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irina [24]
3 years ago
8

4 ≤ y + 2 ≤ -3(y-2)+24​

Mathematics
1 answer:
Bogdan [553]3 years ago
5 0

Answer:

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.

Step-by-step 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

\frac{4y}{4}\le \frac{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.

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Hey there! 

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SO,
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