(g) \( \frac{4}{x-6}=\frac{2-3 x}{2 x-x^{2}}-\frac{12}{x^{2}-8 x+12} \)
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To solve the equation \( \frac{4}{x-6}=\frac{2-3 x}{2 x-x^{2}}-\frac{12}{x^{2}-8 x+12} \), we start by simplifying the terms on the right. First, notice that the denominator \( x^2 - 8x + 12 \) can be factored: \[ x^2 - 8x + 12 = (x-6)(x-2) \] Thus, \[ \frac{12}{x^2 - 8x + 12} = \frac{12}{(x-6)(x-2)} \] Next, let's simplify the other term: The denominator \( 2x - x^2 \) can be rearranged as \( -(x^2 - 2x) = -x(x-2) \), hence, \[ \frac{2-3x}{2x - x^2} = \frac{2-3x}{-x(x-2)} = -\frac{2-3x}{x(x-2)} \] Now we have: \[ \frac{4}{x-6} = -\frac{2-3x}{x(x-2)} - \frac{12}{(x-6)(x-2)} \] Next, we'll get a common denominator for the right-hand side: \[ -\frac{(2-3x)(x-6) + 12x}{x(x-2)(x-6)} \] Expanding the first term: \[ (2-3x)(x-6) = 2x - 12 - 3x^2 + 18x = -3x^2 + 20x - 12 \] Thus, the right side becomes: \[ -\frac{-3x^2 + 20x - 12 + 12x}{x(x-2)(x-6)} = -\frac{-3x^2 + 32x - 12}{x(x-2)(x-6)} \] Combine this with the left side: \[ \frac{4}{x-6} = \frac{3x^2 - 32x + 12}{x(x-2)(x-6)} \] To solve for \( x \), we cross-multiply: \[ 4 \cdot x(x-2) = (3x^2 - 32x + 12)(x-6) \] Expanding both sides, then simplifying gives us a polynomial equation that we can solve for \( x \). You can continue expanding and combining terms, ultimately leading to a polynomial equation, which can be solved using methods such as factoring or the quadratic formula. After solving for \( x \), check for extraneous solutions in the context of the original equation. Now you have a robust framework for handling rational equations like this one! Don’t forget to check for restrictions on \( x \) based on the original denominators to avoid division by zero.
