# One-Variable Compound Inequalities

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Mathematics

### Description

In this discussion, you will be demonstrating your understanding of compound inequalities and the effect that dividing by a negative has on an inequality. Read the following instructions in order and view the example to complete this discussion.

-9 < -1 + 4x < 15                           8 – x > 15 or 6x – 13 > 11

• Solve the compound inequalities as demonstrated in Elementary and Intermediate Algebra and the Instructor Guidance in the left navigation toolbar. 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. This is demonstrated in the Instructor Guidance in the left navigation toolbar.
• 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
Your initial post should be at least 250 words in length. Support your claims with examples from required material(s) and/or other scholarly resources, and properly cite any references. MAT222.W2.DiscussionExample.pdf

### Unformatted Attachment Preview

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 -7 – 5 ≤ 5 – 5 + 3x ≤ 20 – 5 -12 ≤ 3x ≤ 15 -12 ≤ 3x ≤ 15 3 3 3 -4 ≤ x ≤ 5 Subtract 5 from all three parts of the inequality. Divide all three parts by 3 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 6x – 3 > 27 6x – 3 + 3 > 27 + 3 6x > 30 6x > 30 6 6 x>5 This is one part of my “or” compound inequality. Add 3 to both sides. Divide both sides by 6, but it is positive, so no flipping involved. 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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