d) \( \frac{3-x}{(x+3)(x+1)}=\frac{2}{x+1} \)

Solve the equation \( \frac{3-x}{(x+3)(x+1)}=\frac{2}{x+1} \). Solve the equation by following steps: - step0: Solve for \(x\): \(\frac{3-x}{\left(x+3\right)\left(x+1\right)}=\frac{2}{x+1}\) - step1: Find the domain: \(\frac{3-x}{\left(x+3\right)\left(x+1\right)}=\frac{2}{x+1},x \in \left(-\infty,-3\right)\cup \left(-3,-1\right)\cup \left(-1,+\infty\right)\) - step2: Cross multiply: \(\left(3-x\right)\left(x+1\right)=\left(x+3\right)\left(x+1\right)\times 2\) - step3: Simplify the equation: \(\left(3-x\right)\left(x+1\right)=2\left(x+3\right)\left(x+1\right)\) - step4: Calculate: \(2x+3-x^{2}=2\left(x+3\right)\left(x+1\right)\) - step5: Calculate: \(2x+3-x^{2}=2x^{2}+8x+6\) - step6: Move the expression to the left side: \(2x+3-x^{2}-\left(2x^{2}+8x+6\right)=0\) - step7: Calculate: \(-6x-3-3x^{2}=0\) - step8: Factor the expression: \(-3\left(1+x\right)^{2}=0\) - step9: Divide the terms: \(\left(1+x\right)^{2}=0\) - step10: Simplify the expression: \(1+x=0\) - step11: Move the constant to the right side: \(x=0-1\) - step12: Remove 0: \(x=-1\) - step13: Check if the solution is in the defined range: \(x=-1,x \in \left(-\infty,-3\right)\cup \left(-3,-1\right)\cup \left(-1,+\infty\right)\) - step14: Find the intersection: \(x \in \varnothing \) The solution to the equation \( \frac{3-x}{(x+3)(x+1)}=\frac{2}{x+1} \) is that there are no real solutions for \( x \).