Mathematics
Algebra - One-Variable Compound Inequalities Discussion

Question Description

I’m stuck on a Algebra question and need an explanation.

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 (Attached) to complete this discussion:  MAT222.W1.DiscussionExample.pdf 

According to the first letter of your last name, problem is circled in attached document

Week 2 discussion 001.jpg 

  • Solve the compound inequalities as demonstrated in Elementary and Intermediate Algebra and the Instructor Guidance in the left navigation toolbar, being 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 as 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 

Unformatted Attachment Preview

INSTRUCTOR GUIDANCE EXAMPLE: Week One Discussion Domains of Rational Expressions Students, you are perfectly welcome to format your math work just as I have done in these examples. However, the written parts of the assignment MUST be done about your own thoughts and in your own words. You are NOT to simply copy this wording into your posts! Here are my given rational expressions oh which to base my work. 25x2 – 4 5 – 9w 67 9w2 – 4 The domain of a rational expression is the set of all numbers which are allowed to substitute for the variable in the expression. It is possible that some numbers will not be allowed depending on what the denominator has in it. In our Real Number System division by zero cannot be done. There is no number (or any other object) which can be the answer to division by zero so we must simply call the attempt “undefined.” A denominator cannot be zero because in a rational number or expression the denominator divides the numerator. In my first expression, the denominator is a constant term, meaning there is no variable present. Since it is impossible for 67 to equal zero, there are no excluded values for the domain. We can say the domain (D) is the set of all Real Numbers, written in set notation that would look like this: D = {x| x ∈ ℜ} or even more simply as D = ℜ. For my second expression, I need to set the denominator equal to zero to find my excluded values for w. 9w2 – 4 = 0 (3w – 2)(3w + 2) = 0 3w – 2 = 0 or 3w + 2 = 0 3w = 2 or 3w = -2 w = 2/3 or w = -2/3 I notice this is a difference of squares which I can factor. Set each factor equal to zero. Add or subtract 2 from both sides. Divide both sides by 3. These are the excluded values for my second expression. The domain (D) for my second expression is the set of all Reals excluding ±2/3. In set notation, this can be written as D = {w| w ∈ ℜ, w ≠ ±2/3} Now, both of my expressions do not have excluded values. In one expression, I have no excluded values because there is no variable in the denominator and a non-zero number will never just become zero. In the other expression, there are two excluded numbers because both, if inserted in place of the variable, would cause the denominator to become zero and thus the whole expression would become undefined. ...
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Carnegie Mellon University

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