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.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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