91Ó°ÊÓ

When expressing the solutions of inequalities, we often use interval notation. Interval notation allows us to clearly represent the range of values that satisfy an inequality.

Suppose we have a solution set where x can be any value less than a certain number, like in our problem: \(x < \frac{26}{9}\). This is written in interval notation as follows: \( (-\infty, \frac{26}{9}) \). The parenthesis indicates that \(\frac{26}{9}\) is not included in the solution set.

Similarly, for our second part of the solution set, where \( x > \frac{34}{9}\), this is written in interval notation as \( (\frac{34}{9}, \infty) \). Here as well, the parenthesis indicates that \(\frac{34}{9}\) is not included.
To combine these two intervals, we use the union symbol (\cup) to denote that any number falling into either interval is part of the solution. Hence, the final combined solution can be written in interval notation as \( (-\infty, \frac{26}{9}) \cup (\frac{34}{9}, \infty) \).

In the given inequality, \(\left|\frac{5}{3}-\frac{1}{2} x\right|>\frac{2}{9}\), we encounter fractions. One way to simplify solving these inequalities is by eliminating the fractions through multiplication.

To eliminate fractions, we use the Least Common Multiple (LCM) of all denominators involved. In this problem, the denominators are 3, 2, and 9. The LCM of these numbers is 18. This LCM is used as a common factor to multiply each term thus: \( 18 \times \frac{5}{3} - 18 \times \frac{1}{2} x > 18 \times \frac{2}{9} \).

Breaking it down: \( 18 \times \frac{5}{3} = 30\), \( 18 \times \frac{1}{2} x = 9x\), and \( 18 \times \frac{2}{9} = 4\). After multiplication, the inequality changes to: \( 30 - 9x > 4 \). This step helps in making the inequality more manageable.

Given an absolute value inequality like \(\left|\frac{5}{3}-\frac{1}{2} x\right| > \frac{2}{9}\), we need to split it into two separate inequalities to solve.

The general rule for absolute inequalities \( |A| > B \) is to consider the two cases: \( A > B \) or \( -A > B \). Translating this to our specific problem:
\(\frac{5}{3}-\frac{1}{2} x > \frac{2}{9}\) or \(\frac{5}{3}-\frac{1}{2} x < -\frac{2}{9}\).
Solving these two inequalities separately allows us to find the range of values for which the original inequality holds true.
The first inequality simplifies to: \( 30 - 9x > 4 \rightarrow -9x > -26 \rightarrow x < \frac{26}{9} \).
The second inequality simplifies to: \( 30 - 9x < -4 \rightarrow -9x < -34 \rightarrow x > \frac{34}{9} \).

Isolating the variable is a crucial step in solving inequalities. This involves performing operations to get the variable on one side and a constant on the other.

For example, consider the inequality from our problem: \( 30 - 9x > 4 \). Our goal is to isolate x. We first subtract 30 from both sides:
\( 30 - 30 - 9x > 4 - 30 \rightarrow -9x > -26 \).
Next, we divide both sides by -9. Remember, when we divide by a negative, the direction of the inequality sign flips:
\( x < \frac{26}{9} \).
Similarly, for the second inequality: \(30 - 9x < -4\), we subtract 30:
\( -9x < -34 \).
Dividing by -9 and flipping the inequality sign:
\( x > \frac{34}{9} \).
By isolating the variable in each part, we find the solution sets which we then combine into our final solution.