\( \begin{array}{ll}\frac{7 x}{x^{2}+x}-\frac{5}{x}=\frac{3}{x+1} & \text { (6) } \frac{3}{x-2}+\frac{x+3}{4-x^{2}}=\frac{6}{x+2} \\ \text { (7) } \frac{3}{x-6}-\frac{5}{2 x-8}=\frac{x-24}{x^{2}-10 x+24} & \text { (8) } \frac{2 x+1}{x}-\frac{5 x}{x^{2}-4 x}=2\end{array} \)

Solve the equation by following steps: - step0: Solve for \(x\): \(\frac{7x}{x^{2}+x}-\frac{5}{x}=\frac{3}{x+1}\) - step1: Find the domain: \(\frac{7x}{x^{2}+x}-\frac{5}{x}=\frac{3}{x+1},x \in \left(-\infty,-1\right)\cup \left(-1,0\right)\cup \left(0,+\infty\right)\) - step2: Divide the terms: \(\frac{7}{x+1}-\frac{5}{x}=\frac{3}{x+1}\) - step3: Multiply both sides of the equation by LCD: \(\left(\frac{7}{x+1}-\frac{5}{x}\right)x\left(x+1\right)=\frac{3}{x+1}\times x\left(x+1\right)\) - step4: Simplify the equation: \(2x-5=3x\) - step5: Move the variable to the left side: \(2x-5-3x=0\) - step6: Subtract the terms: \(-x-5=0\) - step7: Move the constant to the right side: \(-x=0+5\) - step8: Remove 0: \(-x=5\) - step9: Change the signs: \(x=-5\) - step10: Check if the solution is in the defined range: \(x=-5,x \in \left(-\infty,-1\right)\cup \left(-1,0\right)\cup \left(0,+\infty\right)\) - step11: Find the intersection: \(x=-5\) Solve the equation \( \frac{3}{x-2}+\frac{x+3}{4-x^{2}}=\frac{6}{x+2} \). Solve the equation by following steps: - step0: Solve for \(x\): \(\frac{3}{x-2}+\frac{x+3}{4-x^{2}}=\frac{6}{x+2}\) - step1: Find the domain: \(\frac{3}{x-2}+\frac{x+3}{4-x^{2}}=\frac{6}{x+2},x \in \left(-\infty,-2\right)\cup \left(-2,2\right)\cup \left(2,+\infty\right)\) - step2: Multiply both sides of the equation by LCD: \(\left(\frac{3}{x-2}+\frac{x+3}{4-x^{2}}\right)\left(x+2\right)\left(-x+2\right)=\frac{6}{x+2}\times \left(x+2\right)\left(-x+2\right)\) - step3: Simplify the equation: \(-3-2x=-6x+12\) - step4: Move the expression to the left side: \(-2x+6x=12+3\) - step5: Add and subtract: \(4x=12+3\) - step6: Add and subtract: \(4x=15\) - step7: Divide both sides: \(\frac{4x}{4}=\frac{15}{4}\) - step8: Divide the numbers: \(x=\frac{15}{4}\) - step9: Check if the solution is in the defined range: \(x=\frac{15}{4},x \in \left(-\infty,-2\right)\cup \left(-2,2\right)\cup \left(2,+\infty\right)\) - step10: Find the intersection: \(x=\frac{15}{4}\) Solve the equation \( \frac{2 x+1}{x}-\frac{5 x}{x^{2}-4 x}=2 \). Solve the equation by following steps: - step0: Solve for \(x\): \(\frac{2x+1}{x}-\frac{5x}{x^{2}-4x}=2\) - step1: Find the domain: \(\frac{2x+1}{x}-\frac{5x}{x^{2}-4x}=2,x \in \left(-\infty,0\right)\cup \left(0,4\right)\cup \left(4,+\infty\right)\) - step2: Divide the terms: \(\frac{2x+1}{x}-\frac{5}{x-4}=2\) - step3: Multiply both sides of the equation by LCD: \(\left(\frac{2x+1}{x}-\frac{5}{x-4}\right)x\left(x-4\right)=2x\left(x-4\right)\) - step4: Simplify the equation: \(2x^{2}-12x-4=2x^{2}-8x\) - step5: Cancel equal terms: \(-12x-4=-8x\) - step6: Move the variable to the left side: \(-12x-4+8x=0\) - step7: Add the terms: \(-4x-4=0\) - step8: Move the constant to the right side: \(-4x=0+4\) - step9: Remove 0: \(-4x=4\) - step10: Change the signs: \(4x=-4\) - step11: Divide both sides: \(\frac{4x}{4}=\frac{-4}{4}\) - step12: Divide the numbers: \(x=-1\) - step13: Check if the solution is in the defined range: \(x=-1,x \in \left(-\infty,0\right)\cup \left(0,4\right)\cup \left(4,+\infty\right)\) - step14: Find the intersection: \(x=-1\) Solve the equation \( \frac{3}{x-6}-\frac{5}{2 x-8}=\frac{x-24}{x^{2}-10 x+24} \). Solve the equation by following steps: - step0: Solve for \(x\): \(\frac{3}{x-6}-\frac{5}{2x-8}=\frac{x-24}{x^{2}-10x+24}\) - step1: Find the domain: \(\frac{3}{x-6}-\frac{5}{2x-8}=\frac{x-24}{x^{2}-10x+24},x \in \left(-\infty,4\right)\cup \left(4,6\right)\cup \left(6,+\infty\right)\) - step2: Multiply both sides of the equation by LCD: \(\left(\frac{3}{x-6}-\frac{5}{2x-8}\right)\times 2\left(x-4\right)\left(x-6\right)=\frac{x-24}{x^{2}-10x+24}\times 2\left(x-4\right)\left(x-6\right)\) - step3: Simplify the equation: \(x+6=2x-48\) - step4: Move the expression to the left side: \(x-2x=-48-6\) - step5: Add and subtract: \(-x=-48-6\) - step6: Add and subtract: \(-x=-54\) - step7: Change the signs: \(x=54\) - step8: Check if the solution is in the defined range: \(x=54,x \in \left(-\infty,4\right)\cup \left(4,6\right)\cup \left(6,+\infty\right)\) - step9: Find the intersection: \(x=54\) Here are the solutions for each of the equations provided: 1. **Equation (6)**: \[ \frac{7 x}{x^{2}+x}-\frac{5}{x}=\frac{3}{x+1} \] - **Solution**: \( x = -5 \) 2. **Equation (7)**: \[ \frac{3}{x-2}+\frac{x+3}{4-x^{2}}=\frac{6}{x+2} \] - **Solution**: \( x = \frac{15}{4} \) or \( x = 3.75 \) 3. **Equation (8)**: \[ \frac{3}{x-6}-\frac{5}{2 x-8}=\frac{x-24}{x^{2}-10 x+24} \] - **Solution**: \( x = 54 \) 4. **Equation (9)**: \[ \frac{2 x+1}{x}-\frac{5 x}{x^{2}-4 x}=2 \] - **Solution**: \( x = -1 \) If you need further assistance or explanations for any of these solutions, feel free to ask!