What is the answer of |3x+1|=|x+1|?

 To find the solution to the absolute value equation |3x+1|=|x+1|, we need to consider two possible cases based on the properties of absolute values. The equation implies that either (3x+1) is equal to (x+1), or it is equal to the negation of (x+1). Let's solve these two cases separately:

**Case 1: (3x+1) = (x+1)**

In this case, we set the expressions within the absolute value signs equal to each other and solve for 'x':

\[3x + 1 = x + 1\]

Subtracting 'x' from both sides:

\[2x + 1 = 1\]

Subtracting 1 from both sides:

\[2x = 0\]

Dividing both sides by 2:

\[x = 0\]

So, in this case, the solution is \(x = 0\).

**Case 2: (3x+1) = -(x+1)**

In the second case, we set the expressions within the absolute value signs equal to the negation of each other and solve for 'x':

\[3x + 1 = -(x + 1)\]

Distributing the negative sign on the right side:

\[3x + 1 = -x - 1\]

Combining like terms:

\[4x + 1 = -1\]

Subtracting 1 from both sides:

\[4x = -2\]

Dividing both sides by 4:

\[x = -\frac{1}{2}\]

So, in this case, the solution is \(x = -\frac{1}{2}\).

Now, we need to check if these solutions satisfy the original equation |3x+1|=|x+1|:

1. Substitute \(x = 0\) into the original equation:

\[|3(0) + 1| = |0 + 1|\]

\[|1| = |1|\]

The equation holds true for \(x = 0\).

2. Substitute \(x = -\frac{1}{2}\) into the original equation:

\[|3\left(-\frac{1}{2}\right) + 1| = \left|-\frac{1}{2} + 1\right|\]

\[|-\frac{1}{2} + 1| = |-\frac{1}{2} + \frac{2}{2}|\]

\[\frac{1}{2} = \frac{1}{2}\]

The equation also holds true for \(x = -\frac{1}{2}\).

Therefore, the solutions to the equation |3x+1|=|x+1| are \(x = 0\) and \(x = -\frac{1}{2}\).