Answer:
[tex]3 < x < 5[/tex]
Step-by-step explanation:
Given inequality:
[tex]|2x - 8| < 2[/tex]
[tex]\textsf{Apply absolute rule}: \quad \textsf{If }|u| < a\:\textsf{ when } \: a > 0,\: \textsf{then }-a < u < a[/tex]
[tex]\implies u=2x-8\: \textsf{ and }\:a=2[/tex]
[tex]\implies -2 < 2x-8 < 2[/tex]
Therefore:
[tex]\implies -2 < 2x-8[/tex]
[tex]\implies 6 < 2x[/tex]
[tex]\implies 3 < x[/tex]
And:
[tex]\implies 2x-8 < 2[/tex]
[tex]\implies 2x < 10[/tex]
[tex]\implies x < 5[/tex]
Merge the overlapping intervals: [tex]3 < x < 5[/tex]
Proof
Graph: [tex]y=|2x-8|[/tex]
Add a line at [tex]y=2[/tex]
The interval that satisfies [tex]|2x - 8| < 2[/tex] is the area under the points of intersection (shaded area on the attached graph).
The values of x when [tex]y=|2x-8|[/tex] is less than 2 is more than x = 3 and less than x = 5.
So
Case-1
Case-2
So solution is
