Solve the rational equation \( \frac{1}{x}+\frac{1}{x-7}=\frac{x-6}{x-7} \) a) \( x=1, x=8 \) b) \( x=1, x=7 \) c) \( x=1 \) d) \( x=7 \)

To solve the rational equation \( \frac{1}{x}+\frac{1}{x-7}=\frac{x-6}{x-7} \), we begin by finding a common denominator for the left side, which is \( x(x-7) \). Rewriting the equation with a common denominator gives: \[ \frac{x-7 + x}{x(x-7)} = \frac{x-6}{x-7} \] This simplifies to: \[ \frac{2x - 7}{x(x-7)} = \frac{x - 6}{x - 7} \] Next, we cross multiply to eliminate the fractions: \[ (2x - 7)(x - 7) = (x - 6)x \] Expanding both sides results in: \[ 2x^2 - 14x - 7x + 49 = x^2 - 6x \] Combining terms gives us: \[ 2x^2 - 21x + 49 = x^2 - 6x \] Now, move everything to one side: \[ 2x^2 - 21x + 49 - x^2 + 6x = 0 \] This simplifies to: \[ x^2 - 15x + 49 = 0 \] Next, applying the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), we set \( a = 1, b = -15, c = 49 \): \[ x = \frac{15 \pm \sqrt{(-15)^2 - 4 \cdot 1 \cdot 49}}{2 \cdot 1} \] Calculating the discriminant: \[ (-15)^2 - 4 \cdot 1 \cdot 49 = 225 - 196 = 29 \] This gives us: \[ x = \frac{15 \pm \sqrt{29}}{2} \] Evaluating the roots, we check if these roots produce any contradictions in the original equation, particularly the denominators. Plugging these complex roots back into the initial equation does not yield any integer values. However, if we evaluate boundaries tested for \( x = 1 \) and \( x = 7 \), we find: For \( x = 1 \), it satisfies the original equation. For \( x = 7 \), it results in undefined behavior due to division by zero. Therefore, the only valid solution is: The answer is (c) \( x = 1 \).