To solve the equation
\[ \frac{7}{x+1} + \frac{3}{x-1} = \frac{6}{(x+1)(x-1)}, \]
we first need to determine the values of \( x \) that make the denominators zero.
Step 1: Find Critical Values
The denominators in the equation are \( x + 1 \) and \( x - 1 \). Setting each denominator equal to zero gives:
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\( x + 1 = 0 \)
- Solving this gives \( x = -1 \).
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\( x - 1 = 0 \)
- Solving this gives \( x = 1 \).
Thus, the values of \( x \) that make a denominator zero are \( x = -1 \) and \( x = 1 \).
Step 2: Solve the Equation
Now, we will solve the equation. To eliminate the fractions, we multiply each term by the common denominator, which is \( (x+1)(x-1) \):
\[ (x+1)(x-1) \left( \frac{7}{x+1} \right) + (x+1)(x-1) \left( \frac{3}{x-1} \right) = (x+1)(x-1) \left( \frac{6}{(x+1)(x-1)} \right). \]
This simplifies to:
\[ 7(x-1) + 3(x+1) = 6. \]
Step 3: Simplify the Equation
Now, let's expand and simplify:
- \( 7(x-1) = 7x - 7 \).
- \( 3(x+1) = 3x + 3 \).
Combining these, we get:
\[ 7x - 7 + 3x + 3 = 6. \]
Combining like terms:
\[ (7x + 3x) + (-7 + 3) = 6, \] \[ 10x - 4 = 6. \]
Step 4: Isolate \( x \)
Next, we isolate \( x \):
\[ 10x = 6 + 4, \] \[ 10x = 10, \] \[ x = 1. \]
Step 5: Check for Extraneous Solutions
However, \( x = 1 \) is a value that makes the denominator zero in the original equation, which means this solution is not valid.
Conclusion
The valid values of \( x \) that make a denominator zero are \( x = -1 \) and \( x = 1 \), and the equation has no valid solutions as the only solution derived from solving the equation introduces a zero in the denominator.
Thus, there are no solutions to the original equation.