Question

4 ≤ y + 2 ≤ -3(y-2)+24​

Answer 1

`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}`

The solution graph is also attached.

Explanation:

Given

`4\:\le \:y\:+\:2\:\le -3\left(y-2\right)+24`

as

`\mathrm{If}\:a\le \:u\le \:b\:\mathrm{then}\:a\le \:u\quad \mathrm{and}\quad \:u\le \:b`

so

`4\le \:y+2\quad \mathrm{and}\quad \:y+2\le \:-3\left(y-2\right)+24`

solving the intervals

as

`4\le \:y+2`

`y+2\ge \:4`

subtract 2 from both sides

`y+2-2\ge \:4-2`

`y\ge \:2`

also

`y+2\le -3\left(y-2\right)+24`

`y+2\le \:-3y+30`

subtract 2 from both sides

`y+2-2\le \:-3y+30-2`

`y\le \:-3y+28`

Add 3y to both sides

`y+3y\le \:-3y+28+3y`

`4y\le \:28`

Divide both sides by 4

`(4y)/(4)\le (28)/(4)`

`y\le \:7`

so combining the intervals

`y\ge \:2\quad \mathrm{and}\quad \:y\le \:7`

Merge overlapping intervals

`2\le \:y\le \:7`

Therefore,

`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}`

The solution graph is also attached.