Question 1.35: Solve the following inequalities: (a) x+1/2x-6 >0   (b) ......

(a) Consider the fraction \frac{x+1}{2 x-6}. For the fraction to be positive requires either of the following:

(i)  x+1>0 \text { and } 2 x-6>0

(ii)  x+1<0 \text { and } 2 x-6<0

We consider both cases.

Case (i) 

  \begin{aligned} & x+1>0 \text { and so } x>-1 . \\& 2 x-6>0 \text { and so } x>3 \text {. }\end{aligned}

Both of these inequalities are true only when x > 3. Hence the fraction is positive when x > 3.

Case (ii) 

Both ofthese inequalities are true only when x < — 1. Hence the fraction is positive when x < — 1.

In summary, \frac{x+1}{2 x-6}>0 \text { when } x>3 \text { or } x<-1 \text {. }

(b)

We now consider the fraction \frac{t+1}{t+2}. For the fraction to be negative or zero requires either of the following:

(i) t+1 \leqslant 0 \text { and } t+2>0.

(ii) t+1 \geqslant 0 \text { and } t+2<0.

We consider each case in turn.

Case (i) 

 \begin{aligned} & t+1 \leqslant 0 \text { and so } t \leqslant-1 \\& t+2>0 \text { and so } t>-2\end{aligned}

Hence the inequality is true when t is greater than —2 and less than or equal to — 1.
We write this as —2 < t ≤ -1.

Case (ii) 

 \begin{aligned} & t+1 \geqslant 0 \text { and so } t \geqslant-1 . \\& t+2<0 \text { and so } t<-2 .\end{aligned}

It is impossible to satisfy both t ≥ — 1 and t < —2 and so this case yields no values of t.

In summary, \frac{2 t+3}{t+2} \leqslant 1 \text { when }-2<t \leqslant-1 \text {. }