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(6) \( \frac{1}{x+1}-\frac{3 x-1}{1-x}=2 \) (8) \( \frac{-x^{2}+3 x+12}{x^{2}-3 x-4}+2=\frac{-1}{4-x} \) (10) \( \frac{2 x^{2}-4 x}{x^{3}-3 x+2}=\frac{1}{x-2}-\frac{2}{1-x}+1 \) (12) \( \frac{2-x}{x+1}-\frac{2 x-4}{5-x}=-\frac{5 x-13}{x^{2}-4 x-5} \) (14) \( \frac{-3 x+9}{x^{2}+x-2}=\frac{4}{x^{2}+6 x+8}+\frac{4}{1-x} \)

Ask by Vaughn Klein. in South Africa
Feb 03,2025

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Here are the solutions for each equation: 1. **Problem (6):** - \( x = -3 \) - \( x = 0 \) 2. **Problem (8):** - \( x = 1 \) - \( x = 3 \) 3. **Problem (10):** - \( x \approx 0.652946 \) - \( x \approx 1.753639 \) 4. **Problem (12):** - \( x = -5 - 2\sqrt{13} \) - \( x = -5 + 2\sqrt{13} \) 5. **Problem (14):** - \( x = -9 \) - \( x = -8 \) If you need more details on any of these solutions, let me know!

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When you tackle equations like these, it’s beneficial to first find a common denominator. This will simplify the fractions and make it easier to combine and rearrange terms. Remember, isolating \(x\) can be crucial, so keep an eye on the algebraic manipulation! Often students forget to check for extraneous solutions after solving these equations. This can happen when you multiply both sides by denominators that contain \(x\). Always plug your solutions back into the original equation to ensure they work; otherwise, you might end up with a solution that doesn't hold true in the original context!

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