Algebraic Fractions - University of Plymouth

[Pages:41]Basic Mathematics

Algebraic Fractions

R Horan & M Lavelle

The aim of this document is to provide a short, self assessment programme for students who wish to acquire a basic competence in the use of algebraic fractions.

Copyright c 2004 rhoran@plymouth.ac.uk , mlavelle@plymouth.ac.uk

Last Revision Date: February 25, 2004

Version 1.0

Table of Contents

1. Algebraic Fractions (Introduction) 2. Addition of Algebraic Fractions 3. Simple Partial Fractions 4. Quiz on Algebraic Fractions

Solutions to Exercises Solutions to Quizzes

Section 1: Algebraic Fractions (Introduction)

3

1. Algebraic Fractions(Introduction)

Algebraic fractions have properties which are the same as those for numerical fractions, the only difference being that the the numerator (top) and denominator (bottom) are both algebraic expressions. Example 1 Simplify each of the following fractions.

Solution (a)

2b (a) 7b2 ,

3x + x2 (b) 6x2 .

2b

2?b

2

7b2 = 7 ? b ? b = 7b

(b)

3x + x2

x ? (3 + x)

6x2 = x ? 6x

x ? (3 + x) 3 + x

=

=

x ? 6x 6x

N.B. The cancellation in (b) is allowed since x is a common factor of the numerator and the denominator.

Section 1: Algebraic Fractions (Introduction)

4

Sometimes a little more work is necessary before an algebraic fraction can be reduced to a simpler form.

Example 2 Simplify the algebraic fraction

Solution

x2 - 2x + 1 x2 + 2x - 3

In this case the numerator and denominator can be factored into two terms, thus

x2 - 2x + 1 = (x - 1)2 , and x2 + 2x - 3 = (x - 1)(x + 3) .

( See the package on factorising expressions). With this established the simplification proceeds as follows:

x2 - 2x + 1

(x - 1) ? (x - 1)

x2 + 2x - 3 = (x + 3) ? (x - 1)

x-1

=

(cancelling (x - 1))

x+3

Section 1: Algebraic Fractions (Introduction)

5

Exercise 1. Simplify each of the following algebraic fractions. (Click

on the green letters for solution.)

8y (a) 2y3

2y (b)

4x

7a6b3 (c) 14a5b4

(2x)2 (d)

4x

5y + 2y2 (e)

7y

5ax (f) 15a + 10a2

2z2 - 4z (g) 2z2 - 10z

y2 + 7y + 10

(h)

y2 - 25

w2 - 5w - 14 (i) w2 - 4w - 21

Now try this short quiz.

Quiz Which of the following is a simplified version of t2 + 3t - 4 t2 - 3t + 2 ?

t-4 (a)

t-2

t-4 (b)

t+2

t+4 (c)

t-2

t+4 (d)

t+2

Section 1: Algebraic Fractions (Introduction)

6

So far, simplification has been achieved by cancelling common factors from the numerator and denominator. There are fractions which can be simplified by multiplying the numerator and denominator by an appropriate common factor, thus obtaining an equivalent, simpler expression.

Example 3 Simplify the following fractions.

Solution

(a)

1 4

+

y

1

2

(b)

3x

+

1 x

2

(a) In this case, multiplying both the numerator and the denominator

by 4 gives:

1 4

+y

1 2

=

4

1 4

4

+y

1 2

1 + 4y =

2

(b) To simplify this expression, multiply the numerator and denomi-

nator by x. Thus

3x

+

1 x

x =

3x

+

1 x

3x2 + 1 =

2

2x

2x

Section 1: Algebraic Fractions (Introduction)

7

Now try this exercise on similar examples.

Exercise 2. Simplify each of the following algebraic fractions. (Click

on the green letters for solution.)

(a)

4y

-

3 2

2

(b)

2x

+

1 2

x

+

1 4

(c)

z

-

1 3

z

-

1 2

(d)

2

-

1 x

2

(e)

3t

-

2 t

1

2

(f)

z

-

1 2z

z

-

1 3z

For the last part of this section, try the following short quiz.

Quiz Which of the following is a simplified version of

x

-

1 x+1

?

x-1

x2 - x + 1

x2 - x + 1

x2 - 1

x2 + x - 1

(a) x2 + x + 1 (b) x2 - 1

(c) x2 - x - 1 (d) x2 - 1

Section 2: Addition of Algebraic Fractions

8

2. Addition of Algebraic Fractions

Addition (and subtraction) of algebraic fractions proceeds in exactly the same manner as for numerical fractions.

Example 4 Write the following sum as a single fraction in its simplest

form.

2

1

+

x+1 x+2

Solution The least common multiple of the denominators (see the package on fractions) is (x + 1)(x + 2). Thus

2

1

2 ? (x + 2)

1 ? (x + 1)

+

=

+

x+1 x+2

(x + 1) ? (x + 2) (x + 2) ? (x + 1)

2x + 4

x+1

=

+

(x + 1)(x + 2) (x + 1)(x + 2)

(2x + 4) + (x + 1)

3x + 5

=

=

(x + 1)(x + 2) (x + 1)(x + 2)

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