MHB Can I use the theorem for solving the given inequality?

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The discussion focuses on using a theorem to solve the inequality |[3(x - 2)/4] + 4(x - 1)/3| ≤ 2. The theorem states that if a > 0, then |u| < a is equivalent to -a < u < a. Participants confirm that this theorem can be applied to the given inequality. By multiplying through by 12, the inequality simplifies to |25x - 34| ≤ 24. The final solution indicates that the values of x are within 24/25 units from 34/25 on the number line.
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|[3(x - 2)/4] + 4(x - 1)/3| ≤ 2

Can I use the following theorem?

If a > 0, then | u | < a if and only if -a < u < a
 
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RTCNTC said:
|[3(x - 2)/4] + 4(x - 1)/3| ≤ 2

Can I use the following theorem?

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

yes
 
RTCNTC said:
|[3(x - 2)/4] + 4(x - 1)/3| ≤ 2

Can I use the following theorem?

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

I'd begin by multiplying through by 12 to get:

$$\left|9(x-2)+16(x-1)\right|\le24$$

$$|25x-34|\le24$$

$$\left|x-\frac{34}{25}\right|\le\frac{24}{25}$$

Now it's obvious the solution is the set of all real numbers whose distance on the number line from 34/25 is less than or equal to 24/25...:D
 
Thank you. I can take it from here.
 
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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