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Magazine Chapter 8: Problem 101
Solve the absolute value inequality in part a. Graph the solution set and write it in interval notation. Then use your work from part a to determine the solution set for the absolute value inequality in part \(b\). Graph the solution set and write it in interval notation. a. \(\left|\frac{4 x-4}{3}\right|-1>11\) b. \(\left|\frac{4 x-4}{3}\right|-1 \leq 11\)
Short Answer
Expert verified
Part (a): \\( (-\infty, -8) \cup (10, \infty) \\); Part (b): \\([-8, 10]\\).
Step by step solution
01
Understand the Inequality
We are given \(\left|\frac{4x - 4}{3}\right| - 1 > 11\). First, isolate the absolute value by adding 1 to both sides of the inequality to get \(\left|\frac{4x - 4}{3}\right| > 12\).
02
Set Up the Compound Inequality
The inequality \(\left|A\right| > B\) translates to the compound inequality \(|A| > B \text{ implies either } A > B \text{ or } A < -B\). Thus, set up the two inequalities: \(\frac{4x - 4}{3} > 12\) or \(\frac{4x - 4}{3} < -12\).
03
Solve the First Inequality
Solve \(\frac{4x - 4}{3} > 12\) by first multiplying each side by 3 to clear the fraction: \(4x - 4 > 36\). Next, add 4 to both sides to obtain \(4x > 40\). Finally, divide by 4 to isolate \(x\): \(x > 10\).
04
Solve the Second Inequality
Solve \(\frac{4x - 4}{3} < -12\) by multiplying each side by 3: \(4x - 4 < -36\). Add 4 to both sides to get \(4x < -32\), and then divide by 4: \(x < -8\).
05
Combine Solutions for Part a
The solution set for \(\left|\frac{4x-4}{3}\right| - 1 > 11\) is the union of \(x > 10\) and \(x < -8\). In interval notation, this is \((-\infty, -8) \cup (10, \infty)\).
06
Graph the Solution Set for Part a
Draw a number line. Shade the regions to the left of -8 and to the right of 10, but do not include the points -8 and 10 (open circles).
07
Set Up the Inequality for Part b
Now, consider \(\left|\frac{4x - 4}{3}\right| - 1 \leq 11\). Similarly, isolate the absolute value to obtain \(\left|\frac{4x - 4}{3}\right| \leq 12\).
08
Solve the Compound Inequality for Part b
The inequality \(\left|A\right| \leq B\) is equivalent to \(-B \leq A \leq B\). Thus, solve \(-12 \leq \frac{4x - 4}{3} \leq 12\).
09
Solve the Left Side of the Compound Inequality
Solve \(-12 \leq \frac{4x - 4}{3}\) by multiplying through by 3: \(-36 \leq 4x - 4\). Add 4 to both sides: \(-32 \leq 4x\), then divide by 4 to find \(-8 \leq x\).
10
Solve the Right Side of the Compound Inequality
Solve \(\frac{4x - 4}{3} \leq 12\) as in Step 3, yielding \(x \leq 10\).
11
Combine Solutions for Part b
Combine \(-8 \leq x \leq 10\) giving the solution set \[[-8, 10]\].
12
Graph the Solution Set for Part b
On the number line, shade the segment between -8 and 10, including the endpoints (closed circles).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Interval Notation
Interval notation is a way to describe sets of numbers on a number line. It provides a clear and concise method for indicating the range of solutions in algebraic inequalities. Rather than listing out numbers, interval notation uses parentheses or brackets to denote the inclusion or exclusion of endpoints.
For example:
For example:
- Parentheses "()" are used to indicate that the endpoints are not included in the interval. For example, the interval \((-\infty, -8)\) means all numbers less than -8.
- Brackets "[]" imply that the endpoints are included in the interval. For example, the interval \([-8, 10]\) includes the numbers -8 and 10 along with everything in between.
Compound Inequalities
Compound inequalities involve two separate inequalities that are either connected by "and" (intersection) or "or" (union). These are crucial for representing complex solution sets and help us when dealing with absolute value inequalities.
In the exercise given:
In the exercise given:
- For part a, the inequality \(|A| > B\) is expressed as a compound inequality: either \(A > B\) or \(A < -B\). This translates our absolute value inequality into two separate linear inequalities that must be solved individually. Thus, we get \(x > 10\) or \(x < -8\).
- For part b, the inequality \(|A| \leq B\) becomes the compound inequality \(-B \leq A \leq B\), indicating a range of values. After solving, this gives \(-8 \leq x \leq 10\).
Graphing Inequalities
Graphing inequalities is an effective way to visualize the solution set of an inequality or a system of inequalities. By using a number line, we can clearly see which segments of the number line are included in the solution.
In the steps provided:
In the steps provided:
- For part a, graph the solution set \((-\infty, -8) \cup (10, \infty)\) by drawing a line with open circles at -8 and 10. Shade the regions extending to the left of -8 and to the right of 10 to show all numbers included except endpoints.
- For part b, graph \([-8, 10]\) on a number line by shading the segment between -8 and 10 and using closed circles to include the endpoints. This visually represents that all numbers from -8 to 10 are part of the solution.
