Answer:
[tex]\left|2x-6\right|<4\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:1<x<5\:\\ \:\mathrm{Interval\:Notation:}&\:\left(1,\:5\right)\end{bmatrix}[/tex]
Therefore, the 3rd graph represents the solutions to inequality.
Please also check the graph below.
Step-by-step explanation:
Given the expression
[tex]|2x\:-\:6|\:<\:4[/tex]
[tex]\mathrm{Apply\:absolute\:rule}:\quad \mathrm{If}\:|u|\:<\:a,\:a>0\:\mathrm{then}\:-a\:<\:u\:<\:a[/tex]
[tex]-4<2x-6<4[/tex]
[tex]2x-6>-4\quad \mathrm{and}\quad \:2x-6<4[/tex]
condition 1
[tex]2x-6>-4[/tex]
Add 6 to both sides
[tex]2x-6+6>-4+6[/tex]
[tex]2x>2[/tex]
Divide both sides by 2
[tex]\frac{2x}{2}>\frac{2}{2}[/tex]
[tex]x>1[/tex]
condition 2
[tex]2x-6<4[/tex]
Add 6 to both sides
[tex]2x-6+6<4+6[/tex]
[tex]2x<10[/tex]
Divide both sides by 2
[tex]\frac{2x}{2}<\frac{10}{2}[/tex]
[tex]x<5[/tex]
combine the intervals
[tex]x>1\quad \mathrm{and}\quad \:x<5[/tex]
Merging overlapping intervals
[tex]1<x<5[/tex]
Thus,
[tex]\left|2x-6\right|<4\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:1<x<5\:\\ \:\mathrm{Interval\:Notation:}&\:\left(1,\:5\right)\end{bmatrix}[/tex]
Therefore, the 3rd graph represents the solutions to inequality.
Please also check the graph below.
