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We are tasked with finding the equations of the tangent lines to the curve:

\[
y = \frac{x - 1}{x + 1}
\]

that are parallel to the line \(x - 2y = 4\).

### Step-by-Step Approach:

1. **Find the slope of the given line**:
   Rewrite the line \(x - 2y = 4\) in slope-intercept form:
   \[
   x - 2y = 4 \quad \Rightarrow \quad -2y = -x + 4 \quad \Rightarrow \quad y = \frac{1}{2}x - 2
   \]
   The slope of this line is \(m = \frac{1}{2}\).

2. **Find the slope of the tangent line to the curve**:
   To find the slope of the tangent line at any point on the curve, we need to differentiate the curve equation \(y = \frac{x - 1}{x + 1}\) with respect to \(x\).

   Apply the quotient rule for differentiation. The quotient rule is:
   \[
   \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}
   \]
   where \(u = x - 1\) and \(v = x + 1\).

   First, compute the derivatives:
   \[
   u' = 1 \quad \text{and} \quad v' = 1
   \]
   Now, apply the quotient rule:
   \[
   \frac{dy}{dx} = \frac{(1)(x + 1) - (x - 1)(1)}{(x + 1)^2} = \frac{x + 1 - x + 1}{(x + 1)^2} = \frac{2}{(x + 1)^2}
   \]
   So, the slope of the tangent line at any point \(x\) is:
   \[
   m_{\text{tangent}} = \frac{2}{(x + 1)^2}
   \]

3. **Set the slope of the tangent equal to the slope of the given line**:
   We want the tangent lines to be parallel to the given line, so their slopes must be equal. Set:
   \[
   \frac{2}{(x + 1)^2} = \frac{1}{2}
   \]
   Solve for \(x\):
   \[
   2 = \frac{(x + 1)^2}{2} \quad \Rightarrow \quad 4 = (x + 1)^2 \quad \Rightarrow \quad x + 1 = \pm 2
   \]
   This gives two solutions for \(x\):
   \[
   x + 1 = 2 \quad \Rightarrow \quad x = 1
   \]
   \[
   x + 1 = -2 \quad \Rightarrow \quad x = -3
   \]

4. **Find the corresponding \(y\) values on the curve**:
   Substitute \(x = 1\) and \(x = -3\) into the original curve equation \(y = \frac{x - 1}{x + 1}\).

   - For \(x = 1\):
     \[
     y = \frac{1 - 1}{1 + 1} = 0
     \]
   - For \(x = -3\):
     \[
     y = \frac{-3 - 1}{-3 + 1} = \frac{-4}{-2} = 2
     \]

   So, the points of tangency are \((1, 0)\) and \((-3, 2)\).

5. **Write the equations of the tangent lines**:
   The tangent lines will have the form \(y - y_1 = m(x - x_1)\), where \((x_1, y_1)\) is the point of tangency and \(m = \frac{1}{2}\) (since the lines are parallel to the given line).

   - For the point \((1, 0)\):
     \[
     y - 0 = \frac{1}{2}(x - 1) \quad \Rightarrow \quad y = \frac{1}{2}x - \frac{1}{2}
     \]
   - For the point \((-3, 2)\):
     \[
     y - 2 = \frac{1}{2}(x + 3) \quad \Rightarrow \quad y = \frac{1}{2}x + \frac{7}{2}
     \]

6. **Final answer**:
   The equations of the tangent lines are:
   \[
   y = \frac{1}{2}x - \frac{1}{2}, \quad y = \frac{1}{2}x + \frac{7}{2}
   \]

Would you like further details or clarification on any step?

Here are 5 related questions you could explore:
1. How does the quotient rule work for more complex rational functions?
2. What are some common techniques for finding tangent lines to curves?
3. How can we generalize the method for finding tangents to implicit curves?
4. How does the relationship between parallel lines help in solving these problems?
5. Can you explain the geometric interpretation of the tangent line in this context?

**Tip:** Always ensure your derivative calculations are correct when finding the slope of tangent lines, as even a small error can affect the final result.

Find equations of the tangent lines to the curve y = (x - 1)/(x + 1) that are parallel to the line x - 2y = 4.

Learn how to find the equations of the tangent lines to the curve y = (x - 1)/(x + 1) that are parallel to the line x - 2y = 4. This problem involves differentiation using the quotient rule, finding slopes, and using the tangent line formula. Suitable for students in Grades 11-12.

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