To solve the equation \( |4x + 5| - 2 = |2x - 1| \), we need to consider the different cases for the absolute value expressions. We will break it down into cases based on the critical points where the expressions inside the absolute values change sign. These critical points are \( x = -\frac{5}{4} \) and \( x = \frac{1}{2} \). We will solve the equation for each interval determined by these critical points and check which solutions satisfy the original equation.
The given equation is:
\[
|4x + 5| - 2 = |2x - 1|
\]
To solve this equation, we need to consider the different cases for the absolute values. Absolute value equations can be split into different cases based on the expressions inside the absolute value.
We will consider the following cases based on the expressions inside the absolute values:
- \(4x + 5 \geq 0\) and \(2x - 1 \geq 0\)
- \(4x + 5 \geq 0\) and \(2x - 1 < 0\)
- \(4x + 5 < 0\) and \(2x - 1 \geq 0\)
- \(4x + 5 < 0\) and \(2x - 1 < 0\)
Case 1: \(4x + 5 \geq 0\) and \(2x - 1 \geq 0\)
This implies \(x \geq -\frac{5}{4}\) and \(x \geq \frac{1}{2}\). Therefore, \(x \geq \frac{1}{2}\).
In this case, the equation becomes:
\[
4x + 5 - 2 = 2x - 1
\]
Simplifying:
\[
4x + 3 = 2x - 1
\]
Subtract \(2x\) from both sides:
\[
2x + 3 = -1
\]
Subtract 3 from both sides:
\[
2x = -4
\]
Divide by 2:
\[
x = -2
\]
Since \(x = -2\) does not satisfy \(x \geq \frac{1}{2}\), there is no solution in this case.
Case 2: \(4x + 5 \geq 0\) and \(2x - 1 < 0\)
This implies \(x \geq -\frac{5}{4}\) and \(x < \frac{1}{2}\). Therefore, \(-\frac{5}{4} \leq x < \frac{1}{2}\).
In this case, the equation becomes:
\[
4x + 5 - 2 = -(2x - 1)
\]
Simplifying:
\[
4x + 3 = -2x + 1
\]
Add \(2x\) to both sides:
\[
6x + 3 = 1
\]
Subtract 3 from both sides:
\[
6x = -2
\]
Divide by 6:
\[
x = -\frac{1}{3}
\]
Since \(-\frac{5}{4} \leq -\frac{1}{3} < \frac{1}{2}\), \(x = -\frac{1}{3}\) is a valid solution for this case.
Case 3: \(4x + 5 < 0\) and \(2x - 1 \geq 0\)
This implies \(x < -\frac{5}{4}\) and \(x \geq \frac{1}{2}\). There is no overlap in these conditions, so there is no solution in this case.
Case 4: \(4x + 5 < 0\) and \(2x - 1 < 0\)
This implies \(x < -\frac{5}{4}\) and \(x < \frac{1}{2}\). Therefore, \(x < -\frac{5}{4}\).
In this case, the equation becomes:
\[
-(4x + 5) - 2 = -(2x - 1)
\]
Simplifying:
\[
-4x - 5 - 2 = -2x + 1
\]
Combine like terms:
\[
-4x - 7 = -2x + 1
\]
Add \(4x\) to both sides:
\[
-7 = 2x + 1
\]
Subtract 1 from both sides:
\[
-8 = 2x
\]
Divide by 2:
\[
x = -4
\]
Since \(x = -4\) satisfies \(x < -\frac{5}{4}\), \(x = -4\) is a valid solution for this case.
The solutions to the equation are:
\[
\boxed{x = -\frac{1}{3}}
\]
\[
\boxed{x = -4}
\]
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