Solve for w. \[ \frac{w+3}{w+1}-1=\frac{w-5}{w-7} \] If there is more than one solution, separate them with commas. If there is no solution, click on "No solution".
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To solve the equation \[ \frac{w+3}{w+1}-1=\frac{w-5}{w-7}, \] first, simplify the left side: \[ \frac{w+3}{w+1}-1 = \frac{w+3}{w+1} - \frac{w+1}{w+1} = \frac{(w+3) - (w+1)}{w+1} = \frac{w + 3 - w - 1}{w + 1} = \frac{2}{w+1}. \] Now, the equation becomes: \[ \frac{2}{w + 1} = \frac{w - 5}{w - 7}. \] Next, cross-multiply: \[ 2(w - 7) = (w - 5)(w + 1). \] Expanding both sides gives: \[ 2w - 14 = w^2 - 4w - 5. \] Now, rearranging terms results in: \[ 0 = w^2 - 4w - 5 - 2w + 14, \] which simplifies to: \[ 0 = w^2 - 6w + 9. \] This factors to: \[ 0 = (w - 3)^2. \] Thus, we have a double root: \[ w - 3 = 0 \implies w = 3. \] Finally, we should check that this value does not cause any division by zero in the original equation. When \( w = 3 \): - The denominators \( w + 1 = 4 \) and \( w - 7 = -4 \) are both non-zero. Hence, the solution is: \[ \boxed{3}. \]
