MHB Is This an Example of Absolute Value Inequalities?

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The discussion centers on solving the absolute value inequality |(4 - 5x)/2| > 1. Participants confirm that the theorem stating |u| > a implies u < -a or u > a can be applied. The inequality can be rewritten and simplified by multiplying through by 2, leading to |x - 4/5| > 2/5. The solution is straightforward once the absolute value is isolated. This confirms that the original problem is indeed an example of absolute value inequalities.
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Solve the inequality.

| (4 - 5x)/2 | > 1

Can I use the following theorem?

If a > 0, then | u | > a if and only if u < -a or u > a
 
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Yes, you can: either $$\frac{4- 5x}{2}> 1$$ or [math]\frac{4- 5x}{2}< -1[/math]. Continue, with either equation, by multiplying both sides by the positive number, 2.
 
I can take it from here.
 
RTCNTC said:
Solve the inequality.

| (4 - 5x)/2 | > 1

Can I use the following theorem?

If a > 0, then | u | > a if and only if u < -a or u > a

We are given:

$$\left|\frac{4-5x}{2}\right|>1$$

Multiply through by 2/5:

$$\left|x-\frac{4}{5}\right|>\frac{2}{5}$$

Now the solution is easy to read off...:D
 
Is the question what is known as absolute value inequalities?
 
I have been insisting to my statistics students that for probabilities, the rule is the number of significant figures is the number of digits past the leading zeros or leading nines. For example to give 4 significant figures for a probability: 0.000001234 and 0.99999991234 are the correct number of decimal places. That way the complementary probability can also be given to the same significant figures ( 0.999998766 and 0.00000008766 respectively). More generally if you have a value that...

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