Answer:
[tex]\left|x-3\right|<5\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:-2<x<8\:\\ \:\mathrm{Interval\:Notation:}&\:\left(-2,\:8\right)\end{bmatrix}[/tex]
The graph is also attached.
Step-by-step explanation:
Given the expression
[tex]|x-3|\:<\:5[/tex]
Apply absolute rule:
[tex]\mathrm{If}\:|u|\:<\:a,\:a>0\:\mathrm{then}\:-a\:<\:u\:<\:a[/tex]
so the expression becomes
[tex]-5<x-3<5[/tex]
[tex]x-3>-5\quad \mathrm{and}\quad \:x-3<5[/tex]
solving condition 1
x−3<5
Add 3 to both sides
x−3+3<5+3
x<8
solving condition 2
x−3>−5
Add 3 to both sides
x−3+3>−5+3
x>−2
combining the intervals
[tex]x>-2\quad \mathrm{and}\quad \:x<8[/tex]
Merging overlapping intervals
[tex]-2<x<8[/tex]
Therefore,
[tex]\left|x-3\right|<5\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:-2<x<8\:\\ \:\mathrm{Interval\:Notation:}&\:\left(-2,\:8\right)\end{bmatrix}[/tex]
The graph is also attached.
