To answer the question, let's analyze both compound inequalities.
1. x > -3 and x < 3: This is an intersection of two inequalities where x must be greater than -3 and, at the same time, must be less than 3. The solution set to this compound inequality is all the real numbers between -3 and 3, not including -3 and 3 themselves. This is written in interval notation as (-3, 3).
2. x > -3 or x < 3: This is a union of two inequalities where x can either be greater than -3 or it can be less than 3. If we observe carefully:
- The inequality x < 3 includes all real numbers up to but not including 3.
- The inequality x > -3 includes all real numbers greater than -3.
When we take the union of the two (using "or"), essentially every real number will satisfy at least one part of the compound inequality. If a number is less than 3, it satisfies the second part. If it's not less than 3 (meaning it's equal to 3 or greater than 3), it automatically satisfies the first part (greater than -3). So, every real number falls into one or both parts of this union.
Hence, the new solution set when changing "and" to "or" in the given compound inequality is all real numbers. Therefore, the correct answer to the question is:
a) Yes, the new solution set is all real numbers.