One-Variable Compound Inequalities

In this discussion, you will be demonstrating your understanding of compound inequalities and the effect that dividing by a negative has on an inequality view the example attached.

 

Your “and” compound  inequality is: Your “or” compound inequality is:
–1 < –2x + 4 ≤ 5 x < 0   or   3x + 1 ≥ 7

 

Solve the compound inequalities as demonstrated in the example attached.

  • Be careful of how a negative x-term is handled in the solving process. Show all math work arriving at the solutions.
  • Show the solution sets written algebraically and as a union or intersection of intervals. Describe in words what the solution sets mean, and then display a simple line graph for each solution set.
  • Incorporate the following five math vocabulary words into your discussion. Use bold font to emphasize the words in your writing. Do not write definitions for the words; use them appropriately in sentences describing your math work.
    • Compound inequalities
    • And
    • Or
    • Intersection
    • Union

INSTRUCTOR GUIDANCE EXAMPLE: Week Two Discussion

 

One-Variable Compound Inequalities

 

compound inequality

and

or

intersection

union

This is my “and” compound inequality: -7 ≤ 5 + 3x ≤ 20

What that means is the inequality must fulfill two conditions at the same time. It means 5

+ 3x must be equal to or less than 20 and also at the same time greater than or equal to –

7. I think of these as “between” inequalities because it turns out that the solution set for x

will be between two numbers. Now I will find out what those two numbers are.

 

-7 ≤ 5 + 3x ≤ 20 Subtract 5 from all three parts of the inequality.

-7 – 5 ≤ 5 – 5 + 3x ≤ 20 – 5

-12 ≤ 3x ≤ 15 Divide all three parts by 3

-12 ≤ 3x ≤ 15

3 3 3

-4 ≤ x ≤ 5 So any value of x greater than or equal to -4 and less than

or equal to 5 will make this inequality true.

This -4 ≤ x ≤ 5 is how this compound inequality is written algebraically.

As an intersection of sets it would look like [-4, ) (- , 5] which equals [-4, 5] in

interval notation.

 

<—————-[———–|—————]————–> Here is a number line graph of the

-4 0 5 solution set.

The square brackets mean that the end points are included in the solution set; notice the

green highlighting extends through the square brackets as well.

 

 

This is my “or” compound inequality: 4 – x ≥ 1 or 6x – 3 > 27

What this means is that there are two conditions and one of them must be true with any

given x from the solution set but both cannot be true at the same time. Since the solution

will turn out to be two disjoint intervals, I am going to solve each part of the inequality

separately.

 

4 – x ≥ 1 Subtract 4 from both sides.

4 – 4 – x ≥ 1 – 4

– x ≥ – 3 We must pay close attention to that negative in front of x. To

remove it I must divide both sides of the inequality by -1 which also means I must flip the

inequality symbol over so it points the other direction.

– x ≤ – 3 Symbol is flipped.

-1 -1

 

 

x ≤ 3 This is one part of my “or” compound inequality.

6x – 3 > 27 Add 3 to both sides.

6x – 3 + 3 > 27 + 3

6x > 30 Divide both sides by 6, but it is positive, so no flipping involved.

6x > 30

6 6 x > 5 This is the other part of my “or” compound inequality.

 

The complete solution set written algebraically is

x ≤ 3 or x > 5

 

The solution set written in interval notation is the union of two intervals

(- , 3] (5, )

 

Here is a number line graph of the solution set:

<————–|——-]——(——————————> 0 3 5

Notice that the 3 is included in the solution set but 5 is not.

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