Question

What is the solution set for the inequality |-3-5x|-8<-2

Answer 1

The hard way:


`\left| -3-5x \right| -8<-2\\ \\ \left| -3-5x \right| <6\\ \\ { \left( -3-5x \right) }^( 2 )<{ 6 }^( 2 )\\ \\ \left( -3-5x \right) \left( -3-5x \right) <36`


`\\ \\ 9+15x+15x+25{ x }^( 2 )<36\\ \\ 25{ x }^( 2 )+30x+9<36\\ \\ 25{ x }^( 2 )+30x-27<0\\ \\ Say\quad f\left( x \right) =25{ x }^( 2 )+30x-27,\\ \\ and\quad that\quad f\left( x \right) =0`


`\\ \\ 25{ x }^( 2 )+30x-27=0\\ \\ 25{ x }^( 2 )+30x=27\\ \\ { x }^( 2 )+\frac { 30 }{ 25 } x=\frac { 27 }{ 25 } \\ \\ { x }^( 2 )+\frac { 6 }{ 5 } x=\frac { 27 }{ 25 } \\ \\ { \left( x+\frac { 3 }{ 5 } \right) }^( 2 )-{ \left( \frac { 3 }{ 5 } \right) }^( 2 )=\frac { 27 }{ 25 }`


`\\ \\ { \left( x+\frac { 3 }{ 5 } \right) }^( 2 )=\frac { 36 }{ 25 } \\ \\ x+\frac { 3 }{ 5 } =\pm \frac { 6 }{ 5 } \\ \\ x=-\frac { 3 }{ 5 } \pm \frac { 6 }{ 5 }`


`Therefore:\\ \\ x=\frac { 3 }{ 5 } \quad and\quad x=-\frac { 9 }{ 5 } \quad when\quad f\left( x \right) =0\\ \\ Now:\\ \\ f\left( x \right) <0,\\ \\`

When:

-9/5<x<3/5

Answer 2

`|-3-5x|-8<-2\\ |-3-5x|<6\\ -3-5x<6 \wedge -3-5x>-6\\ -5x<9 \wedge -5x>-3\\ x>-(9)/(5) \wedge x<(3)/(5)\\ x\in\left(-(9)/(5),(3)/(5)\right)`