This pdf includes the following topics:- Cancelling common factors Multiplication and division of algebraic fractions Addition and subtraction of algebraic fractions
1. Arithmetic of Algebraic Fractions 1.4
Introduction Just as one whole number divided by another is called a numerical fraction, so one algebraic expression divided by another is known as an algebraic fraction. Examples are x 3x + 2y x2 + 3x + 1 , , and y x−y x−4 In this Section we explain how algebraic fractions can be simplified, added, subtracted, multiplied and divided.
Prerequisites • be familiar with the arithmetic of numerical fractions Before starting this Section you should . . .
Learning Outcomes • add, subtract, multiply and divide algebraic fractions On completion you should be able to . . .
62 HELM (2008): Workbook 1: Basic Algebra
2. ® 1. Cancelling common factors 10 Consider the fraction . To simplify it we can factorise the numerator and the denominator and then 35 cancel any common factors. Common factors are those factors which occur in both the numerator and the denominator. Thus 10 65×2 2 = = 35 7× 6 5 7 Note that the common factor 5 has been cancelled. It is important to remember that only common 10 2 factors can be cancelled. The fractions and have identical values - they are equivalent fractions 35 7 2 10 - but is in a simpler form than . 7 35 We apply the same process when simplifying algebraic fractions. Example 49 Simplify, if possible, yx x x (a) , (b) , (c) 2x xy x+y Solution yx (a) In the expression , x is a factor common to both numerator and denominator. This 2x common factor can be cancelled to give y6x y = 26x 2 x 1x (b) Note that can be written . The common factor of x can be cancelled to give xy xy 16x 1 = 6 xy y x (c) In the expression notice that an x appears in both numerator and denominator. x+y However x is not a common factor. Recall that factors of an expression are multi- plied together whereas in the denominator x is added to y. This expression cannot be simplified. HELM (2008): 63 Section 1.4: Arithmetic of Algebraic Fractions
3. Task abc 3ab Simplify, if possible, (a), (b) 3ac b+a When simplifying remember only common factors can be cancelled. Your solution abc 3ab (a) = (b) = 3ac b+a b (a) (b) This cannot be simplified. 3 Task 21x3 Simplify , 14x Your solution Factorising and cancelling common factors gives: 21x3 6 7 × 3× 6 x × x2 3x2 = = 14x 6 7 × 2× 6 x 2 Task 36x Simplify 12x3 Your solution Factorising and cancelling common factors gives: 36x 12 × 3 × x 3 3 = 2 = 2 12x 12 × x × x x 64 HELM (2008): Workbook 1: Basic Algebra
4. ® Example 50 3x + 6 Simplify . 6x + 12 Solution First we factorise the numerator and the denominator to see if there are any common factors. 3x + 6 3(x + 2) 3 1 = = = 6x + 12 6(x + 2) 6 2 The factors x + 2 and 3 have been cancelled. Task 12 Simplify . 2x + 8 Your solution 12 = 2x + 8 Answer 6×2 6 Factorise the numerator and denominator, and cancel any common factors. = 2(x + 4) x+4 Example 51 3 3(x + 4) Show that the algebraic fraction and 2 are equivalent. x+1 x + 5x + 4 Solution The denominator, x2 + 5x + 4, can be factorised as (x + 1)(x + 4) so that 3(x + 4) 3(x + 4) = x2+ 5x + 4 (x + 1)(x + 4) Note that (x + 4) is a factor common to both the numerator and the denominator and can be 3 3 3(x + 4) cancelled to leave . Thus and 2 are equivalent fractions. x+1 x+1 x + 5x + 4 HELM (2008): 65 Section 1.4: Arithmetic of Algebraic Fractions
5. Task x−1 1 Show that is equivalent to . x2 − 3x + 2 x−2 First factorise the denominator: Your solution x2 − 3x + 2 = Answer (x − 1)(x − 2) Now identify the factor common to both numerator and denominator and cancel this common factor: Your solution x−1 = (x − 1)(x − 2) Answer 1 . Hence the two given fractions are equivalent. x−2 Example 52 6(4 − 8x)(x − 2) Simplify 1 − 2x Solution The factor 4 − 8x can be factorised to 4(1 − 2x). Thus 6(4 − 8x)(x − 2) (6)(4)(1 − 2x)(x − 2) = = 24(x − 2) 1 − 2x (1 − 2x) Task x2 + 2x − 15 Simplify 2x2 − 5x − 3 First factorise the numerator and factorise the denominator: Your solution x2 + 2x − 15 = 2x2 − 5x − 3 66 HELM (2008): Workbook 1: Basic Algebra
7. Answers 1 1 7 7 1 1. (a) , (b) , (c) , (d) , (e) . 2 2 8 11 4 2 3 1 2. (a) , (b) , (c) , (d) 4 3 8 4 1 1 3. (a) 5, (b) 5, (c) 2 , (d) . 5z 5z 4 15 4 4. (a) , (b) , (c) 2 , (d) 3x 3 x s 1 1 5. (a) , (b) , (c) 2, (d) 3, (e) x − 3, (f) 5 2 2 x+3 x+3 x+3 5(x + 3) 6. (a) , (b) , (c) , (d) 5x + 1 5x 5 25x + 1 x+1 x+3 x−1 3x − 1 5z 7. (a) , (b) , (c) , (d) , (e) x−1 x+7 x+2 x 2 2 1 3 8. (a) , (b) , (c) . x+3 2x + 1 5(3x + 2) x−1 x+2 9. (a) , (b) . x+4 x−2 2. Multiplication and division of algebraic fractions To multiply together two fractions (numerical or algebraic) we multiply their numerators together and then multiply their denominators together. That is Key Point 19 Multiplication of fractions a c ac × = b d bd Any factors common to both numerator and denominator can be cancelled. This cancellation can be performed before or after the multiplication. To divide one fraction by another (numerical or algebraic) we invert the second fraction and then 68 HELM (2008): Workbook 1: Basic Algebra
8. ® Key Point 20 Division of fractions a c a d ad ÷ = × = b 6= 0, c 6= 0, d 6= 0 b d b c bc Example 53 2a 4 2a c 2a 4 Simplify (a) × , (b) × , (c) ÷ c c c 4 c c Solution 2a 4 8a (a) × = 2 c c c 2a c 2ac 2a a (b) × = = = c 4 4c 4 2 (c) Division is performed by inverting the second fraction and then multiplying. 2a 4 2a c a ÷ = × = (from the result in (b)) c c c 4 2 Example 54 1 1 Simplify (a) × 3x, (b) × x. 5x x Solution 3x 1 1 3x 3x 3 (a) Note that 3x = . Then × 3x = × = = 1 5x 5x 1 5x 5 x 1 1 x x (b) x can be written as . Then × x = × = = 1 1 x x 1 x HELM (2008): 69 Section 1.4: Arithmetic of Algebraic Fractions
9. Task 1 y Simplify (a) × x, (b) × x. y x Your solution 1 1 x x (a) ×x= × = y y 1 y y y x yx (b) ×x= × = =y x x 1 x Example 55 2x y Simplify 3x 2y Solution 2x 3x We can write the fraction as ÷ . y 2y Inverting the second fraction and multiplying we find 2x 2y 4xy 4 × = = y 3x 3xy 3 70 HELM (2008): Workbook 1: Basic Algebra
10. ® Example 56 4x + 2 x+3 Simplify × x2 + 4x + 3 7x + 5 Solution Factorising the numerator and denominator we find 4x + 2 x+3 2(2x + 1) x+3 2(2x + 1)(x + 3) × = × = x2 + 4x + 3 7x + 5 (x + 1)(x + 3) 7x + 5 (x + 1)(x + 3)(7x + 5) 2(2x + 1) = (x + 1)(7x + 5) It is usually better to factorise first and cancel any common factors before multiplying. Don’t remove any brackets unnecessarily otherwise common factors will be difficult to spot. Task Simplify 15 3 ÷ 3x − 1 2x + 1 Your solution Answer To divide we invert the second fraction and multiply: 15 3 15 2x + 1 (5)(3)(2x + 1) 5(2x + 1) ÷ = × = = 3x − 1 2x + 1 3x − 1 3 3(3x − 1) 3x − 1 HELM (2008): 71 Section 1.4: Arithmetic of Algebraic Fractions
12. ® 3. Addition and subtraction of algebraic fractions To add two algebraic fractions the lowest common denominator must be found first. This is the simplest algebraic expression that has the given denominators as its factors. All fractions must be written with this lowest common denominator. Their sum is found by adding the numerators and dividing the result by the lowest common denominator. To subtract two fractions the process is similar. The fractions are written with the lowest common denominator. The difference is found by subtracting the numerators and dividing the result by the lowest common denominator. Example 57 State the simplest expression which has x + 1 and x + 4 as its factors. Solution The simplest expression is (x + 1)(x + 4). Note that both x + 1 and x + 4 are factors. Example 58 State the simplest expression which has x − 1 and (x − 1)2 as its factors. Solution The simplest expression is (x − 1)2 . Clearly (x − 1)2 must be a factor of this expression. Also, because we can write (x − 1)2 = (x − 1)(x − 1) it follows that x − 1 is a factor too. HELM (2008): 73 Section 1.4: Arithmetic of Algebraic Fractions
13. Example 59 3 2 Express as a single fraction + x+1 x+4 Solution The simplest expression which has both denominators as its factors is (x + 1)(x + 4). This is the lowest common denominator. Both fractions must be written using this denominator. Note that 3 3(x + 4) 2 2(x + 1) is equivalent to and is equivalent to . Thus writing x+1 (x + 1)(x + 4) x+4 (x + 1)(x + 4) both fractions with the same denominator we have 3 2 3(x + 4) 2(x + 1) + = + x+1 x+4 (x + 1)(x + 4) (x + 1)(x + 4) The sum is found by adding the numerators and dividing the result by the lowest common denomi- nator. 3(x + 4) 2(x + 1) 3(x + 4) + 2(x + 1) 5x + 14 + = = (x + 1)(x + 4) (x + 1)(x + 4) (x + 1)(x + 4) (x + 1)(x + 4) Key Point 21 Addition of two algebraic fractions Step 1: Find the lowest common denominator Step 2: Express each fraction with this denominator Step 3: Add the numerators and divide the result by the lowest common denominator Example 60 1 5 Express + as a single fraction. x − 1 (x − 1)2 Solution The simplest expression having both denominators as its factors is (x − 1)2 . We write both fractions with this denominator. 1 5 x−1 5 x−1+5 x+4 + 2 = 2 + 2 = 2 = x − 1 (x − 1) (x − 1) (x − 1) (x − 1) (x − 1)2 74 HELM (2008): Workbook 1: Basic Algebra
14. ® Task 3 5 Express + as a single fraction. x+7 x+2 First find the lowest common denominator: Your solution Answer (x + 7)(x + 2) Re-write both fractions using this lowest common denominator: Your solution 3 5 + = x+7 x+2 Answer 3(x + 2) 5(x + 7) + (x + 7)(x + 2) (x + 7)(x + 2) Finally, add the numerators and simplify: Your solution 3 5 + = x+7 x+2 Answer 8x + 41 (x + 7)(x + 2) Example 61 5x 3x − 4 Express − as a single fraction. 7 2 Solution In this example both denominators are simply numbers. The lowest common denominator is 14, and both fractions are re-written with this denominator. Thus 5x 3x − 4 10x 7(3x − 4) 10x − 7(3x − 4) 28 − 11x − = − = = 7 2 14 14 14 14 HELM (2008): 75 Section 1.4: Arithmetic of Algebraic Fractions
15. Task 1 1 Express + as a single fraction. x y Your solution The simplest expression which has x and y as its factors is xy. This is the lowest common denom- 1 y 1 x inator. Both fractions are written using this denominator. Noting that = and that = x xy y xy we find 1 1 y x y+x + = + = x y xy xy xy No cancellation is now possible because neither x nor y is a factor of the numerator. Exercises x x 2x x 2x 3x x 2 x+1 3 1. Simplify (a)+ , (b) + , (c) − , (d) − , (e) + , 4 7 5 9 3 4 x+1 x+2 x x+2 2x + 1 x x+3 x x x (f) − , (g) − , (h) − 3 2 2x + 1 3 4 5 2. Find 1 2 2 5 2 3 x+1 x+4 (a) + , (b) + , (c) − , (d) + , x+2 x+3 x+3 x+1 2x + 1 3x + 2 x+3 x+2 x−1 x−1 (e) + . x − 3 (x − 3)2 5 4 3. Find + . 2x + 3 (2x + 3)2 1 11 4. Find s+ 7 21 A B 5. Express + as a single fraction. 2x + 3 x + 1 A B C 6 Express + + as a single fraction. 2x + 5 (x − 1) (x − 1)2 A B 7 Express + as a single fraction. x + 1 (x + 1)2 76 HELM (2008): Workbook 1: Basic Algebra
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