91Ó°ÊÓ

Interval notation is a concise way to describe subsets of the real number line. It's particularly useful when indicating ranges of solutions in inequalities, as students can easily represent infinite sections, finite ranges, and even single points using this notation.

To understand interval notation, one must be familiar with a few symbols. The parenthesis, '(', ')', indicate that the end number is not included, known as 'open interval', while brackets, '[', ']', mean that the end number is included, called 'closed interval'. So, when we describe the solution set of the inequality \(2s > 10 \Rightarrow s > 5\), we use the open interval \(5, \infty)\) to indicate that all numbers greater than 5 are solutions, but 5 itself is not. Similarly, the negative case \(s < 2\) will be represented as \( - \infty, 2\). When combined, as in the given exercise, the disjoint solution sets are united using the union symbol \(\cup\), leading to the interval notation \( ( - \infty, 2 ) \cup ( 5, \infty )\).

Remember, when dealing with absolute value inequalities like the one in our exercise, your solutions may be in two separate intervals, so be prepared to use union to bring them together.

Graphing on a number line is a visual way to represent solutions to inequalities. It's an effective tool that provides a quick visual cue about which numbers satisfy the inequality.

Let's take the results from our earlier step of solving the inequality \( 2s > 10\) and \( s < 2\). Once translated into interval notation, you'll want to bring them to life on a number line. For the inequality \( s > 5\), an open circle is placed at 5 to indicate that 5 is not included, and an arrow drawn to the right to demonstrate that all numbers greater than 5 are part of the solution. On the flip side, \( s < 2\) has an open circle at 2 with an arrow pointing left, showing all numbers less than 2 are solutions.

By representing both on a number line, we effectively visualize the entire solution set. The open circles show that our interval does not include the boundary numbers, adhering to the specifics of interval notation.

An absolute value equation involves the absolute value function, which is the distance of a number from zero on the number line, always non-negative. For instance, \( |x| \), by definition, equals \( x \) if \( x \) is greater than or equals to zero, and \( -x \) if \( x \) is less than zero.

When you encounter an absolute value inequality, such as the one in the exercise \( |2s-7| > 3 \), we recognize that there are two scenarios to consider: either \( 2s - 7\) is greater than 3 or less than -3. Why? Because the absolute value 'erases' the negatives, so we must account for both the positive and the negative cases separately.

To solve such equations, you first remove the absolute value by creating the two individual inequalities that result from considering the 'distance' being both above and below zero. The final outcome will often result in two distinct intervals that represent the range of solutions where the initial absolute value inequality holds true. This process enables us to tackle complex absolute value inequalities with a clear, step-by-step approach.