Question

Solve the inequality $|x+1|+|x-4|>7,$ indicating the least positive integer $x$ satisfying the inequality.

Answer 1

given inequality is $|x+1|+|x-4|>7$

there are 3 cases

$i.)x<-1,ii.)-1<x<4,iii.)x>4$

the respective equations in these cases are

$i.\left)\right(-1-x)+(4-x)>7,ii.)(x+1)+(4-x)>7,iii.\left)\right(x+1)+(x-4)>7$

in first case

$i.)3-2x>7$ implies $x<5$ and we assumed $x<-1$

therefore the solution in this case is $x<-1$

in second case

$ii.)5>7$ which is not possible

therefore there is no solution in this case

in third case

$iii.)2x-3>7$ implies $x>5$ and we assumed $x>4$

therefore the solution in this case is $x>5$

so, the solution is union of all cases which is

$x\in (-\infty ,-1)\cup (4,\infty )$