Question and Answers #5 Fractions

Questions and Answers Sheet 5 Fractions Question #1 Determine if the given problem is an equation or an expression. If it is an equation, then solve. If it is an expression, then simplify. 2𝑥 + 1 3𝑥 − 1 1 + = 3 5 10 Answer: Step 1 Given, The problem is 2𝑥+1 3 + 3𝑥−1 5 = 1 10 Step 2 An expression is a number, a variable, or a combination of numbers and variables and operation symbols. An equation is made up of two expressions connected by an equal sign. Hence it is an equation. Least common denominator ( LCD) is the lowest common multiple of the denominators of a set of fractions. To solve this rational equation the least common denominator is 30. Applying LCD through out the equation. 30 × 2𝑥 + 1 3𝑥 − 1 1 + 30 × = 30 × 3 5 10 10(2𝑥 + 1) + 6(3𝑥 − 1) = 3 20𝑥 + 10 + 18𝑥 − 6 = 3 Combining like terms, 38𝑥 + 4 = 3 38𝑥 = −1 𝑥=− 1 38 Thus, 𝑥 = − 1 38 is solution of the given equation. Question #2 How to simplify this without using fractions in the power? We didn't learn to rationalize with fractions in our school. 3 3 6 (𝑥 √𝑥) ⋅ 3√𝑥 √𝑥 = 𝑥 4 √𝑥 5 Answer: Step 1 Simplify the powers and cancel out the common terms from both the sides. (𝑥 √{𝑥})^{3}3(√{𝑥^{3}√{𝑥}} = 𝑥^{4} √[6]{𝑥^{5}} 𝑥^{3}(√{𝑥})^{3}3(√{𝑥^{3}√{𝑥}} = 𝑥^{4} √[6]{𝑥^{5}} 𝑥^{3}(√{𝑥}√{𝑥}√{𝑥})3(√{𝑥^{3} √{𝑥}} = 𝑥^{4} √[6]{𝑥^{5}} (𝑥^{4} √{𝑥})3(√{𝑥^{3}}√{√{𝑥}} = 𝑥^{4} √[6]{𝑥^{5}} Step 2 Cancel out the common factors from both the sides and simplify the radicals. (𝑥^{4} √{𝑥})3(√{𝑥^{3}}√{√{𝑥}} = 𝑥^{4} √[6]{𝑥^{5}} 6 3(√𝑥)(𝑥 √𝑥) (√√𝑥) = √𝑥 5 6 3(𝑥 2 ) (√√𝑥) = √𝑥 5 Question #3 determine 𝐿−1 [𝐹(𝑠)𝐺(𝑠)] in the following two ways: (a) using the Convolution Theorem, (b) (b) using partial fractions. 𝐹(𝑠) = , 𝐺(𝑠) = 1 1 𝑠 𝑠−2 Answer: Step 1 Given Data The first function is 𝐹(𝑠) = 1/𝑠. The second function is 𝐺(𝑠) = 1/𝑠 − 2. Using the convolution theorem, the expression for the inverse Laplace is, 𝐿−1 [𝐹(𝑠)𝐺(𝑠)] = 𝐿−1 [𝐹(𝑠)] ⋅ 𝐿−1 (𝐺(𝑠)) 1 1 = 𝐿−1 ( ) ⋅ 𝐿−1 ( ) 𝑠 𝑠−2 = 1 ⋅ 𝑒 2𝑡 =𝑓 ⋅𝑔 Step 2 The convolution product of two continuous function f and g is, 𝑡 𝑓 ⋅ 𝑔 = ∫ 𝑓(𝑡 − 𝜏)𝑔(𝜏)𝑑 0 𝑡 1 ⋅ 𝑒 2𝑡 = ∫ 1 ⋅ 𝑒 2𝜏 𝑑 0 1 1 = 𝑒 2𝑡 − 2 2 Hence the Laplace inverse using convolution theorem is 1/2𝑒 2𝑡 − 1/2. Step 3 The expression for the sum of the partial fraction of F(s)G(s) is, 𝐹(𝑠)𝐺(𝑠) = Step 4 1 𝑠(𝑠−2) = 1 2(𝑠−2) − 1 2𝑠 The Laplace inverse of F(s)G(s) using partial fraction is, 𝐿−1 (𝐹(𝑠)𝐺(𝑠)) = 𝐿−1 [ 1 2(𝑠−2) 1 − ] 2𝑠 1 1 𝐿−1 ( ) − 𝐿−1 ( ) 2(𝑠 − 2) 2𝑠 1 1 = 𝑒 2𝑡 − 2 2 Hence the Laplace inverse using partial fraction is 1/2𝑒 2𝑡 − 1/2. Question #4 1 2 2 ⋅3 = 7 3 Please step by step and discuss rules in multiplying with whole numbers & fractions. Answer: Step 1 2+ 3+ 1 7 2 3 2×7+1 can be written as, can be written as, 7 3×3+2 3 = = 14+1 7 = 15 7 11 3 Step 2 So, adding them, we get, 15 7 + 11 3 15 × 3 + 11 × 7 7×3 = 45 + 77 21 = 122 21 =5+ 17 21 Question #5 To multiply two rational expressions, we multiply their ? together and multiply their ? together. So 2 𝑥+1 ⋅ 𝑥 𝑥+3 is the same as ?. Answer: Step 1 Multiplication of rational number is similar to that of fraction. Multiplication of two rational number again produce 𝑝 a rational number itself. We know that rational numbers are the numbers which can be expressed in the form , 𝑞 ≠ 𝑞 0. Step 2 When we have to multiply two rational numbers, we have to multiply the numerator of first rational number with the numerator second rational number and we have to multiply the denominator of first rational number with the denominator of second rational number. Hence we get the required answers as When we multiply two rational numbers we multiply their Numerator together and multiply their denominator together. Hence we have 2 𝑥+1 ⋅ 𝑥 𝑥+3 2𝑥 = (𝑥+1)(𝑥+3) Hence the multiplication two rational numbers are same as that of two fractions. Question #6 Rewrite the equation below so that is does not have fractions 2 3 𝑥+3= 2 7 do not use decimals in your answer Answer: Step 1 If we have an equation of the form 𝑥/𝑎 + 𝑦/𝑏 = 𝑧/𝑐 where a,b,c are non zero then To make a given equation so that it does not have fractions we can multiply the given equation by lcm(a,b,c). Step 2 2 2 𝑥 + 3 = … . (1) 3 7 Multiply equation(1) by 𝑙𝑐𝑚(3,1,7) = 21 then we get 14𝑥 + 63 = 6 Question #7 5 1 1 𝑥+3= + 𝑥 4 4 4 Answer: Step 1 Given, 5 1 1 4 4 4 The equation is, 𝑥 + 3 = + 𝑥 Step 2 5 3 1 1 4 1 4 4 The given equation can be rewrite as, 𝑥 + = + 𝑥 This is an rational equation. So to solve this equation first find the least common denominator (LCD), The least common denominator (LCD) is the smallest number that can be a common denominator for a fractions. So for the least common denominator (LCD) find the least common multiple of the denominators 4, 1,4,4. Thus the least common denominator (LCD) is 4. Multiply by LCD 4 to the both side of the equation, ⇒ 5𝑥 + 12 = 1 + 𝑥 Subtracting 12 from both sides, ⇒ 5𝑥 + 12 − 12 = 1 + 𝑥 − 12 ⇒ 5𝑥 = −11 + 𝑥 Subtracting x from both sides, ⇒ 5𝑥 − 𝑥 = −11 + 𝑥 − 𝑥 ⇒ 4𝑥 = −11 Dividing by 4 to the both sides, ⇒𝑥= −11 4 ⇒ 𝑥 = −2.75 Thus the solution of the given equation is 𝑥 = −2.75 Question #8 I am having trouble with these type of problems. 3 5 (9𝑦 − 4) + 9 16 (15𝑦 + 12) I cant figure out how to simplify with fractions? Answer: Step 1 Given: The expression, 3 5 (9𝑦 − 4) + 9 16 (15𝑦 − 12) Step 2 To add these two fractions we have to make denominator common For this multiply and divide both term by numbers so that we get least common denominator Therefore multiply and divide first term by 16 and second term by 5. 3 ⋅ 16 9⋅5 (9𝑦 − 4) + (15𝑦 + 12) 5 ⋅ 16 16 ⋅ 5 Simplifying, 48 45 (9𝑦 − 4) + (15𝑦 + 12) 80 80 = 432 192 675𝑦 540 𝑦− + + 80 80 80 80 = 1107𝑦 348 + 80 80 = 1107𝑦 + 348 80 Step 3 3 9 5 16 Therefore, given expression simplifies to (9𝑦 − 4) + (15𝑦 + 12) = 1107𝑦+348 80 Question #9 Write the domain of the function in interval notation.Write numbers as integers or simplified fractions. 𝑔(𝑥) = 2𝑥 − 3 over 10𝑥 2 − 33𝑥 − 7 Answer: Step 1 To Determine: Find the domain of given function: 𝑔(𝑥) = 2𝑥 − 3 − 33𝑥 − 7 10𝑥 2 Step 2 Explanation: Domain is the set of all real x, where function is defined. Given a rational function. To find the domain of rational function, we need to find the those x, where function is undefined. If denominator is 0, then the whole fraction is not defined. So, we will set the denominator 0 and find the x 10𝑥 2 − 33𝑥 − 7 = 0 (5𝑥 + 1)(2𝑥 − 7) = 0 5𝑥 + 1 = 0 2𝑥 − 7 = 0 5𝑥 − 1 2𝑥 = 7 1 7 𝑥=− 𝑥= 5 2 So, both are the zeroes of the denominator. 1 7 5 2 It means at 𝑥 = − and 𝑥 = given function is undefined. 1 7 5 2 Domain is all real x except 𝑥 = − and 𝑥 = Step 3 Answer: Domain: 𝐷: {𝑥 ∈ 𝑅: 𝑥 ≠ 𝐷: (−∞, −1 5 7 and 𝑥 ≠ } 2 −1 −1 7 7 ) ∪ ( , ) ∪ ( , ∞) 5 5 2 2 Question #10 If the probability of a certain event not happening is 49 82 , what is the probability of the event happening? Write all answers as simplified fractions. Answer: Step 1 Let's assume, event 𝐴 = Probability of event happening ∵ 𝑃(𝐴′ ) = 49 82 event (A')= probability of event not happening 𝑛 ∑ 𝑃(𝐴𝑖) = 1 𝑙=1 ∵ 𝑃(𝐴) + 𝑃(𝐴′ ) = 1 Answer: ∵ 𝑃(𝐴) = 1 − 49 82 = 33 82 Step 2 Feel free to reach out for any of your queries or concerns. Question #11 Im thinking of two proper fractions whose sum is 19 12 them in simplest form. Answer: Step 1 Let two fractions are a and b. Then according to question: 𝑎+𝑏 = 19 … . . (1) 12 5 And 𝑎𝑏 = … . . (2) 8 Step 2 Now using formula (𝑎 − 𝑏)2 = (𝑎 + 𝑏)2 − 4𝑎𝑏 (𝑎 − 𝑏)2 = 192 5 −4 12 8 (𝑎 − 𝑏)2 = 361 5 361 − 360 − = 144 2 144 (𝑎 − 𝑏)2 = 1 144 (𝑎 − 𝑏) = 1 12 or (𝑎 − 𝑏) = − 1 12 Step 3 If (𝑎 − 𝑏) = 1 12 and (𝑎 + 𝑏) = 19 12 Then a= \frac{5}{6}\ and\ b=\frac{3}{4} If (𝑎 − 𝑏) = − 1 12 and (𝑎 + 𝑏) = 3 5 4 6 Then 𝑎 = and 𝑏 = 19 12 Question #12 𝑥 2 +𝑥−6 Find the partial fraction decomposition. (𝑥 2 Answer: +1)(𝑥−1) and whose product is 5 8 . Determine thefractions, and state Step 1 Formations of Partial fractions is method to separate a fraction into its components such that when the components are added together, they give back the original fraction. This method is very useful in integration or while performing Fourier transforms, Laplace transforms of various functions . Step 2 The given function is represented as 𝑥 2 +𝑥−6 . (𝑥 2 +1)(𝑥−1) Partial fractions for the above function would be of the form 𝐴𝑥+𝐵 𝑥 2 +1 Equating two functions to find the value of A, B and C. 𝑥2 + 𝑥 − 6 𝐴𝑥 + 𝐵 𝐶 = 2 + 2 (𝑥 + 1)(𝑥 − 1) 𝑥 + 1 𝑥 − 1 = (𝐴𝑥 + 𝐵)(𝑥 − 1) + 𝐶(𝑥 2 + 1) (𝑥 2 + 1)(𝑥 − 1) 𝑥 2 + 𝑥 − 6 = (𝐴 + 𝐶)𝑥 2 + (−𝐴 + 𝐵)𝑥 + (−𝐵 + 𝐶) Equating coefficients: 𝐴+𝐶 =1 −𝐴 + 𝐵 = 1 −𝐵 + 𝐶 = −6 𝐵+𝐶 =2 2𝐶 = −4 𝐶 = −2 𝐴=3 𝐵=4 Hence the partial fraction formed is: 𝑥 2 +𝑥−6 (𝑥 2 +1)(𝑥−1) = 3𝑥+4 𝑥 2 +1 Question #13 Solve for the indicated variable. Fully reduce fractions. a.) Solve 2𝐿 + 2𝑊 = 𝑃2𝐿 + 2𝑊 = 𝑃 𝑓𝑜𝑟 𝑊 b.) 2𝑥𝑦𝑧 = 242𝑥𝑦𝑧 = 24 for y Answer: Step 1 Given equation is: 2𝐿 + 2𝑊 = 𝑃 Solve for W. Step 2 Now solve for W, − 2 𝑥−1 + 𝐶 𝑥−1 2𝐿 + 2𝑊 = 𝑃 2𝑊 = 𝑃 − 2𝐿 𝑊= 𝑃 − 2𝐿 2 Thus, the value of W is 𝑊 = 𝑃−2𝐿 2 Step 3 b) Given equation is: 2𝑥𝑦𝑧 = 24 Solve for y. Now solve for y, 2𝑥𝑦𝑧 = 24 𝑦= 24 2𝑥𝑧 𝑦= 12 𝑥𝑧 Thus, the value of W is 𝑦 = 12 𝑥𝑧 Question #14 Please explain the addition property of equality with fractions Answer: Step 1 The additive property of equality says what we add to one side of the equality, we must add the same thing to the other side. Step 2 Suppose we have an equation: 1 3 4 4 𝑥 − = and we need to solve for x Step 3 In order to solve for x, we need to add 1/4 to both sides of the equation according to the additive property of equality as shown on the board 𝑥− 1 1 3 1 + = + 4 4 4 4 Step 4 3 1 4 4 Cancel out -1/4 and +1/4 on the left side and we get the equation as shown on the board 𝑥 = + Step 5 Since both the fractions have a common denominator, we can simply add up their numerators as shown on the board 3+1 𝑥= 4 Step 6 Simplifying the numerator we get 𝑥 = 4 4 On further simplification the 4's get cancelled out on the top and bottom and we get: 𝑥=1 Step 7 Final answer 𝑥=1 Question #15 1) Kareem lives 4 10 of a mile from the mall. Write two equivalent fractions that show what fraction of a mile Kareem lives from the mall. 2) An electronics store sells a large flat screen television for 1,699. Last month, the store sold 8 of these television sets. About how much money did the televisions sell for? Answer: Step 1 Answer(14): Suppose, 𝑎 is any fraction, then the equivalent fraction of 𝑏 Since given that Kareem lives 4 10 𝑎 𝑏 is 𝑎×𝑛 𝑏×𝑛 , where n=1,2,3 ... of a miles from the mall. So, to find the equivalent fractions of 4 10 (distance between Kareem house and mall), multiply it's by 2 in numerator and denominator and also multiply 3 in numerator and denominator of That is, And, 4×2 10×2 4×3 10×3 = = 4 10 . 8 20 12 30 Therefore, two required equivalent fractions that show the fraction of a mile Kareem lives from the mall are 12 30 Step 2 Answer(3): Given that, in the last month, he sold one large screen television for $1,699. So, the price of one large screen television is $1,699. Then, the price of eight large-screen television is $1,699 \𝑡𝑖𝑚𝑒𝑠 8 = $13,592 Therefore, the money for sold 8 large-screen television is $13,592 Question #16 Discuss: Recognizing Partial Fraction Decompositions For each expression, determine whether it is already a partial fraction decomposition or whether it can be decomposed further. a) 𝑥 𝑥 2 +1 + 1 𝑥+1 8 20 and b) c) 𝑥 (𝑥+1)2 1 𝑥+1 2 + (𝑥+1)2 Answer: Step 1 We have to see if the given fractions are partially fraction decomposition or if they can be decomposed further. For partially fraction, we have to factor the denominator and write it separately in the fraction Step 2 Part a We have the fraction 𝑥 1 1 + + 𝑥2 + 1 𝑥 + 1 𝑥 + 1 Here we can see that the fractions are already factored and written separately. Hence this fraction is already partially decomposed. Step 3 Part b Here we have the fraction 𝑥 (𝑥 + 1)2 𝑥 𝐴 𝐵 If we see the denominator, it can further be factored as (𝑥=1)2 = (𝑥+1) + (𝑥+1)2 Therefore, the given fraction was not partially decomposed. Step 4 Part c Here we have the fraction 1 𝑥+1 2 + (𝑥+1)2 The fraction given here is already partially decomposed as the denominator given here is already partially decomposed. Note- since you have posted a question with multiple subparts, we will solve only first three subparts for you. To get the remaining done please resubmit the complete question and mention the subparts to be solved. Question #17 Tell whether each number in the list is a real number, a ratonal number, an irrational number, an integer, or a whole number. Then order the numbers from least to greatest. √64, −5, √9, 2 −5, 2, 9, 64 Answer: Step 1 Since you have asked multiple questions, we are answering the first question for you as of now. If you wish to get the other questions answered as well, kindly submit it again and mention the question numbers to be answered. Whole numbers are basic counting 0,1,2,.. etc Real numbers are the numbers which include both rational and irrational numbers. A rational number is a number that is expressed as the ratio of two integers, where the denominator should not be equal to zero, whereas an irrational number cannot be expressed in the form of fractions. An integer is a number that can be written without a fractional component Step 2 √64, −5, √9, 2 Here, √64 = 8 . It a real, rational, whole number. -5 is a real, rational and an integer. √9 = 3 . It a real, rational, whole number. 2 is a real, rational, whole number. In ascending order: −5,2, √9, √64 Question #18 Mary cut a piece of yarn that is 9 5 feet long into 3 pieces. List 3 different fractions that could show the length of the 3 fractions in ft. Answer: Step 1 Mary cut a piece of yarn that is 9 5 feet long into 3 pieces. Step 2 Since she cut the yarn into 3 pieces, divide the total length of the yarn by 3. 9 ( ) 5 = 9 = 9 =3 3 5 × 3 15 5 The length of a single piece of the yarn is 3 5 feet. Add all the three pieces and check whether the total length of the yarn 3 5 3 3 3+3+3 5 5 5 + + = 9 3 5 5 = Thus, the lengths of three pieces are feet, 3 5 9 5 or not. feet and 3 5 feet. Question #19 Among all pairs of numbers with a sum of 149, find the pair whose product is maximum. Write your answers as fractions reduced to lowest terms. Answer: Step 1 given that sum of all pair of numbers is 149 and product of pair is maximum find the pair and write interms of fractions Step 2 let x and y be two numbers then 𝑥 + 𝑦 = 149 𝑥𝑦 = 𝑚𝑎𝑥 from 𝑥 + 𝑦 = 149 𝑥 = 149 − 𝑦 substitute in product term (149 − 𝑦)𝑦 = 149𝑦 − 𝑦 2 (149 − 𝑦)𝑦 = (𝑦 2 − 149𝑦) (149 − 𝑦)𝑦 = − (𝑦 2 − 149𝑦 + ( (149 − 𝑦)𝑦 = − (𝑦 − 149 2 149 ) + 2 2 the expression is maximum at 𝑦= 149 2 𝑤ℎ𝑒𝑛 (𝑦 − 149 2 2 ) =0 149 2 𝑥 = 149 − 𝑥= 49 2 149 ) )+ 2 2 149 2 149 2 maximum product is 𝑥 = 𝑦 = 149 2 Question #20 Add the fractions. If possible, simplify your answer. 𝑎2 7 8 + 2 =? + 6𝑎 + 9 𝑎 − 9 Answer: Step 1 We have given to add fractions. 𝑎2 7 8 + 2 + 6𝑎 + 9 𝑎 − 9 = 𝑎2 + 6𝑎 + 9 = 𝑎2 + 2 ⋅ 3𝑎 + 32 = (𝑎 + 3)2 = 𝑎2 − 9 = (𝑎 − 3)(𝑎 + 3) = 𝑎2 7 8 7 8 + 2 = + 2 (𝑎 + 3)(𝑎 − 3) + 6𝑎 + 9 𝑎 − 9 (𝑎 + 3) = 7(𝑎 − 3) + 8(𝑎 + 3) (𝑎 + 3)2 (𝑎 − 3) = 7𝑎 − 21 + 8𝑎 + 24 (𝑎 + 3)2 (𝑎 − 3) = 15𝑎 + 3 (𝑎 + 3)2 (𝑎 − 3) = 3(5𝑎 + 1) (𝑎 + 3)2 (𝑎 − 3) = 7 8 3(5𝑎 + 1) + = 𝑎2 + 6𝑎 + 9 𝑎2 = 9 (𝑎 + 3)2 (𝑎 − 3) Step 2 3(5𝑎+1) Hence on adding fractions we get result equal to (𝑎+3)2(𝑎−3)