082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

1.5 Algebraic fractions 31
and
4
x +2 iswritten as 4(x +1)
(x +1)(x +2)
Finally the numerators areadded. Hence we have
2
x +1 + 4
x +2 = 2(x +2)
(x +1)(x +2) + 4(x +1)
(x +1)(x +2)
= 2(x+2)+4(x+1)
(x +1)(x +2)
6x+8
=
(x +1)(x +2)
= 6x+8
x 2 +3x+2
Example1.33 Expressasasingle fraction
x 2 +3x+2
x 2 −1
− 2
2x+6
Solution Each fraction iswritten initssimplest form:
x 2 +3x+2
x 2 −1
=
(x +1)(x +2)
(x −1)(x +1) = x +2
x −1
2
2x+6 = 2
2(x +3) = 1
x +3
Thel.c.d.is (x −1)(x +3).Eachfractioniswritteninanequivalent formwithl.c.d.as
denominator:
So
x +2 (x +2)(x +3)
=
x −1 (x −1)(x +3) , 1
x +3 =
x 2 +3x+2
x 2 −1
x −1
(x −1)(x +3)
− 2
2x+6 = x +2
x −1 − 1
x +3
(x +2)(x +3)
=
(x −1)(x +3) − (x−1)
(x −1)(x +3)
= (x+2)(x+3)−(x−1)
(x −1)(x +3)
= x2 +5x+6−x+1
(x −1)(x +3)
= x2 +4x+7
(x −1)(x +3)