How many solutions exist to the equation `|x|=|2x-1|`?

Options

A. `0`
B. `1`
C. `2`
D. `3`

Answer

To solve the equation \( |x| = |2x - 1| \), we need to consider the different cases for the absolute values.

### Step 1: Set up the cases
The equation \( |x| = |2x - 1| \) can be solved by considering the following cases based on the definitions of absolute value:

1. **Case 1:** \( x \geq 0 \) and \( 2x - 1 \geq 0 \)
2. **Case 2:** \( x \geq 0 \) and \( 2x - 1 < 0 \)
3. **Case 3:** \( x < 0 \) and \( 2x - 1 \geq 0 \)
4. **Case 4:** \( x < 0 \) and \( 2x - 1 < 0 \)

### Step 2: Solve Case 1
In Case 1, both expressions are non-negative, so we have:
\[ x = 2x - 1 \]
Rearranging gives:
\[ x - 2x = -1 \implies -x = -1 \implies x = 1 \]

### Step 3: Solve Case 2
In Case 2, we have:
\[ x = -(2x - 1) \implies x = -2x + 1 \]
Rearranging gives:
\[ x + 2x = 1 \implies 3x = 1 \implies x = \frac{1}{3} \]

### Step 4: Solve Case 3
In Case 3, we have:
\[ -x = 2x - 1 \]
Rearranging gives:
\[ -x - 2x = -1 \implies -3x = -1 \implies x = \frac{1}{3} \]
However, since \( x < 0 \) in this case, this solution is not valid.

### Step 5: Solve Case 4
In Case 4, we have:
\[ -x = -(2x - 1) \implies -x = -2x + 1 \]
Rearranging gives:
\[ -x + 2x = 1 \implies x = 1 \]
Again, since \( x < 0 \) in this case, this solution is not valid.

### Summary of Solutions
From the valid cases, we found:
- From Case 1: \( x = 1 \)
- From Case 2: \( x = \frac{1}{3} \)

Thus, the equation \( |x| = |2x - 1| \) has **two solutions**: \( x = 1 \) and \( x = \frac{1}{3} \).

### Final Answer
The number of solutions to the equation \( |x| = |2x - 1| \) is **2**.

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