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| 10 Theory slides |
| 10 Exercises - Grade E - A |
| Each lesson is meant to take 1-2 classroom sessions |
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.
Add fractions
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.1=aa
Subtract fractions
a/cb=ba⋅c
Substitute values
Add terms
Multiply
Use a calculator
Round to 2 decimal place(s)
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.
Recall the information Ramsha learned about Egyptian fractions.
ba=b⋅xa⋅x
Add fractions
Cancel out common factors
Simplify quotient
Substitute expressions
Add terms
q1=q+11+q(q+1)1
q1=q+11+q(q+1)1
q+11=q+21+(q+1)(q+2)1
q(q+1)1=q(q+1)+11+q(q+1)(q(q+1)+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 qp can be written as an Egyptian fraction.
Find the least common multiples of the expressions.
We want to find the least common multiple (LCM) of the given pair of polynomials. 36u^2c and 30u^2c^4 In this case, we need to find the prime factors of each expression. 36u^2c & = 2^2 * 3^2 * u^2 * c 30u^2c^4 & = 2 * 3 * 5 * u^2 * c^4 Let's list the factors with the highest power. 2^2, 3^2, 5, u^2, c^4 The LCM of the given expressions is the product of these factors.
This expression we obtained is the LCM of the given polynomials.
We want to find the least common multiple of the given pair of polynomials.
12x^2 and 8x^2-16x
To do so, we need to find the prime factors of each expression. Let's start by factoring the first one.
Let's now factor the second expression.
Now we have both expressions written as the product of their prime factors. 12x^2 & = 2^2 * 3 * x^2 8x^2-16x & = 2^3 * x * (x-2) Finally, we multiply the highest power of each factor that occurs in any of the expressions.
This resulting expression is the LCM of the polynomials.
To find the least common multiple of the polynomials, we will follow the same steps that we did in Part B.
10x^2+35x-20
and
4x^2+22x+24
To find the LCM, we need to find the prime factors of each expression. Let's start by factoring the first one. Notice that all of its terms contain a common factor, 5. We will start by factoring it out.
Let's factor the other expression. We can see that 2 is a common factor, so we can start by factoring it out.
We have factored both polynomials. 10x^2 +35x-20 & = 5(2x-1)(x+4) 4x^2 +22x+24 & = 2(2x+3)(x+4) Finally, we can find the LCM by finding the product of the highest power of each factor.
The following expression is the LCM of 10x^2+35x-20 and 4x^2+22x+24. 40x^3+200x^2+130x-120
Find the sum or difference.
We want to add the given rational expressions. a/3a+6+a^2/a^2+2a We will start by factoring the denominators.
Denominator | Factored Form |
---|---|
3a+6 | 3(a+2) |
a^2+2a | a(a+2) |
The least common denominator (LCD) is 3 * a* (a+2). Let's simplify this expression. LCD: 3 * a * (a+2) = 3a^2 + 6a Now we will expand each algebraic fraction such that all the denominators are equal to the LCD. Once the rational expressions have the same denominator, we can add them.
We want to subtract the given rational expressions. 3-x/x^2-x-6+3/x^2-7x+12 We will start by factoring the denominators. Let's start with the first denominator.
We will now factor the second denominator.
Next, we will determine the least common denominator (LCD) considering the factored denominators.
Denominator | Factored Form | LCD |
---|---|---|
x^2-x-6 | (x+2)(x-3) | (x+2)(x-3)(x-4) |
x^2-7x+12 | (x-3)(x-4) |
We will expand each algebraic fraction such that both denominators are equal to the LCD. To do so, we will expand the first expression by (x-4) and the second expression by (x+2). Once the rational expressions have the same denominator, we can add them.
We can multiply the factors in the denominator.
Determine whether the given statement is always, sometimes, or never true.
The least common denominator of two rational expressions is the product of the denominators. |
We are asked to verify if the least common denominator of two rational expressions is the product of the denominators. Let's analyze two rational expressions with exactly the same denominators. 1/x+1 and 2/x+1 The least common multiple of x+1 and x+1 is also x+1. It is not the product of the denominators. Let's analyze two rational expressions with different binomials in denominators. 1/x+1 and 1/x+2 The least common multiple of x+1 and x+2 is (x+1)(x+2), which is the product of denominators. Therefore, the given statement is sometimes true.
We want to determine how we should begin rewriting the function y= 2xx+3 to obtain the form y= ax-h+k. y = 2x/x+3 ⇒ y=a/x-h+k Let's think of the reverse of this process and rewrite the right-hand side of y= ax-h+k as a single fraction.
Therefore, initially, the function should have this form. Since the denominator of the given simple rational function is x+3, we want to rewrite its numerator in the form k( x+3)+ a for some k and a. Let's compare the appropriate expressions of the numerators to each other. 2x= k( x+3)+ a ⇓ 2x= k( x+3)+ a Since the coefficients in front of x must be equal, we know that k= 2. k( x+3)+ a ⇓ 2( x+3)+ a Next, the numerators, 2x and 2( x+3)+ a, should be equal. Let's find a!
Therefore, we should rewrite the numerator of the given function in the following way. 2x= 2( x+3)+ a ⇓ 2x= 2( x+3)+( - 6) Finally, let's find out how we should begin to rewrite the function y.
The last expression corresponds to option C.
We want to add and subtract the rational expressions. 3/t^2-8t-3/3t-24+1/t We can only add and subtract rational expressions that have common denominators. We will start by factoring the denominators so that we can find any common factors.
Denominator | Factored Form |
---|---|
t^2-8t | t(t-8) |
3t-24 | 3(t-8) |
t | Already Factored |
Let's highlight all of the different factors in the denominators. 3/t^2-8t-3/3t-24+1/t ⇕ 3/t (t-8)-3/3 (t-8)+1/t The least common denominator (LCD) is 3 t (t-8). We can subtract and add the expressions by rewriting each of them with the LCD.
Notice that the numerator and denominator have a common factor. Let's simplify it! - 15/3t(t-8) ⇒ - 5/t(t-8)
To identify the restrictions on variable t, we need to find the t-values that make the denominator of the simplified expression and any other denominator 0. For simplicity, we will use their factored forms.
Denominator | Restrictions on the denominator | Restrictions on the variable |
---|---|---|
t(t-8) | t≠ 0 and t-8 ≠ 0 | t≠ 0 and t≠ 8 |
3(t-8) | 3t-8≠ 0 | t≠ 8 |
t | t≠ 0 | t≠ 0 |
t(t-8) | t(t-8)≠ 0 | t≠ 0 and t≠ 8 |
We found two unique restrictions on the variable. t≠ 0 and t≠ 8