Understanding Engineering Mathematics

So, for example,

1

a

+

1

b

=

b

ab

+

a

ab

=

b+a

ab

=

a+b

ab

This is calledputting the fractions over a common denominator.Itiswhatwedid
with numerical fractions in Section 1.2.5. Note that the restrictionsa =0andb =0are
necessary for the left-hand end to exist, and are all that is needed to convert to the final
result.

Examples

(i) 1+

1

x+ 1

=

x+ 1

x+ 1

+

1

x+ 1

=

x+ 1 + 1

x+ 1

=

x+ 2

x+ 1

(ii)

1

x

+

1

x− 1

≡

1

x

×

(x− 1 )

(x− 1 )

+

x

x

×

1

(x− 1 )

=

x− 1

x(x− 1 )

+

x

x(x− 1 )

=

x− 1 +x

x(x− 1 )

=

2 x− 1

x(x− 1 )

(iii)

2

x+ 1

−

3

x+ 2

=

2

x+ 1

×

(x+ 2 )

(x+ 2 )

−

3

(x+ 2 )

×

(x+ 1 )

(x+ 1 )

=

2 (x+ 2 )

(x+ 1 )(x+ 2 )

−

3 (x+ 1 )

(x+ 1 )(x+ 2 )

=

2 x+ 4 − 3 x− 3

(x+ 1 )(x+ 2 )

=

−x+ 1

(x+ 1 )(x+ 2 )

=

1 −x

(x+ 1 )(x+ 2 )

In the above examples it has been pretty clear what the ‘common denominator’ should be.
Consider the example

3

x^2 − 1

+

2 x− 1

(x+ 1 )^2

In this case, you might be tempted to put both fractions over(x^2 − 1 )(x+ 1 )^2. But of
course this ignores the fact that bothx^2 − 1 =(x− 1 )(x+ 1 )and(x+ 1 )^2 have a factor
x+1 in common. The lowest common denominator in this case is thus(x^2 − 1 )(x+ 1 )=
(x− 1 )(x+ 1 )^2 and we write:

3

x^2 − 1

+

2 x− 1

(x+ 1 )^2

=

3

(x− 1 )(x+ 1 )

+

2 x− 1

(x+ 1 )^2

=

3 (x+ 1 )

(x− 1 )(x+ 1 )^2

+

( 2 x− 1 )(x− 1 )

(x− 1 )(x+ 1 )^2