If the line `y=3x+1` and `2y=x+3` are equally inclined to the line `y=mx+4` them `m=(1+-5sqrt(2))/7`.

Answer

To solve the problem, we need to find the value of \( m \) such that the lines \( y = 3x + 1 \) and \( 2y = x + 3 \) are equally inclined to the line \( y = mx + 4 \).

### Step 1: Find the slopes of the given lines
The slope of a line in the form \( y = mx + c \) is given by \( m \).

1. For the line \( y = 3x + 1 \), the slope \( m_1 = 3 \).
2. For the line \( 2y = x + 3 \), we first rewrite it in slope-intercept form:
\[ y = \frac{1}{2}x + \frac{3}{2} \]
Thus, the slope \( m_2 = \frac{1}{2} \).
3. For the line \( y = mx + 4 \), the slope is \( m_3 = m \).

### Step 2: Use the condition of equal inclination
Two lines are equally inclined to a third line if the angles they make with the third line are equal. This can be expressed using the tangent of the angles.

The formula for the tangent of the angle \( \theta \) between two lines with slopes \( m_1 \) and \( m_2 \) is given by:
\[ \tan \theta = \frac{m_1 - m_2}{1 + m_1 m_2} \]

We need to set up the equations for the angles between the lines:
1. For the line \( y = 3x + 1 \) and \( y = mx + 4 \):
\[ \tan \theta_1 = \frac{m - 3}{1 + 3m} \]
2. For the line \( y = \frac{1}{2}x + \frac{3}{2} \) and \( y = mx + 4 \):
\[ \tan \theta_2 = \frac{m - \frac{1}{2}}{1 + \frac{1}{2}m} \]

### Step 3: Set the tangents equal (considering both positive and negative cases)
Since the lines are equally inclined, we have:
\[ \frac{m - 3}{1 + 3m} = \pm \frac{m - \frac{1}{2}}{1 + \frac{1}{2}m} \]

### Step 4: Solve the equations
1. **Using the positive case**:
\[ \frac{m - 3}{1 + 3m} = \frac{m - \frac{1}{2}}{1 + \frac{1}{2}m} \]
Cross-multiplying and simplifying leads to no solution.

2. **Using the negative case**:
\[ \frac{m - 3}{1 + 3m} = -\frac{m - \frac{1}{2}}{1 + \frac{1}{2}m} \]
Cross-multiplying gives:
\[ (m - 3)(1 + \frac{1}{2}m) = -(m - \frac{1}{2})(1 + 3m) \]
Expanding both sides:
\[ m + \frac{1}{2}m^2 - 3 - \frac{3}{2}m = -m - 3m^2 + \frac{1}{2} + \frac{3}{2}m \]
Rearranging terms leads to:
\[ 7m^2 - 2m - 7 = 0 \]

### Step 5: Solve the quadratic equation
Using the quadratic formula \( m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ m = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 7 \cdot (-7)}}{2 \cdot 7} \]
Calculating the discriminant:
\[ = \frac{2 \pm \sqrt{4 + 196}}{14} = \frac{2 \pm \sqrt{200}}{14} = \frac{2 \pm 10\sqrt{2}}{14} \]
Simplifying gives:
\[ m = \frac{1 \pm 5\sqrt{2}}{7} \]

### Final Answer
Thus, the values of \( m \) are:
\[ m = \frac{1 + 5\sqrt{2}}{7} \quad \text{and} \quad m = \frac{1 - 5\sqrt{2}}{7} \]