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

30 Chapter 1 Review of algebraic techniques
Hence
x 2 +5x+6
2x−2
× x2 −x x(x +3)
=
x 2 +3x+2 2(x +1)
Example1.31 Simplify
x 2 +8x+7
x 2 −6x
÷ x +7
x 3 +x 2
Solution The second fraction isinverted togive
x 2 +8x+7
x 2 −6x
× x3 +x 2
x +7
Factorizing numerators and denominators yields
(x +1)(x +7)
x(x −6)
× x2 (x+1)
(x+7)
= (x +1)(x +7)x2 (x +1)
x(x −6)(x +7)
Common factors ofxand (x +7) arecancelled leaving
(x +1)x(x +1)
x −6
which maybe written as
x(x +1) 2
x −6
1.5.5 Additionandsubtractionofalgebraicfractions
Themethodofaddingandsubtractingalgebraicfractionsisidenticaltothatfornumerical
fractions.
Each fraction is written in its simplest form. The denominators of the fractions are
then examined and the lowest common denominator (l.c.d.) is found. This is the simplest
expression that has the given denominators as factors. All fractions are then written
in an equivalent form with the l.c.d. as denominator. Finally the numerators are
added/subtractedand placedover the l.c.d. Consider the following examples.
Example1.32 Expressasasinglefraction
2
x +1 + 4
x +2
Solution Bothfractionsarealreadyintheirsimplestform.Thel.c.d.ofthedenominators, (x+1)
and (x +2), is found. This is (x +1)(x +2). Note that this is the simplest expression
that has bothx+1 andx+2 as factors.
Each fraction iswritten inan equivalent form with the l.c.d. as denominator. So
2
x +1 iswritten as 2(x +2)
(x +1)(x +2)