Answer:
[tex]4\le \:y+2\le \:-3\left(y-2\right)+24\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:2\le \:y\le \:7\:\\ \:\mathrm{Interval\:Notation:}&\:\left[2,\:7\right]\end{bmatrix}[/tex]
The solution graph is also attached.
Step-by-step explanation:
Given
[tex]4\:\le \:y\:+\:2\:\le -3\left(y-2\right)+24[/tex]
as
[tex]\mathrm{If}\:a\le \:u\le \:b\:\mathrm{then}\:a\le \:u\quad \mathrm{and}\quad \:u\le \:b[/tex]
so
[tex]4\le \:y+2\quad \mathrm{and}\quad \:y+2\le \:-3\left(y-2\right)+24[/tex]
solving the intervals
as
[tex]4\le \:y+2[/tex]
[tex]y+2\ge \:4[/tex]
subtract 2 from both sides
[tex]y+2-2\ge \:4-2[/tex]
[tex]y\ge \:2[/tex]
also
[tex]y+2\le -3\left(y-2\right)+24[/tex]
[tex]y+2\le \:-3y+30[/tex]
subtract 2 from both sides
[tex]y+2-2\le \:-3y+30-2[/tex]
[tex]y\le \:-3y+28[/tex]
Add 3y to both sides
[tex]y+3y\le \:-3y+28+3y[/tex]
[tex]4y\le \:28[/tex]
Divide both sides by 4
[tex]\frac{4y}{4}\le \frac{28}{4}[/tex]
[tex]y\le \:7[/tex]
so combining the intervals
[tex]y\ge \:2\quad \mathrm{and}\quad \:y\le \:7[/tex]
Merge overlapping intervals
[tex]2\le \:y\le \:7[/tex]
Therefore,
[tex]4\le \:y+2\le \:-3\left(y-2\right)+24\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:2\le \:y\le \:7\:\\ \:\mathrm{Interval\:Notation:}&\:\left[2,\:7\right]\end{bmatrix}[/tex]
The solution graph is also attached.
