Question
Solve for $x , \frac{|x+3|+x}{x+2} > 1 zZ$
✓
Answer
$\text { We have, } \frac{|x+3|+x}{x+2}>1$
$\Rightarrow \frac{|x+3|+x}{x+2}-1>0$
$\Rightarrow \frac{|x+3|+x-x-2}{x+2}>0$
$\Rightarrow \frac{|x+3|-2}{x+2}>0$
$\Rightarrow x =-3$
$\therefore x =-3 $ is a critical point.
So, here we have two intervals $(-\infty,-3)$ and $[-3, \infty)$
Case $I$ : When $- 3 \leq x<\infty$, then $|x+3|=(x+3)$
$\therefore \frac{|x+3|-2}{x+2}>0$
$\Rightarrow \frac{x+3-2}{x+2}>0$
$\Rightarrow \frac{x+1}{x+2}>0$
$\Rightarrow \frac{(x+1)(x+2)^2}{(x+2)}>0 \times(x+2)^2$
$\Rightarrow(x+1)(x+2)>0$
Product of $(x + 1)$ and $(x + 2)$ will be positive, if both are of same sign.
$\therefore(x+1)>0 $ and $(x+2)>0$
or $(x+1)<0 $ and $(x+2)<0$
$\Rightarrow x>-1 $ and $ x>-2$
or $ x<-1 \text { and } x<-2$
On number line, these inequalities can be represented as,
Thus, $-1< x< \infty$ or $-\infty< x< -2$
But, here - $3 \leq x <\infty$
$\therefore-1< x <\infty$ or $-3 \leq x <-2$
Then, solution set in this case is
$x \in[-3,-2) \cup(-1, \infty)$
Case $II$ : When $x<-3$, then $|x+3|=-(x+3)$
$\therefore \frac{|x+3|-2}{x+2}>0$
$\Rightarrow \frac{-x-3-2}{x+2}>0$
$\Rightarrow \frac{-(x+5)}{x+2}>0$
$\Rightarrow \frac{x+5}{x+2}<0$
$\Rightarrow \frac{(x+5)(x+2)^2}{x+2}<0 \times( x +2)^2$
$\Rightarrow( x +5)( x +2)<0$
Product of $(x + 5)$ and $(x + 2)$ will be negative, if both are of opposite sign.
$\therefore(x+5)>0 $ and $(x+2)<0$
$\text { or }(x+5)<0 $ and $(x+2)>0$
$\Rightarrow x>-5 $ and $ x<-2$
$\text { or } x<-5 $ and $ x>-2$
On number line, these inequalities can be represented as,
Thus, $- 5 < x < - 2$ i.e., solution set in the case is $x\in (- 5, - 2).$
On combining cases $I$ and $II,$ we get the required solution set of given inequality, which is
$x \in(-5,-2) \cup(-1, \infty)$
$\Rightarrow \frac{|x+3|+x}{x+2}-1>0$
$\Rightarrow \frac{|x+3|+x-x-2}{x+2}>0$
$\Rightarrow \frac{|x+3|-2}{x+2}>0$
$\Rightarrow x =-3$
$\therefore x =-3 $ is a critical point.
So, here we have two intervals $(-\infty,-3)$ and $[-3, \infty)$
Case $I$ : When $- 3 \leq x<\infty$, then $|x+3|=(x+3)$
$\therefore \frac{|x+3|-2}{x+2}>0$
$\Rightarrow \frac{x+3-2}{x+2}>0$
$\Rightarrow \frac{x+1}{x+2}>0$
$\Rightarrow \frac{(x+1)(x+2)^2}{(x+2)}>0 \times(x+2)^2$
$\Rightarrow(x+1)(x+2)>0$
Product of $(x + 1)$ and $(x + 2)$ will be positive, if both are of same sign.
$\therefore(x+1)>0 $ and $(x+2)>0$
or $(x+1)<0 $ and $(x+2)<0$
$\Rightarrow x>-1 $ and $ x>-2$
or $ x<-1 \text { and } x<-2$
On number line, these inequalities can be represented as,
Thus, $-1< x< \infty$ or $-\infty< x< -2$
But, here - $3 \leq x <\infty$
$\therefore-1< x <\infty$ or $-3 \leq x <-2$
Then, solution set in this case is
$x \in[-3,-2) \cup(-1, \infty)$
Case $II$ : When $x<-3$, then $|x+3|=-(x+3)$
$\therefore \frac{|x+3|-2}{x+2}>0$
$\Rightarrow \frac{-x-3-2}{x+2}>0$
$\Rightarrow \frac{-(x+5)}{x+2}>0$
$\Rightarrow \frac{x+5}{x+2}<0$
$\Rightarrow \frac{(x+5)(x+2)^2}{x+2}<0 \times( x +2)^2$
$\Rightarrow( x +5)( x +2)<0$
Product of $(x + 5)$ and $(x + 2)$ will be negative, if both are of opposite sign.
$\therefore(x+5)>0 $ and $(x+2)<0$
$\text { or }(x+5)<0 $ and $(x+2)>0$
$\Rightarrow x>-5 $ and $ x<-2$
$\text { or } x<-5 $ and $ x>-2$
On number line, these inequalities can be represented as,
Thus, $- 5 < x < - 2$ i.e., solution set in the case is $x\in (- 5, - 2).$
On combining cases $I$ and $II,$ we get the required solution set of given inequality, which is
$x \in(-5,-2) \cup(-1, \infty)$
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