(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) \( \frac{-3 x+9}{x^{2}+x-2}=\frac{4}{x^{2}+6 x+8}+\frac{4}{1-x} \)

Solve the equation by following steps: - step0: Solve for \(x\): \(\frac{1}{x+1}-\frac{3x-1}{1-x}=2\) - step1: Find the domain: \(\frac{1}{x+1}-\frac{3x-1}{1-x}=2,x \in \left(-\infty,-1\right)\cup \left(-1,1\right)\cup \left(1,+\infty\right)\) - step2: Multiply both sides of the equation by LCD: \(\left(\frac{1}{x+1}-\frac{3x-1}{1-x}\right)\left(x+1\right)\left(1-x\right)=2\left(x+1\right)\left(1-x\right)\) - step3: Simplify the equation: \(2-3x-3x^{2}=-2x^{2}+2\) - step4: Cancel equal terms: \(-3x-3x^{2}=-2x^{2}\) - step5: Move the expression to the left side: \(-3x-3x^{2}-\left(-2x^{2}\right)=0\) - step6: Subtract the terms: \(-3x-x^{2}=0\) - step7: Factor the expression: \(-x\left(3+x\right)=0\) - step8: Separate into possible cases: \(\begin{align}&-x=0\\&3+x=0\end{align}\) - step9: Solve the equation: \(\begin{align}&x=0\\&x=-3\end{align}\) - step10: Check if the solution is in the defined range: \(\begin{align}&x=0\\&x=-3\end{align},x \in \left(-\infty,-1\right)\cup \left(-1,1\right)\cup \left(1,+\infty\right)\) - step11: Find the intersection: \(\begin{align}&x=0\\&x=-3\end{align}\) - step12: Rewrite: \(x_{1}=-3,x_{2}=0\) Solve the equation \( \frac{-x^{2}+3 x+12}{x^{2}-3 x-4}+2=\frac{-1}{4-x} \). Solve the equation by following steps: - step0: Solve for \(x\): \(\frac{-x^{2}+3x+12}{x^{2}-3x-4}+2=\frac{-1}{4-x}\) - step1: Find the domain: \(\frac{-x^{2}+3x+12}{x^{2}-3x-4}+2=\frac{-1}{4-x},x \in \left(-\infty,-1\right)\cup \left(-1,4\right)\cup \left(4,+\infty\right)\) - step2: Rewrite the fraction: \(\frac{-x^{2}+3x+12}{x^{2}-3x-4}+2=-\frac{1}{4-x}\) - step3: Multiply both sides of the equation by LCD: \(\left(\frac{-x^{2}+3x+12}{x^{2}-3x-4}+2\right)\left(-x+4\right)\left(x+1\right)=-\frac{1}{4-x}\times \left(-x+4\right)\left(x+1\right)\) - step4: Simplify the equation: \(-x^{2}+3x-4=-x-1\) - step5: Move the expression to the left side: \(-x^{2}+3x-4-\left(-x-1\right)=0\) - step6: Calculate: \(-x^{2}+4x-3=0\) - step7: Factor the expression: \(\left(-x+3\right)\left(x-1\right)=0\) - step8: Separate into possible cases: \(\begin{align}&-x+3=0\\&x-1=0\end{align}\) - step9: Solve the equation: \(\begin{align}&x=3\\&x=1\end{align}\) - step10: Check if the solution is in the defined range: \(\begin{align}&x=3\\&x=1\end{align},x \in \left(-\infty,-1\right)\cup \left(-1,4\right)\cup \left(4,+\infty\right)\) - step11: Find the intersection: \(\begin{align}&x=3\\&x=1\end{align}\) - step12: Rewrite: \(x_{1}=1,x_{2}=3\) Solve the equation \( \frac{-3 x+9}{x^{2}+x-2}=\frac{4}{x^{2}+6 x+8}+\frac{4}{1-x} \). Solve the equation by following steps: - step0: Solve for \(x\): \(\frac{-3x+9}{x^{2}+x-2}=\frac{4}{x^{2}+6x+8}+\frac{4}{1-x}\) - step1: Find the domain: \(\frac{-3x+9}{x^{2}+x-2}=\frac{4}{x^{2}+6x+8}+\frac{4}{1-x},x \in \left(-\infty,-4\right)\cup \left(-4,-2\right)\cup \left(-2,1\right)\cup \left(1,+\infty\right)\) - step2: Multiply both sides of the equation by LCD: \(\frac{-3x+9}{x^{2}+x-2}\times \left(-x^{2}-x+2\right)\left(x+4\right)=\left(\frac{4}{x^{2}+6x+8}+\frac{4}{1-x}\right)\left(-x^{2}-x+2\right)\left(x+4\right)\) - step3: Simplify the equation: \(-3x\left(-x-4\right)+9\left(-x-4\right)=36+20x+4x^{2}\) - step4: Calculate: \(3x^{2}+12x-9x-36=36+20x+4x^{2}\) - step5: Simplify: \(3x^{2}+3x-36=36+20x+4x^{2}\) - step6: Move the expression to the left side: \(3x^{2}+3x-36-\left(36+20x+4x^{2}\right)=0\) - step7: Calculate: \(-x^{2}-17x-72=0\) - step8: Factor the expression: \(-\left(x+8\right)\left(x+9\right)=0\) - step9: Divide the terms: \(\left(x+8\right)\left(x+9\right)=0\) - step10: Separate into possible cases: \(\begin{align}&x+8=0\\&x+9=0\end{align}\) - step11: Solve the equation: \(\begin{align}&x=-8\\&x=-9\end{align}\) - step12: Check if the solution is in the defined range: \(\begin{align}&x=-8\\&x=-9\end{align},x \in \left(-\infty,-4\right)\cup \left(-4,-2\right)\cup \left(-2,1\right)\cup \left(1,+\infty\right)\) - step13: Find the intersection: \(\begin{align}&x=-8\\&x=-9\end{align}\) - step14: Rewrite: \(x_{1}=-9,x_{2}=-8\) Solve the equation \( \frac{2-x}{x+1}-\frac{2 x-4}{5-x}=-\frac{5 x-13}{x^{2}-4 x-5} \). Solve the equation by following steps: - step0: Solve for \(x\): \(\frac{2-x}{x+1}-\frac{2x-4}{5-x}=-\frac{5x-13}{x^{2}-4x-5}\) - step1: Find the domain: \(\frac{2-x}{x+1}-\frac{2x-4}{5-x}=-\frac{5x-13}{x^{2}-4x-5},x \in \left(-\infty,-1\right)\cup \left(-1,5\right)\cup \left(5,+\infty\right)\) - step2: Multiply both sides of the equation by LCD: \(\left(\frac{2-x}{x+1}-\frac{2x-4}{5-x}\right)\left(-x+5\right)\left(x+1\right)=-\frac{5x-13}{x^{2}-4x-5}\times \left(-x+5\right)\left(x+1\right)\) - step3: Simplify the equation: \(14-5x-x^{2}=5x-13\) - step4: Move the expression to the left side: \(14-5x-x^{2}-\left(5x-13\right)=0\) - step5: Subtract the terms: \(27-10x-x^{2}=0\) - step6: Rewrite in standard form: \(-x^{2}-10x+27=0\) - step7: Multiply both sides: \(x^{2}+10x-27=0\) - step8: Solve using the quadratic formula: \(x=\frac{-10\pm \sqrt{10^{2}-4\left(-27\right)}}{2}\) - step9: Simplify the expression: \(x=\frac{-10\pm \sqrt{208}}{2}\) - step10: Simplify the expression: \(x=\frac{-10\pm 4\sqrt{13}}{2}\) - step11: Separate into possible cases: \(\begin{align}&x=\frac{-10+4\sqrt{13}}{2}\\&x=\frac{-10-4\sqrt{13}}{2}\end{align}\) - step12: Simplify the expression: \(\begin{align}&x=-5+2\sqrt{13}\\&x=\frac{-10-4\sqrt{13}}{2}\end{align}\) - step13: Simplify the expression: \(\begin{align}&x=-5+2\sqrt{13}\\&x=-5-2\sqrt{13}\end{align}\) - step14: Check if the solution is in the defined range: \(\begin{align}&x=-5+2\sqrt{13}\\&x=-5-2\sqrt{13}\end{align},x \in \left(-\infty,-1\right)\cup \left(-1,5\right)\cup \left(5,+\infty\right)\) - step15: Find the intersection: \(\begin{align}&x=-5+2\sqrt{13}\\&x=-5-2\sqrt{13}\end{align}\) - step16: Rewrite: \(x_{1}=-5-2\sqrt{13},x_{2}=-5+2\sqrt{13}\) Solve the equation \( \frac{2 x^{2}-4 x}{x^{3}-3 x+2}=\frac{1}{x-2}-\frac{2}{1-x}+1 \). Solve the equation by following steps: - step0: Solve for \(x\): \(\frac{2x^{2}-4x}{x^{3}-3x+2}=\frac{1}{x-2}-\frac{2}{1-x}+1\) - step1: Find the domain: \(\frac{2x^{2}-4x}{x^{3}-3x+2}=\frac{1}{x-2}-\frac{2}{1-x}+1,x \in \left(-\infty,-2\right)\cup \left(-2,1\right)\cup \left(1,2\right)\cup \left(2,+\infty\right)\) - step2: Multiply both sides of the equation by LCD: \(\frac{2x^{2}-4x}{x^{3}-3x+2}\times \left(x-1\right)\left(x^{2}+x-2\right)\left(-x+2\right)=\left(\frac{1}{x-2}-\frac{2}{1-x}+1\right)\left(x-1\right)\left(x^{2}+x-2\right)\left(-x+2\right)\) - step3: Simplify the equation: \(2x^{2}\left(-x+2\right)-4x\left(-x+2\right)=3x-6-x^{3}+5x^{2}-x^{4}\) - step4: Calculate: \(-2x^{3}+4x^{2}+4x^{2}-8x=3x-6-x^{3}+5x^{2}-x^{4}\) - step5: Simplify: \(-2x^{3}+8x^{2}-8x=3x-6-x^{3}+5x^{2}-x^{4}\) - step6: Move the expression to the left side: \(-2x^{3}+8x^{2}-8x-\left(3x-6-x^{3}+5x^{2}-x^{4}\right)=0\) - step7: Calculate: \(-x^{3}+3x^{2}-11x+6+x^{4}=0\) - step8: Calculate: \(\begin{align}&x\approx 1.753639\\&x\approx 0.652946\end{align}\) - step9: Check if the solution is in the defined range: \(\begin{align}&x\approx 1.753639\\&x\approx 0.652946\end{align},x \in \left(-\infty,-2\right)\cup \left(-2,1\right)\cup \left(1,2\right)\cup \left(2,+\infty\right)\) - step10: Find the intersection: \(\begin{align}&x\approx 1.753639\\&x\approx 0.652946\end{align}\) - step11: Rewrite: \(x_{1}\approx 0.652946,x_{2}\approx 1.753639\) Here are the solutions for each of the equations provided: ### Problem (6) \[ \frac{1}{x+1}-\frac{3 x-1}{1-x}=2 \] **Solutions:** - \( x_{1} = -3 \) - \( x_{2} = 0 \) ### Problem (8) \[ \frac{-x^{2}+3 x+12}{x^{2}-3 x-4}+2=\frac{-1}{4-x} \] **Solutions:** - \( x_{1} = 1 \) - \( x_{2} = 3 \) ### Problem (10) \[ \frac{2 x^{2}-4 x}{x^{3}-3 x+2}=\frac{1}{x-2}-\frac{2}{1-x}+1 \] **Solutions:** - \( x_{1} \approx 0.652946 \) - \( x_{2} \approx 1.753639 \) ### Problem (12) \[ \frac{2-x}{x+1}-\frac{2 x-4}{5-x}=-\frac{5 x-13}{x^{2}-4 x-5} \] **Solutions:** - \( x_{1} = -5 - 2\sqrt{13} \) - \( x_{2} = -5 + 2\sqrt{13} \) ### Problem (14) \[ \frac{-3 x+9}{x^{2}+x-2}=\frac{4}{x^{2}+6 x+8}+\frac{4}{1-x} \] **Solutions:** - \( x_{1} = -9 \) - \( x_{2} = -8 \) If you need further assistance or explanations for any of these problems, feel free to ask!