2. Adding and Subtracting Rational Expressions
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Chapter 4
2.
Adding and Subtracting Rational Expressions
Mathematics often presents us with challenges that mirror real-life scenarios. One such concept is the addition and subtraction of rational expressions. These are essentially fractions where the numerator and denominator can be polynomials. While they may sound complex, understanding them can help in various fields. For instance, in engineering, these expressions can model certain behaviors of systems. In finance, they might help in risk assessment or in breaking down complex transactions. By mastering the technique of adding and subtracting these expressions, individuals open up opportunities to tackle a wide range of problems in diverse domains.
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Lesson Settings & Tools
| | 10 Theory slides |
| | 10 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Adding and Subtracting Rational Expressions
Slide of 10
Mathematical operations such as addition and subtraction can be performed with rational expressions, as well as multiplication and division. The procedures for adding and subtracting rational expressions are the same as for adding and subtracting numerical fractions. These procedures will be developed in this lesson.
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
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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 1N, 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.
External credits: Benoît Stella
The Egyptians were able to write any fraction as a sum of different unit fractions. 3/4 = 1/2 + 1/4 However, the Egyptians did not use a given unit fraction more than once, so they would not have written 34= 14+ 14+ 14.
a How might the Egyptians have expressed 27?
b Find the sum of the rational expressions for any x>0.
1/x+1 + 1/x(x+1)
c Use the result from Part B to show that each unit fraction 1N, with N≥2, can be written as a sum of two or more different unit fractions.
d Describe a procedure for writing any positive rational number, pq with p>0 and q>2, as an Egyptian fraction.
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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 of2:& 2, 4, 6, 8, 10, 12, ... Multiples of3:& 3, 6, 9, 12, 15, ... | 6, 12, 18, 24, ... | LCM(2,3)= 6 |
| 8 and 12 | Multiples of 8:& 8, 16, 24, 32, 40, 48, ... Multiples of12:& 12, 24, 36, 48, ... | 24, 48, 72, 96, ... | 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 |
|---|---|---|---|
| 4x^3 and 6xy^2 | 2^2 * x^3 and 2* 3 * x * y^2 | 12x^3y^2 | 12x^3y^2 is the smallest expression that is divisible by both 4x^3 and 6xy^2. |
| 3x^3+3x^2 and x^2+5x+4 | 3 * x^2 * (x+1) and (x+1)(x+4) | 3x^2(x+1)(x+4) | 3x^2(x+1)(x+4) is the smallest expression that is divisible by both 3x^3+3x^2 and x^2+5x+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: & 12x^2y+48xy+48y Polynomial II: & 3x^3y-18x^2y-48xy The procedure of finding the LCM of the polynomials involves three steps.
1
Factor Each Polynomial and Write Numerical Factors as Product of Primes
expand_more Each polynomial will be factored. Start by factoring the first polynomial. Note that 12y is used in all of its terms. Therefore, it will be factored out.
12x^2y+48xy+48y
SplitIntoFactors
Split into factors
12y * x^2 +12y * 4x + 12y * 4
FactorOut
Factor out 12y
12y (x^2+4x+4)
FacPosPerfectSquare
a^2+2ab+b^2=(a+b)^2
12y (x+2)^2
Then the numerical factor can be written as a product of 2^2 and 3. Polynomial I [0.5em] 12x^2y+48xy+48y ⇕ 2^2 * 3 * y (x+2)^2 Now the other polynomial will be factored. Since the factor 3xy can be seen in each of its terms, start by factoring it out.
3x^3y-18x^2y-48xy
SplitIntoFactors
Split into factors
3xy * x^2-3xy * 6x-3xy * 16
FactorOut
Factor out 3xy
3xy(x^2-6x-16)
3xy(x+2)(x-8)
2
Identify the Factors With the Highest Power
expand_more 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 | 12x^2y+48xy+48y | 2^2 * 3 * y * (x+2)^2 | 2^2 * 3^1 * x^0 * y^() 1 * (x+2)^2 * (x-8)^1 |
| Polynomail II | 3x^3y-18x^2y-48xy | 3 * x * y * (x+2) * (x-8) | 2^0 * 3^1 * x^1 * y^() 1 * (x+2)^1 * (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
3
Multiply the Highest Power of the Factors
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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: & 6x^2y-6x Ramsha: & 3x^3y^2-6x^2y+3x 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.
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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: & 6x^2y-6x Ramsha: & 3x^3y^2-6x^2y+3x
To do so, each expression will be factored completely. Start by factoring the first polynomial. Note that all its terms contain a common factor, 6x.
6x^2y-6x
SplitIntoFactors
Split into factors
6x * xy +6x * (-1)
FactorOut
Factor out 6x
6x(xy-1)
SplitIntoPrimeFactors
Split into prime factors
2* 3 * x(xy-1)
Now the other polynomial will be factored. The factor 3x can be seen in each of its terms, so start by factoring it out.
3x^3y^2-6x^2y+3x
SplitIntoFactors
Split into factors
3x * x^2y^2+3x * (-2xy)+3x * 1
FactorOut
Factor out 3x
3x(x^2y^2-2xy+1)
▼
Factor (x^2y^2-2xy+1)
ProdPow
a^m* b^m=(a * b)^m
3x ((xy)^2-2xy+1 )
SplitIntoFactors
Split into factors
3x ((xy)^2-2(xy) 1 + 1 )
WritePow
Write as a power
3x ((xy)^2-2(xy) 1 + 1^2 )
FacNegPerfectSquare
a^2-2ab+b^2=(a-b)^2
3x(xy-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 | 6x^2y-6x | 2 * 3 * x * (xy-1) | 2^1 * 3^1 * x^1 * (xy-1)^1 |
| Ramsha | 3x^3y^2-6x^2y+3x | 3* x * (xy-1)^2 | 2^0 * 3^1 * x^1 * (xy-1)^2 |
The LCM of these polynomials is the product of the highest power of each factor.
This expression represents the time when Heichi and Ramsha meet again at the starting point.
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Discussion
Using the Least Common Denominator to Add or Subtract Rational Expressions
Concept
Least Common Denominator
The least common denominator (LCD) of two fractions is the least common multiple (LCM) of the denominators of the fractions. In other words, the least common denominator is the smallest of all the common denominators. Some examples are provided in the table below.
| Fractions | Denominators | Multiples of Denominators | Common Denominators | LCM of Denominators (LCD) |
|---|---|---|---|---|
| 2/3 and 1/2 | 3 and 2 | Multiples of3:& 3, 6, 9, 12, 15, ... Multiples of2:& 2, 4, 6, 8, 10, 12, ... | 6, 12 | 6 |
| 5/6 and 1/4 | 6 and 4 | Multiples of6:& 6, 12, 18, 24, 30, ... Multiples of4:& 4, 8, 12, 16, 20, 24, ... | 12, 24 | 12 |
| 1/4 and 5/2 | 4 and 2 | Multiples of4:& 4, 8, 12, ... Multiples of2:& 2, 4, 6, 8, 10, 12, ... | 4, 8, 12 | 4 |
The least common denominator is used when adding or subtracting fractions with different denominators. 2/10 - 1/11 = 22/110_(LCD) - 10/110_(LCD) As with fractions, rational expressions must have a common denominator to be added or subtracted. The least common denominator of two rational expressions is the least common multiple of the denominators of the fractions.
1/x + 1/x+1 = x+1/x(x+1)_(LCD) + x/x(x+1)_(LCD)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.
P(x)/Q(x) ± R(x)/Q(x)=P(x) ± R(x)/Q(x)
Here, P(x), Q(x), and R(x) are polynomials and Q(x)≠ 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.
P(x)/Q(x) ± R(x)/S(x)=P(x) S(x) ± R(x)Q(x)/Q(x)S(x)
Here, P(x), Q(x), R(x), and S(x) are polynomials, Q(x)≠ 0, and S(x)≠ 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. x+2/2x-2 + 2x+1/x^2-4x+3 The result can be found in four steps.
1
Find the Least Common Denominator
expand_more Recall that the least common denominator is the least common multiple of the expressions in the denominators.
x+2/2x-2 + 2x+1/x^2-4x+3
To find the LCM of the polynomials, they need to be factored.
| Denominator | Factored Form |
|---|---|
| 2x-2 | 2(x-1) |
| x^2-4x+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)
2
Rewrite Each Expression with the LCD
expand_more 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.
x+2/2(x-1) + 2x+1/(x-1)(x-3)
▼
Expand by (x-3)
ExpandFrac
a/b=a * (x-3)/b * (x-3)
(x+2) * (x-3)/2(x-1)* (x-3) + 2x+1/(x-1)(x-3)
MultPar
Multiply parentheses
x^2-3x+2x-6/2(x-1)* (x-3) + 2x+1/(x-1)(x-3)
AddTerms
Add terms
x^2-x-6/2(x-1)* (x-3) + 2x+1/(x-1)(x-3)
▼
Expand by 2
ExpandFrac
a/b=a * 2/b * 2
x^2-x-6/2(x-1)* (x-3) + (2x+1) * 2/(x-1)(x-3)* 2
Distr
Distribute 2
x^2-x-6/2(x-1)* (x-3) + 4x+2/(x-1)(x-3)* 2
Multiply
Multiply
x^2-x-6/2(x-1)(x-3) + 4x+2/2(x-1)(x-3)
3
Add the Numerators
expand_more 4
Simplify the Resulting Expression
expand_more Check if the resulting rational expression can be simplified or not.
x^2+3x-4/2(x-1)(x-3)
(x-1)(x+4)/2(x-1)(x-3)
▼
Simplify
x+4/2(x-3)
The result is in simplest form. Note that the excluded values for the sum are 1 and 3. x+4/2(x-3), x ≠ 1 The restrictions on the domain of a sum or difference of rational expressions consist of the restrictions to the domains of each expression.
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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- 20.4t/t^2+36
a Simplify the formula.
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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=6-20.4t/t^2+36 To simplify the formula, 6 will be rewritten as a fraction whose denominator is 1. L= 6/1-20.4t/t^2+36 In order to subtract fractions, they need to have a common denominator. The least common multiple of 1 and t^2+36 is t^2+36, so expand the first fraction by this expression. L=6 (t^2+36)/t^2+36 - 20.4t/t^2+36 Now it is possible to subtract the fractions and simplify the expressions.
L=6(t^2+36)/t^2+36 - 20.4t/t^2+36
Distr
Distribute 6
L=6t^2+216/t^2+36 - 20.4t/t^2+36
SubFrac
Subtract fractions
L=6t^2+216-20.4t/t^2+36
CommutativePropAdd
Commutative Property of Addition
L=6t^2-20.4t+216/t^2+36
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=6t^2-20.4t+216/t^2+36
Substitute
t= 15
L=6( 15)^2-20.4( 15)+216/15^2+36
CalcPow
Calculate power
L=6(225)-20.4(15)+216/225+36
Multiply
Multiply
L=1350-306+216/225+36
AddSubTerms
Add and subtract terms
L=1260/261
UseCalc
Use a calculator
L=4.827586...
RoundDec
Round to 2 decimal place(s)
L ≈ 4.83
The pH of the person's mouth 15 minutes after eating a dessert would be about 4.83.
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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.
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Hint
Let t be the time it takes for the first pipe to fill the pool. Then, the fraction 1t 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 2t 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 * t = 2t |
| Pipe III | 1.5 * 2t = 3t |
| 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 16 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 | 1/t |
| Pipe II | 2t | 1/2t |
| Pipe III | 3t | 1/3t |
| All of Them | 6 | 1/6 |
The sum of the filling rates of all pipes is equal to the filling rate when all of the pipes are open. 1/t + 1/2t + 1/3t = 1/6 To add the rational expressions, all the denominators must be factored.
| Denominator | Factored Form |
|---|---|
| t | t |
| 2t | 2 * t |
| 3t | 3* t |
| 6 | 2* 3 |
The least common denominator (LCD) is 2 * 3 * t, or 6t. Now, expand each fraction such that all the denominators are equal to the LCD.
1/t + 1/2t + 1/3t = 1/6
▼
Expand by LCD
ExpandFrac
a/b=a * 6/b * 6
1/t *6/6+1/2t + 1/3t = 1/6
ExpandFrac
a/b=a * 3/b * 3
1/t *6/6+1/2t * 3/3+1/3t = 1/6
ExpandFrac
a/b=a * 2/b * 2
1/t *6/6+1/2t * 3/3+1/3t * 2/2 = 1/6
ExpandFrac
a/b=a * t/b * t
1/t *6/6+1/2t * 3/3+1/3t * 2/2 = 1/6* t/t
MultFrac
Multiply fractions
6/6t+3/6t+2/6t = t/6t
AddFrac
Add fractions
11/6t = t/6t
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.
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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.
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b Write an expression for the total duration T of trip by using the expressions found in Part A.
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Hint
a Distance traveled is equal to the product of speed and time, d=r* 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* 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 * t ⇒ t = 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 to x will give the time t_1 she spent traveling by bus. t_1 = 90/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 2x+30. With this information, an expression for the length of time that Izabella travels on the train t_2 can be written. t_2 = 210/2x+30 ⇔ t_2 = 105/x+15
b The total duration of the trip is the sum of the rational expressions written in Part A.
T = t_1 + t_2 ⇒ T = 90/x + 105/x+15 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 = 90/x + 105/x+15
▼
Expand by (x+15)
T = 90x+1350/x^2+15x+ 105/x+15
▼
Expand by x
T = 90x+1350/x^2+15x+ 105x/x^2+15x
AddFrac
Add fractions
T = 195x+1350/x^2+15x
This expression represents the total length of time of the trip, written in terms of x, the speed of the bus.
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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 = Pr/1- ( 11+r)^(12t)
a Simplify the formula.
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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.
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Hint
a Start by using the Power of a Quotient Property, ( ab)^n= a^nb^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=Pr/1-( 11+r)^(12t)
PowQuot
(a/b)^m=a^m/b^m
M=Pr/1- 1^(12t)(1+r)^(12t)
BaseOne
1^a=1
M=Pr/1- 1(1+r)^(12t)
Now, to subtract 1(1+r)^(12t) from 1, rewrite 1 as (1+r)^(12t)(1+r)^(12t). The fractions can then be subtracted.
M=Pr/1- 1(1+r)^(12t)
1=a/a
M=Pr/(1+r)^(12t)(1+r)^(12t)- 1(1+r)^(12t)
SubFrac
Subtract fractions
M=Pr/(1+r)^(12t)-1(1+r)^(12t)
DivByFracSL
.a /b/c.=a* c/b
M=Pr(1+r)^(12t)/(1+r)^(12t)-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=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=P r(1+ r)^(12 t)/(1+ r)^(12 t)-1
SubstituteValues
Substitute values
M=( 1500)( 0.005)(1+ 0.005)^(12( 2))/(1+ 0.005)^(12( 2))-1
AddTerms
Add terms
M=(1500)(0.005)(1.005)^(12(2))/(1.005)^(12(2))-1
Multiply
Multiply
M=7.5(1.005)^(24)/(1.005)^(24)-1
UseCalc
Use a calculator
M = 66.480915 ...
RoundDec
Round to 2 decimal place(s)
M ≈ 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.
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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.
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 & Incorrect 3/4 = 1/2 + 1/4 & 34= 14+ 14+ 14
a How might the Egyptians have expressed 27?
b Find the sum of the rational expressions for any x>0.
1/x+1 + 1/x(x+1)
c Use the result from Part B to show that each unit fraction 1N, with N≥2, can be written as a sum of two or more different unit fractions.
d Describe a procedure for writing any positive rational number pq with p>0 and q>2 as an Egyptian fraction.
Answer
a Example Answer: 2/7 = 1/4 + 1/28
b 1/x
c See solution.
d See solution.
Hint
a What is the greatest fraction less than 27?
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 1N.
d Start by writing pq as a sum of multiples of 1q.
Solution
a There are many strategies for writing the given fraction as a sum of unit fractions. One way to write 27 as a sum of unit fractions would be to find the greatest unit fraction less than 27. Note that 14 is less than 27.
1/4 < 2/7 because 7/28 < 8/28 Subtracting 14 from 27 gives 128. Therefore, the given fraction is the sum of 14 and 128. 2/7 = 1/4 + 1/28 This is an Egyptian fraction representation of 27. 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.
1/x+1 + 1/x(x+1) Looking at the denominators, the numerator and denominator of the first expression should be multiplied by x.
1/x+1 + 1/x(x+1)
ExpandFrac
a/b=a * x/b * x
x/x(x+1) + 1/x(x+1)
AddFrac
Add fractions
x+1/x(x+1)
CancelCommonFac
Cancel out common factors
x+1/x(x+1)
SimpQuot
Simplify quotient
1/x
The sum is equal to 1x. 1/x+1 + 1/x(x+1) = 1/x
c In Part B, the sum of the rational expressions was found to be 1x. The reverse of this identity can be applied to find Egyptian fractions of the form 1N, where N> 2.
1/N = 1/N+1 + 1/N(N+1) The expressions of the right hand side can also be rewritten. An equivalent expression for 1N+1 is obtained by replacing N with N+1 in the identity.
1/x = 1/x+1 + 1/x(x+1)
SubstituteExpressions
Substitute expressions
1/N+1 = 1/N+1+1 + 1/( N+1)( N+1+1)
AddTerms
Add terms
1/N+1 = 1/N+2 + 1/(N+1)(N+2)
This equation can be used to rewrite the right-hand side of the first equation. 1/N & = 1/N+1 + 1/N(N+1) & ⇓ 1/N & = 1N+2 + 1(N+1)(N+2) + 1/N(N+1) As it is shown in the equations above, each unit fraction 1N, with N ≥ 2, can be written as a sum of two or more different unit fractions.
d Positive rational numbers of the form pq, where p>0 and q>2, can be written as a sum of unit fractions 1q.
p/q = 1/q+1/q+ ... + 1/q_p If p=1, then it is already an Egyptian fraction. If p=2, then the first 1q is kept and then the second one is replaced with 1q+1+ 1q(q+1). 2/q & = 1/q+ 1/q [0.6em] & = 1/q+ 1/q+1+1/q(q+1) If p>2, then the third 1q is written as 1q+1+ 1q(q+1) and then each of these is replaced using the same idea.
p/q = 1/q + 1/q+ 1/q+ ...
▼
Rewrite
Substitute
1/q= 1/q+1+1/q(q+1)
p/q = 1/q+ 1/q+1+1/q(q+1) + 1/q + ...
Substitute
1/q= 1/q+1+ 1/q(q+1)
p/q = 1/q+1/q+1+1/q(q+1) + 1/q+1+ 1/q(q+1) + ...
Substitute
1/q+1= 1/q+2 + 1/(q+1)(q+2)
p/q = 1/q+1/q+1+1/q(q+1) + 1/q+2 + 1/(q+1)(q+2) + 1/q(q+1) + ...
Substitute
1/q(q+1)= 1/q(q+1)+1 + 1/q(q+1)(q(q+1)+1)
p/q = 1/q+ 1/q+1+1/q(q+1)+ 1/q+2 + 1/(q+1)(q+2) +1/q(q+1)+1 + 1/q(q+1)(q(q+1)+1) + ...
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 pq can be written as an Egyptian fraction.
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Adding and Subtracting Rational Expressions
Exercise 2.1
The actual distance d_o of the object, the distance d_i of the image , and the focal length f of the lens satisfy the thin-lens equation. 1/f = 1/d_o+ 1/d_i
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We are asked to find a simplified form of the thin-lens equation. When we look at the choices, we see that the variable f is on the left side for each equation, so our first step will be to solve the thin-lens equation for f.
The result corresponds to option D.
We found a simplified equation for variable f in Part A.
f = d_i d_o/d_o+d_i
We can use this equation to find the focal length f of a lens when we know the actual distance d_o of the object from the lens and the distance d_i of the image from the lens. Let's show what we know on a diagram.
We can substitute d_o=50 and d_i=20 into our equation and find the focal length of the lens.
The focal length is about 14.3 centimeters.
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Simplify the complex fraction. 1x^2y+ 1xy^2/1x+ 1y
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A complex fraction is a rational expression that has at least one fraction in its numerator, denominator, or both. We want to simplify the given complex fraction. 1x^2y+ 1xy^2/1x+ 1y To do so, we will combine the expressions in the numerator and the denominator separately. Then, we will multiply the new numerator by the reciprocal of the new denominator. Let's start by simplifying the numerator.
Now let's simplify the denominator.
Next, we can rewrite the complex fraction using the simplified components. 1x^2y+ 1xy^2/1x+ 1y ⇔ y+xx^2y^2/y+xxy Finally, we will multiply the numerator by the reciprocal of the new denominator. y+xx^2y^2/y+xxy ⇔ y+x/x^2y^2 * xy/y+x We can simplify this expression.
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External credits: TUBS
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Let's analyze the given equation. T=d/s+j+d/s-j The denominators of the rational expressions do not share any common factors. The least common denominator is therefore their product. Let's multiply each fraction by the denominator of the other fraction. We will be able to add them together afterward.
We are given the following information about the round-trip flight between Seattle and Miami.
- The distance d is 2730 miles.
- The average airplane speed s is 520 miles per hour.
- The average speed of the jet stream j is 110 miles per hour.
We are asked to find the total time T of the flight for the given conditions. Let's substitute the values into the simplified formula from Part A and find T.
Therefore, the total time is about 11 hours.
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We are given that the amount of the two medications in the bloodstream are determined by A(t) and B(t). A(t) & =5t/3t^2+12t+9 [0.7em] B(t) & =7t/5t^2+20t+15 Let's add these two functions together to determine a function for the total amount of the medication in a person's bloodstream after t hours. A(t) + B(t) = 5t/3t^2+12t+9 + 7t/5t^2 + 20t + 15 Remember, in order to add rational expressions, they must have the same denominator. For this reason, we need to factor the denominators of the rational expressions and find their least common multiple. This will be a candidate for a common denominator.
| Factor | |
|---|---|
| 3t^2+12t+9 | 5t^2+20t+15 |
| 3t^2+ 3t+9t+9 | 5t^2+ 5t+15t+15 |
| 3t(t+1)+9(t+1) | 5t(t+1)+15(t+1) |
| (3t+9)(t+1) | (5t+15)(t+1) |
| 3(t+3)(t+1) | 5(t+3)(t+1) |
We factored the denominators of the fractions. The least common multiple (LCM) must contain each factor in both denominators. LCM: 3* 5* (t+1)* (t+3) For the fractions to have a denominator as above, let's expand each of them by the factor of LCM their denominators do not have.
Now that the fractions have common denominators, we are able to add them.
We have found the simplified version of the given function!
From Part A, we know that the total amount of medication in LaShay's bloodstream is determined by the following equation.
A(t)+B(t)=46t/15t^2+60t+45
Here, t represents the time in hours after the medication is taken. Since we are asked to find the amount of medication in her system after 6 hours, let's substitute 6 for t into the equation.
We conclude that 6 hours after the medications are taken, the amount in LaShay's bloodstream is approximately 0.29 milligrams.
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Adding and Subtracting Rational Expressions
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