^{12 t}}

PowQuot

$(\frac{a}{b})^{m} = \frac{a ^{m}}{b ^{m}}$

M = \frac{P r}{1 - \frac{1 ^{12 t}}{( 1 + r ) ^{12 t}}}

BaseOne

$1^{a} = 1$

M = \frac{P r}{1 - \frac{1}{( 1 + r ) ^{12 t}}}

Now, to subtract

\frac{1}{( 1 + r ) ^{12 t}}

from

1,

rewrite

1

\frac{( 1 + r ) ^{12 t}}{( 1 + r ) ^{12 t}} .

The fractions can then be subtracted.

M = \frac{P r}{1 - \frac{1}{( 1 + r ) ^{12 t}}}

$1 = \frac{a}{a}$

M = \frac{P r}{\frac{( 1 + r ) ^{12 t}}{( 1 + r ) ^{12 t}} - \frac{1}{( 1 + r ) ^{12 t}}}

SubFrac

Subtract fractions

M = \frac{P r}{\frac{( 1 + r ) ^{12 t} - 1}{( 1 + r ) ^{12 t}}}

DivByFracSL

$a / \frac{b}{c} = \frac{a \cdot c}{b}$

M = \frac{P r ( 1 + r ) ^{12 t}}{( 1 + r ) ^{12 t} - 1}

The formula cannot be simplified further because the numerator and denominator do not share a common factor.

b Consider the simplified formula written in Part A.

M = \frac{P r ( 1 + r ) ^{12 t}}{( 1 + r ) ^{12 t} - 1}

Here,

P

is the amount of money that is borrowed,

t

is the number of years,

r

is the monthly interest rate, and

M

is the monthly payment. Ramsha's mother wants to find her monthly payment

M

when

P = 1500,

t = 2,

and

r = 0.5 % = 0.005 .

Substitute the known values and find

M .

M = \frac{P r ( 1 + r ) ^{12 t}}{( 1 + r ) ^{12 t} - 1}

SubstituteValues

Substitute values

M = \frac{( 1 5 0 0 ) ( 0 . 0 0 5 ) ( 1 + 0 . 0 0 5 ) ^{12 (2)}}{( 1 + 0 . 0 0 5 ) ^{12 (2)} - 1}

AddTerms

Add terms

M = \frac{( 1 5 0 0 ) ( 0 . 0 0 5 ) ( 1 . 0 0 5 ) ^{12 (2)}}{( 1 . 0 0 5 ) ^{12 (2)} - 1}

Multiply

M = \frac{7 . 5 ( 1 . 0 0 5 ) ^{24}}{( 1 . 0 0 5 ) ^{24} - 1}

UseCalc

Use a calculator

M = 66.480915 \dots

RoundDec

Round to $2$ decimal place(s)

M \approx 66.48

Therefore, Ramsha's mother would need to pay about

$ 66.48

per month if she repays a loan of

$ 1500

over

2

years.

Closure

Egyptian Fractions

Rational expressions are basically fractions that have variables in their numerators or denominators. As with numeric fractions, rational expressions can be added and subtracted. To clarify the connection between fractions and rational expressions, the challenge presented at the beginning of the lesson will be solved.

$Eye of Horus and Egyptian fractions$

External credits: Benoît Stella

Recall the information Ramsha learned about Egyptian fractions.

Egyptians used unit fractions to represent all other fractions.
The Egyptians were able to express any fraction as a sum of unit fractions where all the unit fractions were different.

Correct \frac{3}{4} = \frac{1}{2} + \frac{1}{4} Incorrect \frac{3}{4} = \frac{1}{4} + \frac{1}{4} + \frac{1}{4}

a How might the Egyptians have expressed

\frac{2}{7} ?

b Find the sum of the rational expressions for any

x > 0 .

\frac{1}{x + 1} + \frac{1}{x ( x + 1 )}

c Use the result from Part B to show that each unit fraction

\frac{1}{N},

with

N \geq 2,

can be written as a sum of two or more different unit fractions.

d Describe a procedure for writing any positive rational number

\frac{p}{q}

with

p > 0

and

q > 2

as an Egyptian fraction.

Answer

a Example Answer:

\frac{2}{7} = \frac{1}{4} + \frac{1}{2 8}

\frac{1}{x}

c See solution.

d See solution.

Hint

a What is the greatest fraction less than

\frac{2}{7} ?

b Find a common denominator. Use it to rewrite the rational expression with a common denominator.

c Consider the answer to Part B and write an equation for

\frac{1}{N} .

d Start by writing

\frac{p}{q}

as a sum of multiples of

\frac{1}{q} .

Solution

a There are many strategies for writing the given fraction as a sum of unit fractions. One way to write

\frac{2}{7}

as a sum of unit fractions would be to find the greatest unit fraction less than

\frac{2}{7} .

Note that

\frac{1}{4}

is less than

\frac{2}{7} .

\frac{1}{4} < \frac{2}{7} because \frac{7}{2 8} < \frac{8}{2 8}

Subtracting

\frac{1}{4}

from

\frac{2}{7}

gives

\frac{1}{2 8} .

Therefore, the given fraction is the sum of

\frac{1}{4}

and

\frac{1}{2 8} .

\frac{2}{7} = \frac{1}{4} + \frac{1}{2 8}

This is an Egyptian fraction representation of

\frac{2}{7} .

Note that there are different representations of the given fraction. Here, only one of them is shown.

b In order to find the sum of the rational expressions, both must have a common denominator.

\frac{1}{x + 1} + \frac{1}{x ( x + 1 )}

Looking at the denominators, the numerator and denominator of the first expression should be multiplied by

x .

\frac{1}{x + 1} + \frac{1}{x ( x + 1 )}

ExpandFrac

$\frac{a}{b} = \frac{a \cdot x}{b \cdot x}$

\frac{x}{x ( x + 1 )} + \frac{1}{x ( x + 1 )}

AddFrac

Add fractions

\frac{x + 1}{x ( x + 1 )}

CancelCommonFac

Cancel out common factors

\frac{x + 1}{x ( x + 1 )}

SimpQuot

Simplify quotient

\frac{1}{x}

The sum is equal to

\frac{1}{x} .

\frac{1}{x + 1} + \frac{1}{x ( x + 1 )} = \frac{1}{x}

c In Part B, the sum of the rational expressions was found to be

\frac{1}{x} .

The reverse of this identity can be applied to find Egyptian fractions of the form

\frac{1}{N},

where

N > 2 .

\frac{1}{N} = \frac{1}{N + 1} + \frac{1}{N ( N + 1 )}

The expressions of the right hand side can also be rewritten. An equivalent expression for

\frac{1}{N + 1}

is obtained by replacing

N

with

N + 1

in the identity.

\frac{1}{x} = \frac{1}{x + 1} + \frac{1}{x ( x + 1 )}

SubstituteExpressions

Substitute expressions

\frac{1}{N + 1} = \frac{1}{N + 1 + 1} + \frac{1}{( N + 1 ) ( N + 1 + 1 )}

AddTerms

Add terms

\frac{1}{N + 1} = \frac{1}{N + 2} + \frac{1}{( N + 1 ) ( N + 2 )}

This equation can be used to rewrite the right-hand side of the first equation.

\frac{1}{N} \frac{1}{N} = \frac{1}{N + 1} + \frac{1}{N ( N + 1 )} ⇓ = \frac{1}{N + 2} + \frac{1}{( N + 1 ) ( N + 2 )} + \frac{1}{N ( N + 1 )}

As it is shown in the equations above, each unit fraction

\frac{1}{N},

with

N \geq 2,

can be written as a sum of two or more different unit fractions.

d Positive rational numbers of the form

\frac{p}{q},

where

p > 0

and

q > 2,

can be written as a sum of unit fractions

\frac{1}{q} .

\frac{p}{q} = p \frac{1}{q} + \frac{1}{q} + \dots + \frac{1}{q}

p = 1,

then it is already an Egyptian fraction. If

p = 2,

then the first

\frac{1}{q}

is kept and then the second one is replaced with

\frac{1}{q + 1} + \frac{1}{q ( q + 1 )} .

\frac{2}{q} = \frac{1}{q} + \frac{1}{q} = \frac{1}{q} + \frac{1}{q + 1} + \frac{1}{q ( q + 1 )}

p > 2,

then the third

\frac{1}{q}

is written as

\frac{1}{q + 1} + \frac{1}{q ( q + 1 )}

and then each of these is replaced using the same idea.

\frac{p}{q} = \frac{1}{q} + \frac{1}{q} + \frac{1}{q} + \dots

Rewrite

Substitute

$\frac{1}{q} = \frac{1}{q + 1} + \frac{1}{q ( q + 1 )}$

\frac{p}{q} = \frac{1}{q} + \frac{1}{q + 1} + \frac{1}{q ( q + 1 )} + \frac{1}{q} + \dots

Substitute

$\frac{1}{q} = \frac{1}{q + 1} + \frac{1}{q ( q + 1 )}$

\frac{p}{q} = \frac{1}{q} + \frac{1}{q + 1} + \frac{1}{q ( q + 1 )} + \frac{1}{q + 1} + \frac{1}{q ( q + 1 )} + \dots

Substitute

$\frac{1}{q + 1} = \frac{1}{q + 2} + \frac{1}{( q + 1 ) ( q + 2 )}$

\frac{p}{q} = \frac{1}{q} + \frac{1}{q + 1} + \frac{1}{q ( q + 1 )} + \frac{1}{q + 2} + \frac{1}{( q + 1 ) ( q + 2 )} + \frac{1}{q ( q + 1 )} + \dots

Substitute

$\frac{1}{q ( q + 1 )} = \frac{1}{q ( q + 1 ) + 1} + \frac{1}{q ( q + 1 ) ( q ( q + 1 ) + 1 )}$

\frac{p}{q} = \frac{1}{q} + \frac{1}{q + 1} + \frac{1}{q ( q + 1 )} + \frac{1}{q + 2} + \frac{1}{( q + 1 ) ( q + 2 )} + \frac{1}{q ( q + 1 ) + 1} + \frac{1}{q ( q + 1 ) ( q ( q + 1 ) + 1 )} + \dots

By repeating this procedure, replacing every duplicate unit fraction with two unit fractions with larger denominators, all repeating unit fractions will eventually disappear. Therefore, any positive rational number $\frac{p}{q}$ can be written as an Egyptian fraction.

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Hint

Solution

Adding and Subtracting With Like Denominators

Adding and Subtracting With Unlike Denominators

Hint

Solution

Hint

Solution

Hint

Solution

Hint

Solution

Answer

Hint

Solution