Question and Answers #19 Fractions

Questions and Answers Sheet 19 Fractions Question #1 A bicyclist rides for 4 h at a speed of 12 mph and then returns at a speed of 16 mph. Find the cyclist's average speed for the trip. (Enter your answer as a fraction. Reduce all fractions completely.) Answer: Step 1 Given that a bicyclist rides for 4 h at a speed of 12 mph so distance covers by a bicyclist are 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 = 𝑠𝑝𝑒𝑒𝑑 × 𝑡𝑖𝑚𝑒 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 = 12 × 4 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 = 48 𝑚𝑖𝑙𝑒𝑠 ...........(1) Step 2 Now the time is taken in returning by the speed of 16 mph 𝑡𝑖𝑚𝑒 = 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑠𝑝𝑒𝑒𝑑 𝑡𝑖𝑚𝑒 = 48 16 𝑡𝑖𝑚𝑒 = 3 ℎ𝑜𝑢𝑟𝑠 ..............(2) Step 3 So, the cyclist's average speed for the trip average speed = total distance covers total time taken average speed = 48 + 48 4+3 average speed = 96 𝑚𝑝ℎ 7 Question #2 In the following exercises, simplify each fraction. Do not convert any improper fractions to mixed numbers. 15𝑎 15𝑏 Answer: Concept used: 1) A fraction is considered simplified if there are no common factors in the numerator and denominator. a) The process of simplifying a fraction is called reducing the fraction. b) For reducing the fraction, we use the equivalent fractions property in reverse. We know that the equivalent fractions property is, 𝑎 𝑏 = 𝑎⋅𝑐 𝑏⋅𝑐 , 𝑤ℎ𝑒𝑟𝑒𝑏 ≠ 0, 𝑐 ≠ 0 and a,b,c are numbers. Now, the reverse form of equivalent fractions property is, 𝑎⋅𝑐 𝑏⋅𝑐 𝑎 = , where 'c' is the common factor in the numerator and denominator. 𝑏 Calculation: We are the given fraction with 15𝑎 15𝑏 . From the above concept: Notice that 15 is factor of both 15a and 15b. Factor the numerator and denominator. Remove the common factors. 15𝑎 15𝑏 𝑎⋅15 𝑏⋅15 𝑎 𝑏 Hence the simplified form of the given fraction 15𝑎 15𝑏 is, as follows: 𝑎 𝑏 . Question #3 Set up partial fractions decompasition for the following fractions. Do not find A,B,C, etc. 1 𝑥3 − 𝑥2 Answer: Step 1 Since you have asked multiple questions in single request so we will be answering only first question 1 𝑥3 − 𝑥2 Step 2 ⇒ 1 1 = 𝑥 3 − 𝑥 2 𝑥 2 (𝑥 − 1) Partial fraction decomposition ⇒ 1 𝑥 3 −𝑥 2 𝐴 𝐵 𝑥 𝑥2 = + + 𝐶 𝑥−1 Question #4 Subtract fractions. Simplify the answer. 5 7 − =? 3𝑦 2 2𝑦 Answer: Step 1 Given expression Step 2 Simplifying. 5 3𝑦 2 − 7 2𝑦 1 [ 5 𝑦 3𝑦 7 − ] 2 1 10 − 21𝑦 [ ] 𝑦 6𝑦 (10 − 21𝑦) 6𝑦 2 Question #5 List two equivalent fractions. 1 2 3 3 , a) 4 6 2 3 , b) 4 6 3 5 , c) 6 12 2 4 , d) 3 8 Answer: Step 1 Fraction is equivalent to 1 2 2 1 = 4 2 3 1 = 6 2 Step 2 B option is correct Question #6 Write 𝑥2 𝑥+2 and 𝑥+4 2𝑥−1 in terms of the LCD of the fractions. Answer: Step 1 LCD of there two term is 𝐿𝐶𝐷 = (𝑥 + 2)(2𝑥 − 2) write in terms of LCD 𝑥2 𝑥2 2𝑥 − 1 = × 𝑥 + 2 𝑥 + 2 2𝑥 − 1 = 𝑥 2 (2𝑥 − 1) (𝑥 + 2)(2𝑥 − 1) 𝑥2 2𝑥 3 − 𝑥 2 = 𝑥 + 2 (𝑥 + 2)(2𝑥 − 1) (𝑥 + 4)(𝑥 + 2) 𝑥+4 = 2𝑥 − 1 (2𝑥 − 1)(𝑥 + 2) 𝑥+4 𝑥 2 + 6𝑥 + 0 = 2𝑥 − 1 (2𝑥 − 1)(𝑥 + 2) Question #7 1 1 4 3 10 − 5 =? Solve and leave answer in reduced fractional format, no improper fractions. Answer: Step 1 1 1 10 − 5 4 3 = 40 + 1 15 + 1 41 16 − = − 4 3 4 3 Take 𝐿𝐶𝑀 = = (41)(3)−(16)(4) 12 123 − 64 59 = = 4.91 12 12 Step 2 Answer is 4.91 Question #8 Subtract fractions. Simplify the answer. 𝑥−9 𝑥+9 − =? 6𝑥 6𝑥 Answer: Given by 𝑥−9 𝑥+9 − 6𝑥 6𝑥 = 𝑥−9−𝑥−9 6𝑥 = −18 6𝑥 =− 3 𝑥 This is the solution. Question #9 Perform the indicated operations, and express your answers in simplest form. 𝑛 𝑛+3 12𝑛 + 26 + + 𝑛 − 6 𝑛 + 8 𝑛2 + 2𝑛 − 48 Answer: Step 1 Perform the indicted operation. To add or subtract algebraic fractions having different Denominators, first find a common denominator, change each fraction to an equivalent fraction with common denominator and then combine each numerator. Step 2 𝑛 𝑛+3 12𝑛 + 26 + + 2 𝑛 − 6 𝑛 + 8 𝑛 + 2𝑛 − 48 ⇒ 𝑛 𝑛+3 12𝑛 + 26 + + 𝑛 − 6 𝑛 + 8 (𝑛 − 6)(𝑛 + 8) ⇒ (𝑛 + 3)(𝑛 − 6) 𝑛(𝑛 + 8) 12𝑛 + 26 + + =0 (𝑛 − 6)(𝑛 + 8) (𝑛 + 8)(𝑛 − 6) (𝑛 − 6)(𝑛 + 8) ⇒ 𝑛2 + 8𝑛 + 𝑛2 − 6𝑛 + 3𝑛 − 18 + 12𝑛 + 26 =0 (𝑛 − 6)(𝑛 + 8) ⇒ 2𝑛2 + 17𝑛 + 8 = 0 ⇒ 2𝑛2 + 1𝑛 + 16𝑛 + 8 = 0 Step 3 ⇒ 𝑛(2𝑛 + 1) + 8(2𝑛 + 1) = 0 ⇒ (𝑛 + 8)(2𝑛 + 1) = 0 ⇒ 𝑛 = −8, 𝑛 = − 1 2 Question #10 Use the distributive property to write each expression without parentheses. Then simplify the result. 5(𝑥 + 4𝑚 + 2) Answer: Step 1 The distributive property is one of the important property in Algebra . The distributive property is used to simplify the arithmetic operation . According to distributive property , the operation on numbers which are in the parenthesis can be distributed for each number outside the parenthesis . By using this method , we can solve algebraic expressions and fractions involving this form of operation . Step 2 The distributive property can be used to multiply two or more values (variables) inside a parenthesis with addition symbol in between them . In general for two values we can write the distributive property as , 𝐴(𝐵 + 𝐶) = 𝐴𝐵 + 𝐴𝐶 . We have to use the distributive property of the form 𝑎(𝑏 + 𝑐 + 𝑑) = 𝑎 × 𝑏 + 𝑎 × 𝑐 + 𝑎 × 𝑑 . Consider the expression , 5(𝑥 + 4𝑚 + 2) . By using using the distributive property of the form 𝑎(𝑏 + 𝑐 + 𝑑) = 𝑎 × 𝑏 + 𝑎 × 𝑐 + 𝑎 × 𝑑 , we can write. 5(𝑥 + 4𝑚 + 2) = 5 × 𝑥 + 5 × 4𝑚 + 5 × 2 Multiply the values , we get, 5(𝑥 + 4𝑚 + 2) = 5𝑥 + 20𝑚 + 10 Hence we have the simplified value 5𝑥 + 20𝑚 + 10 . Question #11 Find all the zeros for: 𝑦 = 𝑥 3 − 2𝑥 2 − 5𝑥 + 6 Answer: Step 1 Given that, 𝑦 = 𝑥 3 − 2𝑥 2 − 5𝑥 + 6 and we have to calculate the zeros of the function. We know that, for zeros of the function we find the fractions of the function. Step 2 If we put 𝑥 = −2 in given function then the value of function is zero, thus we can say that, (𝑥 + 2) is a fraction of given function. And when we divide the function by (𝑥 − 2) then, quotient will be (𝑥 2 − 4𝑥 + 3) . Thus we can write the given function as, (𝑥 − 2)(𝑥 2 − 4𝑥 + 3) = 0 (𝑥 − 2)(𝑥 2 − 3𝑥 − 𝑥 + 3) = 0 (𝑥 − 2)[𝑥(𝑥 − 3) − (𝑥 − 3)] = 0 (𝑥 − 2)(𝑥 − 3)(𝑥 − 1) = 0 𝑥 = 1, 2, 3 Answer: Thus zeros of the given function are (1, 2, 3). Question #12 Find the equation of the line passing through the point (5, −8)5, −8 with a slope of −17 − 17. You may enter your equation in either the slope-intercept or point-slope form. Make sure to enclose 9 5 2 2 fractions in parentheses, for example: 𝑦 = ( ) ⋅ 𝑥 + ( ) . Answer: Step 1 Given, The line passing through the point (5,-8) with a slope of -17 We have to write the equation of a straight line in point-slope form. Step 2 We have Point = (𝑥1 , 𝑦1 ) = (6, −8) Slope = 𝑚 = −17 Equation of a straight line in point-slope form 𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 ) ⇒ 𝑦 − (−8) = −17(𝑥 − 5) (substituted the values) ⇒ 𝑦 − (−8) = −17(𝑥 − 5) ⇒ 𝑦 + 8 = −17𝑥 + 85 ⇒ 𝑦 = −17𝑥 + 85 − 8 Answer: ⇒ 𝑦 = −17𝑥 + 77 Question #13 What is the value of 1 − 4 + 9 − 16 + 25 − 36 + 49 − 64 + 81 − 100 =? Answer: Step 1 Given, 1 − 4 + 9 − 16 + 25 − 36 + 49 − 64 + 81 − 100 Step 2 Now, ∴ 1 − 4 + 9 − 16 + 25 − 36 + 49 − 64 + 81 − 100 = 1 + 9 + 25 + 49 + 81 − 4 − 16 − 36 − 64 − 100 = 1 + 9 + 25 + 49 + 81 − (4 + 16 + 36 + 64 + 100) = 165 − (220) = 165 − 220 = −55 ∴ The value of1 − 4 + 9 − 16 + 25 − 36 + 49 − 64 + 81 − 100 = −55 Question #14 Put the following equation of a line into slope-intercept form, simplifying all fractions. 8𝑥 − 6𝑦 = 54 Answer: Step 1 From the given data: The equation is: 8𝑥 − 6𝑦 = 54 Step 2 The equation of the slope intercept form is 𝑦 = 𝑚𝑥 + 𝑏 From the given data: 8𝑥 − 6𝑦 = 54 −6𝑦 = 54 − 8𝑥 Simplifying the equation we get: 6𝑦 = 8𝑥 − 54 𝑦= 8 54 𝑥− 6 6 𝑦= 4 𝑥−9 3 Comparing the above equation with slope intercept form: 𝑦 = 𝑚𝑥 + 𝑏 𝑦= 4 𝑥−9 3 Slope = 𝑚 = 4 3 y-intercept = 𝑏 = −9 Question #15 Rewrite the equation below so that it does not have fractions. 2 3 5− 𝑥 = 3 4 Do not use decimals in your answer. ? −? 𝑥 =? Answer: Step 1 2 3 5− 𝑥 = 3 4 Step 2 5= 3 2 + 𝑥 4 3 In the denominates we have 3 and 4 so multiply full equation with LCM, 3 × 4 = 12 . 12 × 5 = 3 × 12 2 × 12 + 𝑥 4 3 60 = 9 + 12𝑥 Answer: 60 − 12𝑥 = 9 Question #16 For the following equation, determine the values of the missing entries. Reduce all fractions to lowest terms. 12𝑥 − 10𝑦 = 12 Note: Each column in the table represents an ordered pair. If multiple solutions exist, you only need to identify one. Answer: Step 1 The table must be provide with the problem. Given, The equation is 12𝑥 − 10𝑦 = 12 . Step 2 If you are given a value of x, just plug it into the equation and solve for y. If you are given a value of y, just plug it into the equation and solve for x. Let 𝑦 = 0 , 12𝑥 − 10(0) = 12. 12𝑥 = 12 Dividing both sides by 12, 𝑥 = 1. Thus the solution of this equation is (1, 0). Question #17 Solve the equation below with factoring. Separate multiple solutions with a comma any reduce and fractions. 20𝑥 2 + 9𝑥 − 18 = 0 𝑥 =? Answer: Step 1 Given 20𝑥 2 + 9𝑥 − 18 = 0 Step 2 20𝑥 2 + 9𝑥 − 18 = 0 20𝑥 2 + 24𝑥 − 15𝑥 − 18 = 0 4𝑥(5𝑥 + 6) − 3(5𝑥 + 6) = 0 ⇒ (5𝑥 + 6)(4𝑥 − 3) = 0 ⇒𝑥= −6 5 3 ,𝑥 = . 4 Question #18 Put the following equation of a line into slope-intercept form, simplifying all fractions. 𝑥−𝑦=3 Answer: Step 1 The equation of a line in slope intercept form is written as 𝑦 = 𝑚𝑥 + 𝑏 where m is the slope and b is the y intercept. Step 2 So 𝑥 − 𝑦 = 3 can be written as 𝑦 = 𝑥 − 3 Where 1 is the slope and -3 is y-intercept. Therefore equation of line in y-intercept form is 𝑦 = 𝑥 − 3 Question #19 Solve the equation below with factoring. Separate multiple solutions with a comma any reduce and fractions. 4𝑥 2 − 2𝑥 = 𝑥 2 + 8 𝑥 =? Answer: Step 1 4𝑥 2 − 2𝑥 = 𝑥 2 + 8 4𝑥 2 − 𝑥 2 − 2𝑥 − 8 = 0 Step 2 3𝑥 2 − 2𝑥 − 8 = 0 3𝑥 2 − 6𝑥 + 4𝑥 − 8 = 0 3𝑥(𝑥 − 2) + 4(𝑥 − 2) = 0 (𝑥 − 2)(3𝑥 + 4) = 0 4 𝑥 = − ,𝑥 = 2 3 Question #20 A parabola opening up or down has vertex (0,4) and passes through (−12,8). Write its equation in vertex form. Simplify any fractions. 𝑦 =? Answer: Step 1 Given: vertex (0,4) and point (−12, −8) Step 2 Calculation: 𝑦 = 𝑎(𝑥 − ℎ)2 + 𝑘 −8 = 𝑎(−12 − 0)2 + 4 −8 = 144𝑎 + 4 144𝑎 = −8 − 4 𝑎= −12 144 𝑎= −1 12 𝑦= −1 (𝑥 − 0)2 + 4 12 𝑦=− 𝑥2 12 +4