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Solve For \( x \) \( \frac{4 x-14}{x+4}-\frac{2 x-4}{x+2}=0 \)

Ask by Sandoval O'Quinn. in South Africa
Jan 22,2025

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The solutions are \( x=-1 \) and \( x=6 \).

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The Deep Dive

To solve for \( x \) in the equation \[ \frac{4 x - 14}{x + 4} - \frac{2 x - 4}{x + 2} = 0, \] we start by combining the fractions: \[ \frac{4 x - 14}{x + 4} = \frac{2 x - 4}{x + 2}. \] Cross-multiply to eliminate the denominators: \[ (4x - 14)(x + 2) = (2x - 4)(x + 4). \] Now, expand both sides: \[ 4x^2 + 8x - 14x - 28 = 2x^2 + 8x - 4x - 16. \] This simplifies to: \[ 4x^2 - 6x - 28 = 2x^2 + 4x - 16. \] Next, move all terms to one side: \[ 4x^2 - 2x^2 - 6x - 4x - 28 + 16 = 0, \] which simplifies to: \[ 2x^2 - 10x - 12 = 0. \] Now divide the entire equation by 2 to simplify: \[ x^2 - 5x - 6 = 0. \] Next, we factor the quadratic equation: \[ (x - 6)(x + 1) = 0. \] Setting each factor to zero gives us: \[ x - 6 = 0 \quad \text{or} \quad x + 1 = 0. \] Thus, the solutions are: \[ x = 6 \quad \text{or} \quad x = -1. \] Finally, we should verify that these solutions do not make the original denominators zero: For \( x = 6 \): \( x + 4 = 10 \) (not zero), \( x + 2 = 8 \) (not zero). For \( x = -1 \): \( x + 4 = 3 \) (not zero), \( x + 2 = 1 \) (not zero). Both values are valid solutions. Therefore, the solutions for \( x \) are: \[ \boxed{6} \quad \text{and} \quad \boxed{-1}. \]

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