Solving Absolute Value Inequalities
To solve the inequality \(|2x - 3| > |x + 3|\), you need to consider different cases based on the values of \(x\) which change the expressions inside the absolute value brackets.
There are two main cases to consider for each absolute value expression:
1. When the expression inside the absolute value brackets is non-negative;
2. When the expression inside the absolute value brackets is negative.
Let's deal with these step by step:
For \(|2x - 3|\):
- Case 1: \(2x - 3 \geq 0\), which simplifies to \(x \geq \frac{3}{2}\).
- Case 2: \(2x - 3 < 0\), which simplifies to \(x < \frac{3}{2}\).
For \(|x + 3|\):
- Case 1: \(x + 3 \geq 0\), which simplifies to \(x \geq -3\).
- Case 2: \(x + 3 < 0\), which simplifies to \(x < -3\).
Combining these cases, you have to solve the inequality under the following scenarios:
1. \(x \geq \frac{3}{2}\) and \(x \geq -3\) (since \(x \geq \frac{3}{2}\) covers \(x \geq -3\)).
This simplifies the inequality to \(2x - 3 > x + 3\):
\(x > 6\).
2. \(x < \frac{3}{2}\) and \(x \geq -3\):
Here we have to take two subcases because of the range of \(x\):
a. \( -3 \leq x < \frac{3}{2} \), therefore the inequality becomes \(3 - 2x > x + 3\):
\(-3x > 0\), which simplifies to \(x < 0\).
b. \( x < -3 \), therefore the inequality becomes \(3 - 2x > 3 - x\):
\(x < 0\), which always holds true for \(x < -3\).
Putting it all together, there are two intervals for \(x\) that solve the inequality:
- \(x < 0\) for the range between \(-3\) and \(\frac{3}{2}\),
- \(x > 6\).
The final solution to the inequality \(|2x - 3| > |x + 3|\) is:
\[x < 0 \text{ or } x > 6.\]
