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Here are a few recommended readings before getting started with this lesson.
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 N1, 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.
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,… 3,6,9,12,15,…
|
6,12,18,24,… | LCM(2,3)=6 |
8 and 12 | Multiples of 8:Multiples of 12: 8,16,24,32,40,48,… 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 |
---|---|---|---|
4x3and6xy2
|
22⋅x3and2⋅3⋅x⋅y2
|
12x3y2 | 12x3y2 is the smallest expression that is divisible by both 4x3 and 6xy2. |
3x3+3x2andx2+5x+4
|
3⋅x2⋅(x+1)and(x+1)(x+4)
|
3x2(x+1)(x+4) | 3x2(x+1)(x+4) is the smallest expression that is divisible by both 3x3+3x2 and x2+5x+4. |
Finding the LCM of polynomials requires identifying the factors with the highest power that appear in each polynomial.
Split into factors
Factor out 12y
a2+2ab+b2=(a+b)2
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 | 12x2y+48xy+48y | 22⋅3⋅y⋅(x+2)2 | 22⋅31⋅x0⋅y1⋅(x+2)2⋅(x−8)1 |
Polynomail II | 3x3y−18x2y−48xy | 3⋅x⋅y⋅(x+2)⋅(x−8) | 20⋅31⋅x1⋅y1⋅(x+2)1⋅(x−8)1 |
What is the least common multiple of the given polynomials?
Split into factors
Factor out 6x
Split into prime factors
Split into factors
Factor out 3x
am⋅bm=(a⋅b)m
Split into factors
Write as a power
a2−2ab+b2=(a−b)2
Given Form | Factored Form | All Related Factors | |
---|---|---|---|
Heichi | 6x2y−6x | 2⋅3⋅x⋅(xy−1) | 21⋅31⋅x1⋅(xy−1)1 |
Ramsha | 3x3y2−6x2y+3x | 3⋅x⋅(xy−1)2 | 20⋅31⋅x1⋅(xy−1)2 |
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) |
---|---|---|---|---|
32 and 21 | 3 and 2 | Multiples of 3:Multiples of 2: 3,6,9,12,15,… 2,4,6,8,10,12,…
|
6, 12 | 6 |
65 and 41 | 6 and 4 | Multiples of 6:Multiples of 4: 6,12,18,24,30,… 4,8,12,16,20,24,…
|
12, 24 | 12 |
41 and 25 | 4 and 2 | Multiples of 4:Multiples of 2: 4,8,12,… 2,4,6,8,10,12,…
|
4, 8, 12 | 4 |
When adding and subtracting rational expressions, the same rules apply as when adding and subtracting fractions.
If the rational expressions have a common denominator, the numerators can be added or subtracted directly.
Q(x)P(x)±Q(x)R(x)=Q(x)P(x)±R(x)
Here, P(x), Q(x), and R(x) are polynomials and Q(x)=0.
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.
Q(x)P(x)±S(x)R(x)=Q(x)S(x)P(x)S(x)±R(x)Q(x)
Denominator | Factored Form |
---|---|
2x−2 | 2(x−1) |
x2−4x+3 | (x−1)(x−3) |
ba=b⋅(x−3)a⋅(x−3)
Multiply parentheses
Add terms
ba=b⋅2a⋅2
Distribute 2
Multiply
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.Distribute 6
Subtract fractions
Commutative Property of Addition
t=15
Calculate power
Multiply
Add and subtract terms
Use a calculator
Round to 2 decimal place(s)
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.
Let t be the time it takes for the first pipe to fill the pool. Then, the fraction t1 represents the part of the pool that the first pipe fills in one hour.
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 61 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 | t1 |
Pipe II | 2t | 2t1 |
Pipe III | 3t | 3t1 |
All of Them | 6 | 61 |
Denominator | Factored Form |
---|---|
t | t |
2t | 2⋅t |
3t | 3⋅t |
6 | 2⋅3 |
ba=b⋅6a⋅6
ba=b⋅3a⋅3
ba=b⋅2a⋅2
ba=b⋅ta⋅t
Multiply fractions
Add fractions
Izabella is planning a trip that involves a 90-kilometer bus ride and a high-speed train ride. The entire trip is 300 kilometers.
The average speed of the high-speed train is 30 kilometers per hour more than twice the average speed of the bus.