Good Evening,

I need help I am not sure what I did wrong.

THis is the comment from instructor

My assigned number is 20. My "**and**" **compound inequality** is 0 ≤ 6x + 3 < 12 and my "**or**" **compound inequality** is -x + 7 ≥ 10 **or** 3x - 3 ≤ 12.

I will first work with my "**and**" **compound** **inequality**, 0 ≤ 6x + 3 < 12. When a **compound** **inequality**is joined by the word ** and**, I need to look for the values that make both inequalities true. In my given

**compound inequality**this means that 6x+3 must be less than "12" and must be greater than

**or**equal to 0. In other words, I want to find numbers that when graphed satisfy both inequalities.

First I need to solve the **compound inequality**, meaning that I need *x* alone I need to subtract 3 from each piece of the inequality. This means I need to subtract 3 from 0, subtract 2 from 6x+3, and subtract 3 from 12. Which gives me the answer -3 ≤ 6x < 9. The inequality can then be divided by 6 to get x alone and simplified further by dividing -3 over 6 and dividing 9 over 6 by 3. I am left with the algebraic **compound inequality** . Therefore, any value of *x* that is greater than **or**equal to **and** less than will satisfy the inequality **and** make it true.

0 ≤ 6x + 3 < 12

__-3 -3 -3__

-3 ≤ 6x < 9

As an** intersection** of sets the **compound inequality** is which is in interval notation. The set notation is . Graphing the **compound inequality** on a number line would look like this;

ß------------------------------------- [--------I-----)---------------------------------------------------------à 0

A square bracket means that number fraction is included and parentheses means that **fraction **is not included.

For my *or***compound** **inequality**, -x + 7 ≥ 10 **or** 3x - 3 ≤ 12 I first want to solve each inequality. To solve -x + 7 ≥ 10 I will subtract 7 from each piece. My result is then -x ≥ 3. Now, because there is a negative sign in front of the x I want to divide each side by -1. Thus, my **compound** **inequality** is x ≤ -3. Notice that I changed the sign because the positive 3 changed to a negative 3.

-x + 7 ≥ 10

__ -7 -7__

-x ≥ 3

x -3

To solve 3x - 3 ≤ 12, I will add 3 to each piece. My result is then 3x ≤ 15, whereby I will divide each part by 3 to get *x* alone. The inequality is then x ≤ 5.

3x - 3 ≤ 12

__ +3 +3__

3x ≤ 15

x ≤ 5

The complete solution set written algebraically is x -3 **or** x ≤ 5. The solution set as an **intersection** of sets would be [-3,)(. Which would equal [-3,5] in interval notation. Written in interval notation the answer will be separated by a __union__

The graphed number line would of the solution set would look like this;

ß---------------------[---------------I--------------------]--------------------------à

-3 0

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#### Top Answer

"and" compound inequality solution: [ − 2 ... View the full answer