Understanding Engineering Mathematics

=

3 (x+ 1 )+( 2 x− 1 )(x− 1 )

(x− 1 )(x+ 1 )^2

=

3 x+ 3 + 2 x^2 − 3 x+ 1

(x− 1 )(x+ 1 )^2

=

2 x^2 + 4

(x− 1 )(x+ 1 )^2

Here we have really used the LCM of the denominators.

Solution to review question 2.18

(i) We first of all expresseachfraction over a common denominator and

thencombine the numerators. The common denominator in this case

is simply the product of the denominators and we write

2

x+ 1

−

3

x− 2

≡

2 (x− 2 )

(x+ 1 )(x− 2 )

−

3 (x+ 1 )

(x+ 1 )(x− 2 )

where the factors shown are introduced on top and bottom – essentially

we are multiplying by 1 in each case.

≡

2 (x− 2 )− 3 (x+ 1 )

(x+ 1 )(x− 2 )

=

−x− 7

(x+ 1 )(x− 2 )

This approach extends naturally to three or more fractions as shown

in the next solution.

(ii) The common denominator in this case is(x− 1 )(x+ 1 )(x+ 2 )and

we have

1

x− 1

+

1

x+ 1

−

1

x+ 2

≡

(x+ 1 )(x+ 2 )

(x− 1 )(x+ 1 )(x+ 2 )

+

(x− 1 )(x+ 2 )

(x− 1 )(x+ 1 )(x+ 2 )

−

(x− 1 )(x+ 1 )

(x− 1 )(x+ 1 )(x+ 2 )

≡

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

(x− 1 )(x+ 1 )(x+ 2 )

≡

x^2 + 3 x+ 2 +x^2 +x− 2 −(x^2 − 1 )

(x^2 − 1 )(x+ 2 )

≡

x^2 + 4 x+ 1

(x^2 − 1 )(x+ 2 )