Solve |6x – 7| < = 25 and write interval notation for the solution set.

Undo the absolute value:

6x - 7 <= 25

and 6x - 7 >= -25

Now solve for x:
6x - 7 <= 25
6x <= 32
x <= 16/3

6x-7 >= -25
6x >= -18
6x >= -3

-3 <= x <= 16/3

Therefore, interval notation is [-3, 16/3]. Brackets are used because x can equal -3 and 16/3.

|6x – 7| ≤ 25

so 6x-7 ≤ 25 AND -6x+7 ≤ 25
6x ≤ 32 AND -6x ≤ 18
x ≤ 16/3 AND x ≥ -3

-3 ≤ x ≤ 16/3

What would be the answer in interval notation?

A. [-16/3,3]
B. [-3,16/3]
C. (-infinity, -16/3 U [3, infinity)
D. (-infinity, -3] U [16/3,infinity)

Based on the answer you provided I am goin to say B. Am I correct? Thanks!

"Interval notation is [-3, 16/3]."

Yes, B is correct.

Thank you!

To solve this inequality, we need to isolate the absolute value expression and consider two cases: when the expression inside the absolute value is positive, and when it is negative.

Case 1: 6x - 7 ≥ 0
In this case, the absolute value becomes |6x - 7| = 6x - 7. Therefore, we can rewrite the inequality as follows:
6x - 7 ≤ 25

Solving this inequality:
6x ≤ 32
x ≤ 32/6
x ≤ 16/3

Case 2: 6x - 7 < 0
In this case, the absolute value becomes |6x - 7| = -(6x - 7), which gives us the opposite sign. Therefore, we can rewrite the inequality as follows:
-(6x - 7) ≤ 25

Solving this inequality:
-6x + 7 ≤ 25
-6x ≤ 18
x ≥ -18/6
x ≥ -3

Combining the solutions from both cases, we have:
x ≤ 16/3 and x ≥ -3

Now we can represent the solution set using interval notation:
Interval Notation: (-∞, -3] ∪ [16/3, ∞)