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THis is the comment from instructor My assigned number is 20.

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THis is the comment from instructor





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

I will first work with my "andcompound inequality, 0 ≤ 6x + 3 < 12. When a compound inequalityis 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 orequal 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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