# 1.4. Arithmetic of Algebraic Fractions. Introduction. Prerequisites. Learning Outcomes

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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 y, x + 2y x y, and x 2 + x + 1 x 4 In this Section we explain how algebraic fractions can be simplified, added, subtracted, multiplied and divided. Prerequisites Before starting this Section you should... Learning Outcomes On completion you should be able to... be familiar with the arithmetic of numerical fractions add, subtract, multiply and divide algebraic fractions 62 HELM (2006): Workbook 1: Basic Algebra

2 1. Cancelling common factors Consider the fraction 10. To simplify it we can factorise the numerator and the denominator and then 5 cancel any common factors. Common factors are those factors which occur in both the numerator and the denominator. Thus Note that the common factor 5 has been cancelled. It is important to remember that only common factors can be cancelled. The fractions 10 5 and 2 have identical values - they are equivalent fractions 7 - but 2 10 is in a simpler form than 7 5. We apply the same process when simplifying algebraic fractions. Example 49 Simplify, if possible, (a) yx 2x x xy, (c) x x + y (a) In the expression yx, x is a factor common to both numerator and denominator. This 2x common factor can be cancelled to give y x 2 x y 2 (b) Note that x xy 1 x xy 1 y 1x can be written. The common factor of x can be cancelled to give xy 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 multiplied together whereas in the denominator x is added to y. This expression cannot be simplified. HELM (2006): Section 1.4: Arithmetic of Algebraic Fractions 6

3 Task Simplify, if possible, (a) abc ac ab b + a When simplifying remember only common factors can be cancelled. (a) abc ac (b) ab b + a (a) b (b) This cannot be simplified. Task Simplify 21x 14x, Factorising and cancelling common factors gives: 21x 14x 7 x x2 7 2 x x2 2 Task Simplify 6x 12x Factorising and cancelling common factors gives: 6x 12x 12 x 12 x x 2 x 2 64 HELM (2006): Workbook 1: Basic Algebra

4 Example 50 Simplify x + 6 6x First we factorise the numerator and the denominator to see if there are any common factors. x + 6 6x + 12 (x + 2) 6(x + 2) The factors x + 2 and have been cancelled. Task Simplify 12 2x x + 8 Factorise the numerator and denominator, and cancel any common factors (x + 4) 6 x + 4 Example 51 Show that the algebraic fraction x + 1 and (x + 4) x 2 + 5x + 4 are equivalent. The denominator, x 2 + 5x + 4, can be factorised as (x + 1)(x + 4) so that (x + 4) x 2 + 5x + 4 (x + 4) (x + 1)(x + 4) Note that (x + 4) is a factor common to both the numerator and the denominator and can be cancelled to leave x + 1. Thus x + 1 and (x + 4) are equivalent fractions. x 2 + 5x + 4 HELM (2006): Section 1.4: Arithmetic of Algebraic Fractions 65

5 Task Show that x 1 x 2 x + 2 is equivalent to 1 x 2. First factorise the denominator: x 2 x + 2 (x 1)(x 2) Now identify the factor common to both numerator and denominator and cancel this common factor: x 1 (x 1)(x 2) 1. Hence the two given fractions are equivalent. x 2 Example 52 6(4 8x)(x 2) Simplify 1 2x The factor 4 8x can be factorised to 4(1 2x). Thus 6(4 8x)(x 2) 1 2x (6)(4)(1 2x)(x 2) (1 2x) 24(x 2) Task Simplify x2 + 2x 15 2x 2 5x First factorise the numerator and factorise the denominator: x 2 + 2x 15 2x 2 5x 66 HELM (2006): Workbook 1: Basic Algebra

6 (x + 5)(x ) (2x + 1)(x ) Then cancel any common factors: (x + 5)(x ) (2x + 1)(x ) x + 5 2x Simplify, if possible, Exercises (a) , (c) , (d) 7 14, (e) Simplify, if possible, (a) Simplify (a) 5z z, 4. Simplify (a) 4x x, 15x (b) x, 2 5. Simplify, if possible, (a) x + 1 2(x + 1) x + 1 2x Simplify, if possible, (a) 5x x , (c), (d) z (b) 5z, (c) 5 25z, (d) 5z 2 25z 2 5x x 4s (c) s,, (c) 2(x + 1) x x4 (d) 7x, (d) x + x + 1 5x + 15, (c), (d) 25 5x x Simplify (a) x2 + 10x + 9 x 2 + 8x 9 x 2 9 x 2 + 4x 21, (c) 2x2 x 1 2x 2 + 5x + 2, (d) x2 4x + 1, (e) 5z2 20z x 2 x 2z 8 8. Simplify (a) 9. Simplify (a) 6 x + 9 2x 4x 2 + 2x, (c) x 2 15x + 10x 2 x 2 1 x 2 + 5x + 4 x2 + 5x + 6 x 2 + x 6. 5x 15, (e), (f) 5 5x 15 x. HELM (2006): Section 1.4: Arithmetic of Algebraic Fractions 67

7 s 1. (a) , (c) 7 8, (d) 7 11, (e) (a) 2 8, (c) 1 4, (d) 4. (a) 5 5, (c) 1 5z 2, (d) 1 5z. 4. (a) 4 15 x, (c) 4, (d) x s2 5. (a) 1 2 1, (c) 2, (d), (e) x, (f) (a) x + 5x + 1 x + 5x, (c) x +, (d) 5 7. (a) x + 1 x 1 x + 8. (a) x + 7, (c) x 1 x x + 1 2x + 1, (c) 5(x + 2). 9. (a) x 1 x + 4 x + 2 x 2. 5(x + ) 25x + 1 x 1, (d), (e) 5z 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 b c d ac 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 multiply. 68 HELM (2006): Workbook 1: Basic Algebra

8 Key Point 20 Division of fractions a b c d a b d c ad b 0, c 0, d 0 bc Example 5 Simplify (a) 2a c 4 c 2a c c 4, (c) 2a c 4 c (a) (b) (c) 2a c 4 c 8a c 2 2a c c 4 2ac 4c 2a 4 a 2 Division is performed by inverting the second fraction and then multiplying. 2a c 4 c 2a c c 4 a 2 (from the result in (b)) Example 54 Simplify (a) 1 5x x 1 x x. (a) Note that x x 1. Then 1 5x x 1 5x x 1 x 5x 5 (b) x can be written as x 1. Then 1 x x 1 x x 1 x x 1 HELM (2006): Section 1.4: Arithmetic of Algebraic Fractions 69

9 Task Simplify (a) 1 y x y x x. 1 (a) y x 1 y x 1 x y (b) y x x y x x 1 yx x y Example 55 2x y Simplify x 2y We can write the fraction as 2x y x 2y. Inverting the second fraction and multiplying we find 2x y 2y x 4xy xy 4 70 HELM (2006): Workbook 1: Basic Algebra

10 Example 56 4x + 2 Simplify x 2 + 4x + x + 7x + 5 Factorising the numerator and denominator we find 4x + 2 x 2 + 4x + x + 7x + 5 2(2x + 1) (x + 1)(x + ) x + 7x + 5 2(2x + 1)(x + ) (x + 1)(x + )(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 x 1 2x + 1 To divide we invert the second fraction and multiply: 15 x 1 2x x 1 2x + 1 (5)()(2x + 1) (x 1) 5(2x + 1) x 1 HELM (2006): Section 1.4: Arithmetic of Algebraic Fractions 71

11 Exercises 1. Simplify (a) , (c) , (d) Simplify (a) , (c) , (d) Simplify (a) 2 x + y 1 2(x + y), (c) 2 (x + y) 4. Simplify (a) x (x + 4), (c) 7 (x + 4), (d) x y x + 1 y + 1, (f) πd2 4 Q πd 2, (g) Q πd 2 /4 (e) 1 y x2 + x y + 1, 5. Simplify 6/7 s + 6. Simplify 7. Simplify s x + 2 x 2x x + 1 x x 1 1. (a) , (c) 9 16, (d) (a) , (c) 8 11, (d) 49. (a) 2(x + y) (x + 4) 4. (a) 7 (g) 4Q πd (s + ) 6 x 5(x 1) x(2x + 1) 2(x + y), (c) (x + 4), (c) 7 2(x + y) (x + 4), (d) 7 x(x + 1) x(x + 1), (e) y(y + 1) y(y + 1), (f) Q/4, 72 HELM (2006): Workbook 1: Basic Algebra

12 . 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. 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. 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 (2006): Section 1.4: Arithmetic of Algebraic Fractions 7

13 Example 59 Express as a single fraction x x + 4 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 x + 1 is equivalent to (x + 4) (x + 1)(x + 4) and 2 x + 4 is equivalent to 2(x + 1). Thus writing (x + 1)(x + 4) both fractions with the same denominator we have x x + 4 (x + 4) (x + 1)(x + 4) + 2(x + 1) (x + 1)(x + 4) The sum is found by adding the numerators and dividing the result by the lowest common denominator. (x + 4) (x + 1)(x + 4) + 2(x + 1) (x + 1)(x + 4) (x + 4) + 2(x + 1) (x + 1)(x + 4) 5x + 14 (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 : Add the numerators and divide the result by the lowest common denominator Example 60 1 Express x (x 1) 2 The simplest expression having both denominators as its factors is (x 1) 2. We write both fractions with this denominator. 1 x (x 1) x 1 2 (x 1) (x 1) x (x 1) x (x 1) 2 74 HELM (2006): Workbook 1: Basic Algebra

14 Task Express x x + 2 First find the lowest common denominator: (x + 7)(x + 2) Re-write both fractions using this lowest common denominator: x x + 2 (x + 2) (x + 7)(x + 2) + 5(x + 7) (x + 7)(x + 2) Finally, add the numerators and simplify: x x + 2 8x + 41 (x + 7)(x + 2) Example 61 Express 5x 7 x 4 2 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 7 x x 14 7(x 4) 14 10x 7(x 4) x 14 HELM (2006): Section 1.4: Arithmetic of Algebraic Fractions 75

15 Task Express 1 x + 1 y The simplest expression which has x and y as its factors is xy. This is the lowest common denominator. Both fractions are written using this denominator. Noting that 1 x y xy and that 1 y x xy we find 1 x + 1 y y xy + x xy y + x xy No cancellation is now possible because neither x nor y is a factor of the numerator. Exercises 1. Simplify (a) x 4 + x 7 2x 5 + x 9, (c) 2x x 4, (d) x x x + 2, (e) x + 1 x + x + 2, (f) 2x + 1 x 2, (g) x + 2x + 1 x, (h) x 4 x 5 2. Find 1 (a) x x + 2 x x + 1, (c) 2 2x + 1 x + 2, (d) x + 1 x + + x + 4 x + 2, (e) x 1 x + x 1 (x ) 2.. Find 5 2x (2x + ) Find 1 7 s Express 6 Express 7 Express A 2x + + B x + 1 A 2x B (x 1) + C (x 1) 2 A x B (x + 1) 2 76 HELM (2006): Workbook 1: Basic Algebra

16 8 Express Ax + B x 2 + x C x 1 9 Express Ax + B + C x Show that x 1 1 x 1 x 2 is equal to x 1 x 2 x x 2 x. 11 Find (a) x 4 x 5 + x x 4 ( x 5 + x ). s 1. (a) 11x 2x 28 45, (c) x 12, (d) x 2 2 (x + 1)(x + 2), (e) x2 + 6x + 2, x(x + 2) 2. (a) (f) x + 2, (g) x 2x2, (h) x (2x + 1) 20 x + 7 (x + 2)(x + ) 7x + 17 (x + )(x + 1), (c) 1 (2x + 1)(x + 2), (d) 2x2 + 10x + 14 (x + )(x + 2), (e) x2 x + 2 (x ) 2 10x + 19 (2x + ) 2 s A(x + 1) + B(2x + ) (2x + )(x + 1) A(x 1) 2 + B(x 1)(2x + 5) + C(2x + 5) (2x + 5)(x 1) 2 A(x + 1) + B (x + 1) 2 (Ax + B)(x 1) + C(x 2 + x + 10) (x 1)(x 2 + x + 10) 9. (Ax + B)(x + 1) + C x (a) 5x 1x HELM (2006): Section 1.4: Arithmetic of Algebraic Fractions 77

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