Rational Expressions Module
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Rational Expressions Module
- 2. Ms. Lorie Jane L Letada Rational Expressions is an algebraic expression where both numerator and denominator are polynomials. During elementary years, you have learned that a fraction is a ratio of a numerator and denominator where the denominator is not equal to zero, but in this module, we extend the concept of fractions to algebraic expressions. Here is the map of the lessons that will be covered in this module. Module 2 Rational Expressions At the end of this module, the learners will: 1. illustrate rational expressions; 2. simplify rational expressions; 3. perform operations on rational algebraic expressions; 4. solve problems involving rational expressions. Rational Expressions Simplifying Rational expressions Operation of Rational Expressions Adding and Subtracting Rational Expressions Multiplying and Dividing Rational Expressions
- 3. Module 2: Rational Expressions Read and analyze each item carefully and encircle the letter of the best answer. Pre-test!
- 4. Ms. Lorie Jane L Letada
- 5. Module 2: Rational Expressions
- 6. Ms. Lorie Jane L Letada A Rational Expressions is reduced to its simplest form if the numerator and denominator have no common factors. Putting a rational expression in its simplest form is important as this will make operations easier. Simplify the following in its simplest form. 1. 4π 12π 2. 9π 21π 3. 4π€2 6π€β8π€2 Lesson 1 Simplifying Rational Expressions Step 1 Factor the numerator and denominator and get the GCF. 4a = (2) (2) a 12b= (3) (2) (2) b GCF: (2) (2) GCF: 4 Thus, the common factor is 4. Step 2: Divide out the common factor. 4π 12π = 4 ( π ) 4 (3π) = π 3π Thus, the simplify is π 3π . Step 1 Factor the numerator and denominator and get the GCF. 9m = 21n = GCF: GCF: Thus, the common factor is __. Step 2: Divide out the common factor. Step 1 Get the GCF of each term. 4π€2 = (2) (2) (w) (w) 6w = (3) (2) (w) 8π€2 = (2)(2) (2)(w) GCF: (2) (w) GCF: 2w Thus, the common factor is 2w. Step 2: Factor the common numerator and denominator. 4π€2 6π€β8π€2 = 2π€ ( 2) 2π€ (3β4π€ ) Step 3: Divide out the common factor. 4π€2 6π€β8π€2 = 2π€ ( 2) 2π€ (3β4π€ ) = 2 3β4π€ Thus, the simplify is 2 3β4π€ .
- 7. Module 2: Rational Expressions 4. 6π2 2πβ4π3 5. 2π₯β2π¦ π¦2β π₯2 6. 3π₯+ 3π¦ π₯2β π¦2 7. 4π+ 4π π2β π2 Step 1 Get the GCF of each term. 6π2 = 2p = 4π3 = GCF: GCF: Step 2: Factor the common numerator and denominator. 6π2 2πβ4π3 = Step 3: Divide out the common factor. Step 1 Factor the common numerator and denominator. 2π₯β2π¦ π¦2β π₯2 = 2 (π¦βπ₯) ( π¦βπ₯ )(π¦+ π₯) Step 2: Divide out the common factor. 2π₯β2π¦ π¦2β π₯2 = 2 (π¦βπ₯) ( π¦βπ₯ )(π¦+ π₯) = 2 (π₯+ π¦) You can only apply the difference of two squares if: *The two terms are both perfect squares. *The operation is subtraction. Example: π¦2 β π₯2 = (y β x ) ( y + x ) Step 1 Factor the common numerator and denominator. 3π₯+ 3π¦ π₯2β π¦2 = Step 2: Divide out the common factor. To factor π₯2 β π¦2 , get the square root of each term. First Term: βx2 = x Second Term: βy2 = y Using x and y, form the sum (x + y) & the difference (x - y). Thus, x2 - y2 = (x + y) (x β y). Step 1 Factor the common numerator and denominator. 4π+ 4π π2β π2 = Step 2: Divide out the common factor. A variable is perfect square if its exponent is even number.
- 8. Ms. Lorie Jane L Letada 8. π2+πβ6 π2β7π+10 9. π2β7πβ30 π2β5πβ24 10. π2β49 π2β2πβ35 Simplify each rational expression. 1. 5ππ2 20ππ = 5. π₯β1 5π₯β5 = 2. 5πβ 10π = 6. π2+2πβ15 π2βπβ6 = 3. 15ππ2 45ππ3 π = 7. π2+7π+10 π2+5π = 4. π₯2β9 π₯2βπ₯β6 = 8. π2+8π+12 π2+3πβ18 = Skill Booster! Step 1 Factor the common numerator and denominator. π2+πβ6 π2β7π+10 = ( π+3)(πβ2 ) (πβ5)(πβ2) Step 2: Divide out the common factor. π2+πβ6 π2β7π+10 = ( π+3)(πβ2 ) (πβ5)(πβ2) = (π+3) (πβ5 ) Step 1 Factor the common numerator and denominator. π2β7πβ30 π2β5πβ24 = Step 2: Divide out the common factor. Step 1 Factor the common numerator and denominator. π2β49 π2β2πβ35 = (πβ7 (π+7 ) (πβ7 )(π+5 ) Step 2: Divide out the common factor. π2β49 π2β2πβ35 = (πβ7 (π+7 ) (πβ7 )(π+5 ) = ( π+7 ) (π+5 ) To factor π2 + π β 6 , find any factors of -6 that when you add the result is 1, Hence, the correct pair is 3 and -2. So, the factors of π2 + π β 6 = (m +3) (m -2). * When you multiply positive & negative numbers, the result is negative. Example: (5 )(-3) = -15 *When you add positive & negative numbers, the sign used in the result will be based which absolute num. is greater. Examples: 5 + -3 = 2 *When you add positive & negative numbers, the sign used in the result will be based which absolute num. is greater. Examples: 5 + -3 = 2
- 9. Module 2: Rational Expressions To multiply rational expressions, recall the rules for multiplying fractions. If the denominators are not equal to zero, then we simply multiply the numerators and denominators. The same rule applies to rational expressions. If a, b, c, and d represent polynomials where b β 0 and d β 0. Then, Multiply each rational expression and simplify. 1. π ππ β π ππ = 2. π ππ β π ππ = 3. ππ ππ π β π π πππ π = Lesson 2 Multipication of Rational Expressions π π β π π = ππ ππ Step 1 Factor each expression. π ππ β π ππ = π (π)(π) β ( π) (π) ( π) (π) Step 2 Cancel all common factors. π (π)(π) * (π) (π) (π) (π) Step 3: Write what is left. π ππ β π ππ = π (π) (π) = π π Step 1 Factor each expression. Step 2 Cancel all common factors. Step 3: Write what is left. Step 1: Factor each expression. ππ ππ π β π π πππ π = ( π )( π) ( π)(π π) β (π π)(π π) ( π)( π)(π π) Step 2 Cancel all common factors. ππ ππ π β π π πππ π = ( π ) ( π) ( π) (π π) β (π π) (π π) ( π) ( π) (π π) Step 3: Write what is left. ππ ππ π β π π πππ π = π (π)(π) = π π
- 10. Ms. Lorie Jane L Letada 4. ππ ππ π β ππ π πππ π = π. ππ πβπ π πβππ+π β ππβπ ππ+ππ = π. π+π π β πππ π π π+ππ+ππ = π. ππ ππ+π β ππ+π ππ π = π . π πβπ β (πβπ ) π π = Step 1: Factor each expression. Step 2 : Cancel all common factors. Step 3: Write what is left. Step 1: Factor each expression. ππ πβπ π πβππ+π β ππβπ ππ+ππ = (ππβπ )(ππ+π) (πβπ)(πβπ) * π ( πβπ ) π(ππ+π) Step 2 : Cancel all common factors. ππ πβπ π πβππ+π β ππβπ ππ+ππ = (ππβπ )(ππ+π) (πβπ)(πβπ) * π ( πβπ ) π(ππ+π) Step 3: Write what is left. ππ πβπ π πβππ+π β ππβπ ππ+ππ = ( ππβπ ) ( πβπ) β π π = π ( ππβπ ) π ( πβπ ) Show your solution here .β¦. Step 1: Factor each expression. ππ ππ+π β ππ+π ππ π = ππ π(ππ+π) β (ππ+π) π ( ππ ) Step 2 : Cancel all common factors. ππ ππ+π β ππ+π ππ π = ππ π(ππ+π) β (ππ+π) π ( ππ ) ππ ππ+π β ππ+π ππ π = π π ( π) = π ππ Show your solution here .β¦. Cancelation involving rational expressions should be applied carefully. In the expression 2π₯β3 π₯β3 ,you cannot cancel x in the numerator and denominator even though π₯ π₯ is equal to 1 because it is not a factor of 2π₯β3 π₯β3 . πΌπ ππππ‘, 2π₯β3 π₯β3 is already in simplest form. Factor, in mathematics, a number or algebraic expression that divides another number or expression evenlyβi.e., with no remainder. For example, 3 and 6 are factors of 12 because 12 Γ· 3 = 4 exactly and 12 Γ· 6 = 2 exactly.
- 11. Module 2: Rational Expressions Math Focus To multiply rational expressions: β’ Write each numerator and denominator in factored form. β’ Divide out any numerator factor with any matching denominator factor. β’ Multiply numerator by numerator and denominator by denominator. β’ Simplify as needed. Multiply each rational expression and simplify. 1. ππ ππ β π π = 5. (π π+ππ+π ) πβπ * π+π π π+ππ+π = 2. πππ ππ π β ππ π πππ π = 6. (π πβπ ) ππ π * (ππ π ππ+π = 3. πππ ππ π β ππ π πππ π = 7. (π π+ππ+π ) π+π * ππ+ππ π+π = = 4. ππ+π ππβπ * (π π+πβπ ) π = Skill Booster!
- 12. Ms. Lorie Jane L Letada In dividing rational expressions, observe the same rules as when dividing fractions. If the denominators are not equal to zero, then Find the quotient of each rational expression. 1. π π Γ· ππ π 2. π ππ Γ· ππ ππ 3. ππ π π πππ π π π Γ· πππ π πππ π Lesson 3 Division of Rational Expressions π π Γ· π π = π π β π π = ππ ππ Step 1 : Get the reciprocal of the divisor and proceed to multiplication. 2 3 Γ· 10 9 = 2 3 β 9 10 Step 2 : Factor each expression and cancel all common factors. 2 3 Γ· 10 9 = 2 3 β 9 10 = 2 3 β ( 3 )(3) (5)(2) Step 3 : Write what is left 2 3 Γ· 10 9 = 3 5 Write your solution hereβ¦ Step 1 : Get the reciprocal of the divisor and proceed to multiplication. ππ π π πππ π π π Γ· πππ π πππ π = ππ π π πππ π π π * πππ π πππ π Step 2 : Factor each expression and cancel all common factors . ππ π π πππ π π π * πππ π πππ π = (π)(π)(π)(π) (π)(π)(π)(π)(π)(π)(π) * (π)(π)(π)(π)(π) (π)(π)(π)(π) Step 3 : Write what is left. π ππ π * π π = ππ ππ π
- 13. Module 2: Rational Expressions 4. πππ π ππ π Γ· πππ π πππ π 5. πβπ π πβ π π Γ· π+π π π+ πππ+ π π 6. π+π πβπ Γ· π π+ππ π πβπ π 7. ππ πβππ π+ π π ππ π π Γ· ππ βπ ππ π π Write your solution hereβ¦ Step 1 : Get the reciprocal of the divisor and proceed to multiplication. πβπ π πβ π π Γ· π+π π π+ πππ+ π π = πβπ π πβ π π β π π+ πππ+ π π π+π Step 2 and 3 : Factor each expression and cancel all common factors. Write, what is left. πβπ π πβ π π Γ· π+π π π+ πππ+ π π = πβπ (πβπ)(π+π ) β (π+π)(π+π) π+π πβπ π πβ π π Γ· π+π π π+ πππ+ π π = 1 Write your solution hereβ¦ Solution: ππ πβππ π+ π π ππ π π Γ· ππ βπ ππ π π = ππ πβππ π+ π π ππ π π * ππ π π ππ βπ = (ππ βπ)(ππ βπ) π(π )(π) * (π)(π)(π )(π )(π )(π) (ππ βπ) = (ππ βπ) π * π (π )(π ) π =3π π (ππ β π) Any number or variables divided by itself is always equal to 1. Example: (πβπ)(π+π) (πβπ)(π+π ) = 1 Any number or variables divided by 1 is always equal to itself. Examples: (ππ βπ) π = (3d-e) π (π )(π ) π = 3π π
- 14. Ms. Lorie Jane L Letada Find the quotient and simplify. 1. πππ π π π ππππ Γ· πππ π πππ = 4. π πβπβπ π+π Γ· ππ+π ππ+π = 2. π ππ Γ· π ππ = 5. ππ+ππ ππ Γ· ππ ππ+ππ = 3. π πβπ Γ· ππ π πβπ = 6. π πβππβππ π π+ππ+π Γ· π πβππ+ππ π πβπ = In your lower grade math, you usually learn how to add and subtraction fraction before you are taught multiplication and division. However, with rational expressions, multiplication and division are sometimes taught first because these operations are easier to perform than addition and subtraction. Addition and subtraction of rational expressions are not as easy to perform as multiplication because, as with numeric fractions, the process involves finding common denominators. By working carefully and writing down the steps along the way, you can keep track of all of the numbers and variables and perform the operations accurately. Skill Booster!
- 15. Module 2: Rational Expressions In adding and subtracting rational expressions with same denominators, perform the operation in the numerator and keep the denominator. The resulting sum and difference must be reduced to lowest terms whenever possible. Provided the denominator is not equal to zero, then Perform the indicated operations. 1. π π + π π 2. π π + π π 2. ππ+π π+ π + ππ π+ π 3. . ππ+π π+ π + ππ+π π+ π Lesson 4 Addition and Subtraction of Rational Expressions (Same Denominators ) π π + π π = π + π π π π β π π = π β π π Step 1 : Since the denominator is the same, add the numerator and copy the denominator. π π + π π = π+π π = π π Step 1 : Step 1 : Since the denominator is the same, combine the numerator and keep the denominator. ππ+π π+ π + ππ π+ π = ππ+π+ππ π+ π Step 2 : Combine like terms. ππ+π π+ π + ππ π+ π = ππ+ππ+π π+ π = ππ+π π+ π Solution:
- 16. Ms. Lorie Jane L Letada 4. ππ +π π+ π β πππ π+ π 5. π βπ πβ π β ππ πβ π 6. ππ+ π π πβππ+π β πβπ π πβππ+π 7. π + π π π+π β ππ+π π π+π Step 1 : Since the denominator is the same, combine the numerator and keep the denominator. ππ +π π+ π β πππ π+ π = ππ +π βπππ π+ π Step 2 : Combine like terms. ππ +π π+ π β πππ π+ π = ππ βπππ + π π+ π = βππ+π π+ π Solution: When you subtract two numbers, just follow the KCC Rules ; Example: 5r - 10r *Keep the Subtrahend 5r *Change the operation from minus to add (5r + * Change the sign of the minuend to its opposite sign from +10r to -10r 5r + -10r, and proceed to the rules of addition. 5r + -10r = -5r Solution: ππ+ π π πβππ+π β πβπ π πβππ+π = ππ+ π β (πβπ) π πβππ+π = ππ+ πβπ+π π πβππ+π = π+ π π πβππ+π Multiply negative to each term Combine like terms Solution: π + π π π+π β ππ+π π π+π = π + π β (ππ + π) π π +π = ---------------- = ---------------- Multiply negative to each term Combine like terms * When you multiply unlike terms, the result is negative. Example: (-1 )(w) = -w *When you multiply like terms, the result is positive. Example: (-1)(-1) = 1 *When you add unlike terms, the sign used in the result will be based which absolute num. is greater. Examples: e + -6e = -5e We use -5e since -6e has the greater absolute value and its sign is negative.
- 17. Module 2: Rational Expressions Perform the indicated operation for rational expressions with like denominators. 1. ππ πππ β ππ πππ 5. ππβ π π πβ π β ππβπ π πβ π 2. πππ ππ π + π ππ π 6. ππ+π (π+π) + π π (π+π) 3. π π π + π β π +ππ π+π 7. π π π+π + πππ+π π π+π 4. π + ππ π + ππβπ π π. ππ+ π π πβ ππ + πβππ π πβ ππ Skill Booster!
- 18. Ms. Lorie Jane L Letada When the denominators of the two rational algebraic expressions are different, we need to find the least common denominator (LCD) of the given expressions. The least common denominator must contain all prime factor of each denominator raised to the highest power. It is the common multiple of the denominator. Provided that the denominators are not equal to zero, then, Perform the indicated operation and simplify if possible. 1. π ππ + π ππ Lesson 5 Addition and Subtraction of Rational Expressions (Different Denominators ) π π + π π = ππ + ππ ππ π π β π π = ππ β ππ ππ Step 1 : Find the least common denominator (LCD). ππ = ( 2 ) ( r ) 3s = (3) (s) LCD = (2) (r ) (3)(s) LCD = (2) (3) ( r ) (s) LCD = 6rs Step 2 : Express each fraction with the LCD as the denominator. π ππ + π ππ = π (ππ)+π (ππ) πππ πππ ππ = 3s πππ ππ = 3s πππ ππ = 2r Step 3 : Add the numerator and simplify when possible. π ππ + π ππ = π (ππ)+π (ππ) πππ = ππ+ ππ πππ Observe how to add dissimilar fraction.
- 19. Module 2: Rational Expressions 2. π ππ + π ππ 3. π π+π + π πβπ Step 1 . Combine the denominator. π π+π + π πβπ = (π+π) ( πβπ) Step 2 . Cross Multiply. π π+π + π πβπ = π ( πβπ ) +π (π+π ) (π+π) ( πβπ) Step 1 : Find the least common denominator (LCD). ππ = 2y = LCD = LCD = LCD = Step 2 : Express each fraction with the LCD as the denominator. Step 3 : Add the numerator and simplify when possible. Step 3 . Multiply the numerator. π π+π + π πβπ = π ( πβπ ) +π (π+π ) (π+π) ( πβπ) = π πβπ + ππ +π (π+π) ( πβπ) = π π+π+ππβπ (π+π) ( πβπ) You can only add or subtract if they have the same variables and exponents regardless of its numerical coefficients Example: ππ π and ππ π
- 20. Ms. Lorie Jane L Letada 4. ππ π +π - π πβπ 5. π+π π +π - πβπ πβπ n Put your solution hereβ¦.. Step 1 . Combine the denominator. π+π π +π - πβπ πβπ = ( π+π)β(πβπ) ( π +π) (πβπ ) Step 2 . Cross Multiply. π+π π +π - πβπ πβπ = [( π+π)( πβπ ) ]β[ ( πβπ)( π+π )] ( π +π)( πβπ) Step 3 . Multiply the numerator. π+π π +π - πβπ πβπ = [( π+π)( πβπ ) ]β[ ( πβπ)( π+π )] ( π +π)( πβπ) π+π π +π - πβπ πβπ = (π πβππβπ) β( π πβππβπ ) π πβπ Use FOIL method to simplify these the numerator. Use FOIL method to simplify these the denominator. Step 4. Simplify the numerator. π+π π +π - πβπ πβπ = (π πβππβπ) β( π πβππβπ ) π πβπ π+π π +π - πβπ πβπ = π πβππβπ β π π+ ππ +π π πβπ π+π π +π - πβπ πβπ = π π πβπ = 0 Multiply negative to each terms. Combine like terms,
- 21. Module 2: Rational Expressions 6. π+π π + π - π βπ πβ π 7. π +π π β π - π βπ π β π 8. π +π πβ π + π βπ π+ π Put your solution hereβ¦. Solution: π +π π β π - π βπ π β π = [( π +π)( πβπ)]β[( π βπ)(π βπ) ( π β π)(πβπ ) = (π πβππβπ ) β(π πβππ +π) ππ βππβππ +π = π πβππβπ β π π+ππ βπ ππ βππβππ +π = π πβ π πβππ +ππ βπ ππ βππβππ +π Cross multiply Get the LCD Use FOIL Method Combine like terms Solution: Like Terms - both terms have the same variable group abc; ex. abc and 22abc Unlike Terms are terms that are not in group ;example β4π¦3 and 5π₯3 because the variables are different
- 22. Ms. Lorie Jane L Letada Perform the indicated operation and simplify. π. π ππ + π ππ 4. ππ π +π - ππ π βπ π. π ππ + π ππ 5. π +π πβ π + π βπ π+ π π. ππ π +π - π πβπ 6. π +π πβ π + ππ π + π Skill Booster! Math Guro Math Guro Rational Expressions β nothing more than a fraction in which the numerator and/or the denominator are polynomials. Here are some examples of rational expressions. Fractions βrepresents a part of a whole or, more generally, any number of equal parts. Numerator βthe number above the line in a common fraction showing how many of the parts indicated by the denominator are taken, for example, 2 in 2/3. Denominator βthe number below the line in a common fraction; a divisor. Like Terms β Like terms - both terms have the same variable group abc; ex. abc and 22abc Unlike Terms β examples β4π¦3 and 5π₯3 because the variables are different Similar Fraction β fractions that have the same denominator; Examples: π π , π π and 4 3 Dissimilar Fraction β fractions that have different denominators; Examples: π π , π π and π π Variables β A symbol for a number we don't know yet. It is usually a letter like x or y. Example: in x + 2 = 6, x is the variable. Literal Coefficients β literal coefficient is a variable used to represent a number
- 23. Module 2: Rational Expressions The only application of rational equations that we will cover in this class is shared work problems, which means two or more people (or objects) working together on a job to share the work. βͺ Person A does a job in 2 hours, so their rate is 1 2 the job per hour βͺ Person B does the same job in 3 hours, so their rate is 1 3 of the job per hour βͺ Persons A and B do the job together in π₯ hours, so their rate together is 1 π₯ of the job per hour 1 2 + 1 3 = 1 π₯ These are the steps on how to solve work problems. 1. Make sure the units of time are consistent. For example, if one person does a job in 45 minutes, and another person does the same job in 2 hours, convert both rates to either minutes or hours. (60 mins. x 2 hrs. = 120 mins.) In terms of minutes, the equation would be 1 45 + 1 120 = 1 π₯ 2. Make sure your answer makes sense . If one person does a job in 45 minutes, and another person does the same job in 2 hours, it shouldnβt take both people working together more than 45 minutes to do the job. 6 Lesson Problem Solving Involving Rational Expressions
- 24. Ms. Lorie Jane L Letada 1. It takes a boy 90 minutes to mow the lawn, but his sister can mow it in an hour. How long (in minutes) would it take them to mow the lawn if they worked together using two lawn mowers? Let us look at the following problems. Step 1: Number of minutes that the boy and his sister work together Step 2: Let 1 π₯ = number of mins. work together π ππ + π ππ = π π Step 3: π ππ + π ππ = π π 5400π [ π ππ + π ππ = π π ] 5400x 60x + 90x = 5400 150x = 5400 πππ πππ π = ππππ πππ X = 36 Step 4: Therefore, there are 36 minutes to mow the lawn.
- 25. Module 2: Rational Expressions 2. Two pipes can be used to fill a swimming pool. When the first pipe is closed, the second pipe can fill the pool in 9 hours. When the second pipe is closed, the first pipe can fill the pool in 7 hours. How long (in hours) will it take to fill the pool if both pipes are open? Round your answer to two decimal places. Step 1: Number of hours that both pipes will fill the pool. Step 2: Let 1 π₯ = number of hours that the pipes will fill together the pool π π + π π = π π Step 3: π π + π π = π π 63π [ π π + π π = π π ] 63x 7x + 9x = 63 16x = 63 ππ ππ π = ππ ππ X = 3.94 X = 3 hrs. , 56 mins (0.94 x 60) and 15 sec (.25 x 60) Step 4: Therefore, there are 3.94 hours( to fill the pool together. LCD (9,7, X) 63x Γ· 9 63π₯ π₯ LCD (9,7, X) When you want to get the LCD in a rational equation and perform the operation, make sure to divide both side with the LCD. Example: 63π [ π π + π π = π π ] 63x
- 26. Ms. Lorie Jane L Letada 3. Walter and Helen are asked to paint a house. Walter can paint the house by himself in 12 hours and Helen can paint the house by herself in 16 hours. How long would it take to paint the house if they worked together? 4. On average, Mary can do her homework assignments in 100 minutes. It takes Frank about 2 hours to complete a given assignment. How long (in minutes) will it take the two of them working together to complete an assignment? Step 1: Step 2: Step 3: Step 4: Step 1: Number of minutes that both of them will finish the assignment together. Step 2: Let x = num. of minutes Finish the task together 1 100 + 1 120 = 1 π₯ Step 3: π πππ + π πππ = π π 12000π [ π πππ + π πππ = π π ] 12000π 120x + 100x = 12000 220x = 12000 πππ πππ π = πππππ πππ x = 54.55 x = 54.55 mins Step 4: Therefore, there are 54.55 minutes to finish the assignment together.
- 27. Module 2: Rational Expressions Read and analyze each item carefully and solve the problem. 1. Working together, Bill and Tom painted a fence in 4 hours. If Tom has painted the same fence before by himself in 7 hours, how long (in hours) would it take Bill on his own? Round your answer to one decimal place. 2. Tom and Jerry have to stuff and mail 1000 envelopes for a new marketing campaign. Jerry can do the job alone in 6 hours. If Tom helps, they can get the job done in 4 hours. How long would it take Tom to do the job by himself? 3. One pipe can fill a swimming pool in 10 hours, while another pipe can empty the pool in 15 hours. How long would it take to fill the pool if both pipes were accidentally left open? 4. One roofer can put a new roof on a house three times faster than another. Working together they can roof a house in 5 days. How long would it take the faster roofer working alone? 5. Triplets, Justin, Jason, and Jacob are working on a school project. Justin can complete the project by himself in 6 hours, Jason can complete the project by himself in 9 hours, and Jacob can complete the project by himself in 8 hours. How long would it take the triplets to complete the project if they work together? Skill Booster!
- 28. Ms. Lorie Jane L Letada Lesson 1: Simplifying Rational Expression. A Rational Expressions is reduced to its simplest form if the numerator and denominator have no common factors. Lesson 2: Multiplying Rational Expressions To multiply rational expressions, recall the rules for multiplying fractions. If the denominators are not equal to zero, then we simply multiply the numerators and denominators. The same rule applies to rational expressions. If a, b, c, and d represent polynomials where b β 0 and d β 0. Then, Lesson 3: Dividing Rational Expressions In dividing rational expressions, observe the same rules as when dividing fractions. If the denominators are not equal to zero, then Sum It Up ! Step 1 Factor the numerator and denominator and get the GCF. 4a = (2) (2) a 12b= (3) (2) (2) b GCF: (2) (2) GCF: 4 Thus, the common factor is 4. Step 2: Divide out the common factor. 4π 12π = 4 ( π ) 4 (3π) = π 3π Thus, the simplify is π 3π . π π β π π = ππ ππ π π Γ· π π = π π β π π = ππ ππ
- 29. Module 2: Rational Expressions Lesson 4: Adding and Subtracting Rational with Same Denominator In adding and subtracting rational expressions with same denominators, perform the operation in the numerator and keep the denominator. The resulting sum and difference must be reduced to lowest terms whenever possible. Provided the denominator is not equal to zero, then Lesson 5: Adding and Subtracting Rational with Different Denominator When the denominators of the two rational algebraic expressions are different, we need to find the least common denominator (LCD) of the given expressions. Provided that the denominators are not equal to zero, then, Lesson 6: Work Problems Involving Rational Equations Shared work problems mean two or more people (or objects) working together on a job to share the work. βͺ Persons A and B do the job together in π₯ hours, so their rate together is 1 π₯ of the job per hour 1 2 + 1 3 = 1 π₯ π π + π π = π + π π π π β π π = π β π π π π + π π = ππ + ππ ππ π π β π π = ππ β ππ ππ
- 30. Ms. Lorie Jane L Letada A. Direction: Read and analyze each questions carefully and encircle the letter of the correct answer. 1. Which of the following expressions is a rational algebraic expression? 2. What is the value of a non-zero polynomial raised to 0? a. constant c. undefined b. zero d. cannot be determined 3. What rational algebraic expression is the same as π₯2β1 π₯β1 ? a. x + 1 c. 1 b. x β 1 d. -1 4. You own 3 hectares of land and you want to mow it for farming. What will you do to finish it at the very least time? a. Rent a small mower. c. Do kaingin. b. Hire 3 efficient laborers. d. Use germicide. 6. Angelo can complete his school project in x hours. What part of the job can be completed by Angelo after 3 hours? a. x + 3 b. x β 3 c. π₯ 3 d. 3 π₯ Post-Test!
- 31. Module 2: Rational Expressions 9. Your father, a tricycle driver, asked you regarding the best motorcycle to buy. What will you do to help your father? a. Look for the fastest motorcycle. b. Canvass for the cheapest motorcycle. c. Find an imitated brand of motorcycle. d. Search for fuel-efficient type of motorcycle. 10. Your friend asked you to make a floor plan. As an engineer, what aspects should you consider in doing the plan? a. Precision c. Appropriateness b. Layout and cost d. Feasibility 11. As a contractor in number 19, what is the best action to do in order to complete the project on or before the deadline but still on the budget plan? a. All laborers must be trained workers. b. Rent more equipment and machines. c. Add more equipment and machines that are cheap. d. There must be equal number of trained and amateur workers.
- 32. Ms. Lorie Jane L Letada 15. Your SK Chairman planned to construct a basketball court. As a contractor, what will you do to realize the project? a. Show a budget proposal. b. Make a budget plan. c. Present a feasibility study. d. Give a financial statement.
- 33. Module 2: Rational Expressions B. Direction: Classify the following expressions if it either Rational Expression or not. Put each expression on the correct box. Rational Algebraic Expressions Not Rational Algebraic Expressions
- 34. Ms. Lorie Jane L Letada C.Direction: Match each rational algebraic expression to its equivalent simplified expression from choices A to E. Write the rational expression in the appropriate column. If the equivalent is not among the choices, write it in column F. D. Hours and Prints!
- 35. Module 2: Rational Expressions
- 36. Ms. Lorie Jane L Letada References Diaz, Z., Mojica M. (2013) . Next Century Mathematics 8; Quezon City ; Phoenix Publishing House , Inc Escaner, J., Sepida, M., Catalla, D. (2013)., Spiral Math 8; Quezon Ciity; Trinita Publishing , Inc Mathematics 8 Learnerβs Module K-12; DepEd K-12 Modified Curriculum Guide and Teacherβs Guide for Mathematics http://www.mesacc.edu/~scotz47781/mat120/notes/work/work_probs.htmlhttps://web.ics.purdue.edu/ ~pdevlin/Traditional%20Class/Lesson%2011/Applications%20of%20Rational%20Equations.pdf https://www.google.com/search?safe=active&sxsrf=ALeKk02mur0lCuIoGdtBBGuCWadLdxUynw%3A1594786462276&e i=noIOX7ivEJTdmAW-iY- IBQ&q=adding+and+subtracting+rational+expressions&oq=ADDING+AND+SUBTR&gs_lcp=CgZwc3ktYWIQARgAMgQIAB BDMgQIABBDMgQIABBDMgQIABBDMgQIABBDMgQIABBDMgQIABBDMgQIABBDMgQIABBDMgQIABBDOgQIIxAnOgUIA BCRAjoHCAAQsQMQQzoFCAAQsQM6CggAELEDEIMBEEM6AggAOgQIABAKUOM6WIVxYN58aANwAHgAgAGvA4gBsyaSA QkwLjkuOC4zLjGYAQCgAQGqAQdnd3Mtd2l6&sclient=psy-ab https://www.youtube.com/watch?v=uVpsz-xpnPo https://www.youtube.com/watch?v=1KGVwaUhR-s https://www.youtube.com/watch?v=RROSgr4oXjU https://www.youtube.com/watch?v=WbvLjmK4Kmc