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

1.6 Solution of inequalities 35
Example1.35 Solve the following inequalities:
(a)
x +1 2t+3
>0 (b)
2x−6 t +2 1
x +1
Solution (a) Consider the fraction . For the fraction to be positive requires either of the
following:
2x−6
(i) x+1>0and2x−6>0.
(ii) x+1<0and2x−6<0.
We consider both cases.
Case(i)
x+1>0andsox>−1.
2x−6>0andsox>3.
Both of these inequalities are true only whenx > 3. Hence the fraction is positive
whenx > 3.
Case(ii)
x+1<0andsox<−1.
2x−6<0andsox<3.
(b)
Bothoftheseinequalitiesaretrueonlywhenx < −1.Hencethefractionispositive
whenx < −1.
x +1
Insummary,
2x−6 >0whenx>3orx<−1.
2t+3
t +2 1
2t+3
t +2 −10
t +1
t +2 0
Wenowconsiderthefraction t +1
t +2 .Forthefractiontobenegativeorzerorequires
either ofthe following:
(i) t+10andt+2>0.
(ii)t+1 0andt+2<0.
We consider each case inturn.
Case(i) t+10andsot −1.
t+2>0andsot>−2.
Hencetheinequalityistruewhent isgreaterthan −2andlessthanorequalto −1.
We write thisas −2 <t −1.
Case(ii) t+10andsot −1.
t+2<0andsot<−2.