Solve x in the rational equation: \( \frac{x^{2}-4 x}{x-2}=\frac{14-9 x}{x-2} \) Answer:

To solve the equation \( \frac{x^{2}-4 x}{x-2}=\frac{14-9 x}{x-2} \), we first note that for \( x \neq 2 \), we can multiply both sides by \( x-2 \) to eliminate the denominator: \[ x^2 - 4x = 14 - 9x \] Next, we rearrange the equation: \[ x^2 - 4x + 9x - 14 = 0 \] This simplifies to: \[ x^2 + 5x - 14 = 0 \] Now, we can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 5 \), and \( c = -14 \): \[ b^2 - 4ac = 5^2 - 4 \cdot 1 \cdot (-14) = 25 + 56 = 81 \] Now substitute this back into the quadratic formula: \[ x = \frac{-5 \pm \sqrt{81}}{2 \cdot 1} = \frac{-5 \pm 9}{2} \] Using the positive and negative cases: 1. \( x = \frac{-5 + 9}{2} = \frac{4}{2} = 2 \) 2. \( x = \frac{-5 - 9}{2} = \frac{-14}{2} = -7 \) However, since we originally eliminated the denominator by assuming \( x \neq 2 \), we discard \( x = 2 \). Thus, the solution is: \[ \boxed{-7} \]