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Q.
The solution set of the inequality $|x+1|+|x|>3$ is
Linear Inequalities
Solution:
On LHS of the given inequality, we have two terms both containing modulus. By equating the expression with in the modulus to zero, we get $x=-1,0$ as critical points. These critical points divide the real line in three parts as $(-\infty,-1),[-1,0),[0, \infty)$
Case I When $-\infty< x< -1$,
$|x+1|+|x|>3 \Rightarrow-x-1-x >3 \Rightarrow x< -2$
Case II When $-1 \leq x<0$,
$|x+1|+|x| >3 \Rightarrow x+1-x>3 \Rightarrow 1>3 \text { (not possible) }$
Case III When $0 \leq x< \infty$,
$|x+1|+|x|>3 \Rightarrow x+1+x>3 \Rightarrow x>1 \text {. }$
Combining the results of cases (I), (II) and (III), we get
$x \in(-\infty,-2) \cup(1, \infty)$