Mathematics

Key unit Competence: By the end of the unit, the learner should be able to perform
 operations on rational expressions and use them in different situations.
 

Unit Outline

• Definition of algebraic fraction

• Simplification of algebraic fractions

• Subtraction and addition of algebraic
    fractions with linear denominator

• Multiplication of algebraic fractions

• Division of algebraic fractions

• Solving of rational equations.

Introduction

Unit Focus Activity

(a) Consider a fraction such as 2÷2x – 4 .

(i) Find the value of the fraction
when x = 0, 1, 2, 3, 4.

(ii) Is there any value of x for which
there cannot be any meaningful
value for the fraction in (i)
above? If your answer is yes,
explain.

(b) Given group of numbers such as
(i) 2, 3, 4 (ii) x, 2x, 2x + 6, find the
LCM of each group.
Express each of the following as a
single fraction under a common
denominator:

(ii) Given that a number divided
by itself equals 1, evaluate:
1/2 ÷ 1/2  ; 1/3 ÷ 1/3; 2/5 ÷ 2/5 ; a/b ÷ a/b

(iii) Now evaluate also;
1/2 × 2/1  ; 1/3 × 3/1 ; 2/5 × 5/2 ; a/b × b/a
What can you say about the
answers in part c(ii) and (iii)?

(iv) Create a multiplication question
which gives the same answer as
each of the following:
1/4 ÷ 1/4 ; 1/5 ÷ 1/5; 2/3 ÷ 2/3.

(d) Work out the following:
(i) a ÷ b/c            (ii) b/c ÷ a

Introduction

Unit Focus Activity

(a) Consider a fraction such as 2/2x – 4 .

(i) Find the value of the fraction
when x = 0, 1, 2, 3, 4.

(ii) Is there any value of x for which
there cannot be any meaningful value for the fraction in (i)
above? If your answer is yes,explain.

(b) Given group of numbers such as
(i) 2, 3, 4 (ii) x, 2x, 2x + 6, find the
LCM of each group.
Express each of the following as a
single fraction under a common
denominator:

1/2 + 1/3 + 1/4; 1/2 – 1/3

1/x + 1/2x + 2/2x + 6 ; 1/x + 1/2x

(c) (i) Evaluate the following;

1/2 × 1/2 ; 1/3 × 1/3; 3/4 × 3/4

Consider the following expressions:

In each of these expressions, the numerator
or the denominator or both contain a
variable or variables. These are examples of algebraic fractions.
Since the letter used in these fractions and for real numbers, we deal with
algebraic fractions in the same way as we do with fractions in arithmetic.

3.1 Definition of algebraic fraction

 Activity 3.1

Consider the fractions: 3y/1 – x , 5/y + 4 , 2/7
5x – 6/x + 3, x + y/2, 3/4
.
1. Identify the algebraic fractions.

2. Find the value of the variable.

that makes each of the following
expressions zero:

(i) x + 3         (ii) y + 4

(iii) 1 – x

3. What do your answer in (2) above
reveal to you about the fractions
such as:

5x – 6/x + 3 , 5/y + 4 and 3y/1 – x ?

Now consider the following fractions:

(i) 2/x        (ii) x + 3/x – 1
 

(iii) y – 4y – 6

The expressions 2/x,  x + 3/x – 1 and y – 4/2y – 6 are
 all algebraic fractions.

(i) 2/x , the fraction is valid for all real
   numbers except when x = 0

(ii) x + 3/x – 1 exists only if x – 1 ≠ 0
   x – 1 ≠ 0 if x ≠ 1.
∴ the fraction is not defined when
     x = 1.

(iii) y – 4/2y – 6 , the fraction is defined
  (exists) only if the denominator
   is not equal to zero.
Thus if 2y – 6 = 0, then 2y = 6,
y = 6
2 = 3.
∴ y – 4/2y – 6 exists for all real values of
  y except when y = 3.
Note:

(a) If x = 0 (in (i) above), it means dividing
     2 by zero which is not defined.

(b) If x = 1 (in (ii) above), the denominator
     or divisor becomes zero which is not
     defined/which does not exist.

(c) Similarly, in (iii) if y = 3, then 2y – 6
    = 0 (the divisor)which is not defined/
    which does not exist.

In general, an algebraic fraction exists
only if the denominator is not equal to zero. The values of the variable that
make the denominator zero is called a restriction on the variable(s). An
algebraic fraction can have more than one restriction.

Example 3.1

 Identify the restriction on the variable in
 the fraction 3xy/(x + 3) (x – 2) .

Exercise 3.1

1. Identify the restrictions on the
variables of each of the following
fraction.

2. Find the restrictions on the variables
    in:-

3. Find the restrictions on the variables
   in the following fractions:

3.2 Simplification of Fractions

  Activity 3.2

Write the following fractions in the
 simplest form:

A fraction is in its simplest form if its
numerator and denominator do not have common factors. To simplify means to
divide both numerator and denominator by the common factor or factors.
A numerator and a denominator can be divided by the same factor without altering
the value of the fraction.
For example, in 8/12, the numerator and denominator
have a common factor 4.

∴ 8/12 = 8 ÷ 4/12 ÷ 4

            = 2/3

We say that 8/12 and 2/3 are equivalent
  fractions.

∴ 9/12 is equivalent to 3/4 .

If both the numerator and denominator
 of a fraction have more than one term, we
 simplify the fraction by:

(i) Factorising both numerator and
 denominator where necessary.

(ii) Cancelling by the common factor.

Remember: In Senior 2, you learnt to
 factorise algebraic expressions.

Both the numerator and the denominator contain two terms
  each.

The Activity 3.3 introduces fractions
involving quantratic terms which can be expressed as products of linear
 expressions.

Activity 3.3

Factorise the following expressions:-
(a) x² – 81          (b) 3x² – 3

(c) x² – x –12      (d) 2x² – 9x +10

Now consider the following expressions:
(a) x2 – 144           (b) 2x² – 2

(c) x² – 11x + 28     (d) 2x² + 11x + 12
Factorise completely:
(a) x² – 144 is a difference of two

squares.
∴ x² – 144 = x2 – 122
                  = (x – 12) (x + 12)
(factors of a difference of two squares)
(b) 2x² – 2 = 2(x² – 1)
(2 is a common factor)
∴ 2x² – 2 = 2(x –1) (x +1)
     is a difference of two
       squares.
(c) x² – 11x + 28 is a quadratic
     expression.
x² – 11x + 28 = ^x2 – 7x – 4x + 28
 (split the middle term)

x² – 11x + 28 = x(x – 7) – 4 (x – 7)
      (factorise by grouping)
x² – 11x + 28 = (x – 7) (x – 4)

(d) 2x² + 11x + 12
     = 2x² + 8x + 3x + 12 (Split middle term)
     = 2x(x + 4) + 3 (x + 4)
         (factorise by grouping)
    = (x + 4) (2x + 3)

Exercise 3.2

Simplify the following fractions:

For each of the following fractions in
question 8 and 9 below;

(i) write the expression in factor form

(ii) note the restrictions on the variables

(iii) simplify the fractions.

3.3 Addition and subtraction of algebraic fraction with linear
     denominator

Activity 3.4

Remember: To add or to subtract simple fractions, first, find the LCM
of the denominator then convert each fraction into a fraction having this LCM
as denominator. Add or subtract the numerators and simplify your answer.
Now let us consider numbers 3a, 4b, 5c,

To find the LCM of two or more numbers,
first express each number as a product of
its prime factors.
3a = 3 ×1 × a
4b = 2 × 2 × b = 22b
5c = 5 × 1 × c
LCM of 3a, 4b and 5c is 3a × 22b × 5c =
60 abc

Finding equivalent factions means expressing with common denominator.

Note:

1. Addition of algebraic fractions is
    performed in the same way in the
    activity 3.4 above.

2. You can only add fractions if their
    denominators are the same.

3. The basic rule governing fractions
    is that numerator and denominator  can be multiplied by the same factor
    without altering the value of the fraction.
   Your skills in arithmetic should extend to
    skills in algebra.

Exercise 3.3

1. Find the least common multiple of
each of the following:-

Simplify the fractions in the following
 questions.

3.4 Multiplication of algebraic fractions

Activity 3.5

In order to multiply fractions, we identify
common factors, or possible factors of given expression and divide both the
numerator and denominator by the common factors.
For example, to simply an expression such as:

have no common factor is evident.
But it is possible to factorise the expressions
in both numerators and denominators. In Senior 2 we learned how

to factorise.

(a) Now factorise the following:

(i) a² – 4a          (ii) a² + 5a

(iii) 3a – 12

(iv) a² + 7a + 10

a² – 4a is a quadratic expression with only 2 terms, a is a common factor.
∴ a² – 4a = a(a – 4)
  a² + 5a is another quadratic expression
with two terms whose common factor
 is a.
∴ a² + 5a = a(a + 5)
 3a – 12 is a linear expression 3 is a common factor.
∴ 3a – 12 = 3 (a – 4)
 a² + 7a + 10 is a quadratic expression with three terms.
 a² + 7a + 10 = (a + 2)(a + 5)

Note that there are some common factors
 in both numerator and denominator
 which can cancel out.

Exercise 3.4

3.5 Division of algebraic fractions

 Activity 3.6

In general,
if a and b are whole numbers, then
1/a is the reciprocal of a and b /a is

the reciprocal of a/b .
In order to divide a fraction, we must
be able to identify the reciprocal of the
divisor.

Exercise 3.5

1. Write down the reciprocal of each of
the following:

3.6 Solving rational equations

Activity 3.7

Find the LCM of the following.
(a) 12, 16, 24                (b) a, b

(c) (a – 3), 2a² – 18     (d) a², (a + 1)

(e) b, 6, 3b²

Now let us repeat activity 3.7 using:

(a) 6, 8, 16              (b) x, 2

(c) x – 3, x² – 9       (d) x, (x + 1)

(e) y, 4, 3y

The LCM of a group of numbers is the
least or smallest number divided by the
given numbers.
Thus, we start expressing each number or
expression as a product of prime factors.
(a) 6, 8, 16:        6 = 2 × 3
                         8 = 2 × 2 × 2 = 23
                        16 = 2 × 2 × 2 × 2 = 24
                  LCM = 3 × 24 = 48

(b) x and 2 are prime numbers.
     ∴ LCM of x and 2 is 2x.
(c) x + 3 and x2 – 9 are algebraic
      expressions.
    x + 3 is prime, but x2 – 9 is a difference
     of squares.
  ∴ x2 – 9 = (x – 3) (x + 3)

The prime factors involved are
(x + 3), (x – 3).
∴ the LCM = (x + 3) (x – 3).
(d) x and x + 1 are prime expressions
  therefore LCM of x and (x + 1) is
  x(x + 1).
(e) y, 4, 3y:
  y is prime,
4 = 2 × 2 = 2²
3y = 3 × y     ∴ LCM = 22 × 3 × y
In order to be able to solve rational
equations;

(i) Start by finding the LCM of the denominators in each equation.

(ii) Use the LCM to eliminate the denominators by multiplying each
     fraction or term by the LCM.

(iii) Then solve the resulting equation.

Exercise 3.6

    Unit Summary
• An algebraic fraction is defined or said to exist only if its denominator
is not equal to zero. For example, a fuction such as 3/x is valid for all
values of x except when x = 0. The value of a variable that makes the
denominator of a fraction zero is called a restriction on the variable.

• Two algebraic fractions are said to be equivalent if both can be reduced
or simplified to the same simpleast fraction. For example, 2/4 , 5/10 and 20/40, …
   are equivalent, and all are reducible
to 1/2 .

• To add or subtract algebraic fractions, we must first express
them with a common denominator, which represents the LCM of the
denominators of the individual fractions.

• To multiply algebraic fractions,we begin by identifying common
factors in both numerators and denominators. The factors may not
be obvious in such a case, factorise all the algebraic expressions involved if
possible, then proceed to cancel andmultiply.

•Division by a fraction, algebraic or atherimise, means multiplying the
divided by the receiprocal of the divisor. Remember the product of a
fraction and its reciprocal equal to 1. For example, reciprocal of 1/2 is 2,
that of a is 1/a , that of a/b is b/a and 30 on.

• Algebraic equations involving fractions are also called rational equations. To
 solve rational equations, we begun by eliminating the denominations by
 multiplying all the terms by the LCM of the denominators. Then proceed
 to solve the resulting equation.

Unit 3 Test

Simplify each of the following algebraic fractions by expressing them as single
fractions in their lowest terms.

In each case list the restrictions that  apply.