im sorry but this problem is part of a major grade i have in my pre-ap algerbrea 2 class....

I will look at it later, check in the am.

i'll try... arigato, sorry for all the trouble

Of course! I can definitely help you with those problems.

1. To solve the inequality |ax + b| ≤ c for x in terms of a, b, and c, where a > 0 and c > 0, we need to consider two cases:

Case 1: ax + b ≥ 0
In this case, the absolute value of ax + b is equal to ax + b itself, so we have ax + b ≤ c. To isolate x, we can subtract b from both sides, which gives us ax ≤ c - b. Finally, divide both sides by a (since a > 0) to get x ≤ (c - b) / a.

Case 2: ax + b < 0
In this case, the absolute value of ax + b is equal to -(ax + b), so we have -(ax + b) ≤ c. To isolate x, we can multiply both sides by -1 (which changes the direction of the inequality), resulting in ax + b ≥ -c. Now, subtract b from both sides to get ax ≥ -c - b. Finally, divide both sides by a to get x ≥ (-c - b) / a.

Therefore, the solution to the inequality |ax + b| ≤ c in terms of a, b, and c, where a > 0 and c > 0, is x ≤ (c - b) / a or x ≥ (-c - b) / a.

2. To solve the inequality |ax + b| > c for x in terms of a, b, and c, where a > 0 and c > 0, we also need to consider two cases:

Case 1: ax + b ≥ 0
In this case, the absolute value of ax + b is equal to ax + b itself, so we have ax + b > c. To isolate x, we can subtract b from both sides, which gives us ax > c - b. Finally, divide both sides by a (since a > 0) to get x > (c - b) / a.

Case 2: ax + b < 0
In this case, the absolute value of ax + b is equal to -(ax + b), so we have -(ax + b) > c. To isolate x, we can multiply both sides by -1, resulting in ax + b < -c. Now, subtract b from both sides to get ax < -c - b. Finally, divide both sides by a to get x < (-c - b) / a.

Therefore, the solution to the inequality |ax + b| > c in terms of a, b, and c, where a > 0 and c > 0, is x > (c - b) / a or x < (-c - b) / a.

I hope this helps you solve the problems! If you have any further questions, feel free to ask.