(2) Calcule \( \frac{2}{y+3}-\frac{1}{y^{2}-9}=\frac{3}{5 y-15} \) para \( y \neq-3 \) e para \( y \neq 3 \)

Solve the equation \( \frac{2}{y+3}-\frac{1}{y^{2}-9}=\frac{3}{5 y-15} \). Solve the equation by following steps: - step0: Solve for \(y\): \(\frac{2}{y+3}-\frac{1}{y^{2}-9}=\frac{3}{5y-15}\) - step1: Find the domain: \(\frac{2}{y+3}-\frac{1}{y^{2}-9}=\frac{3}{5y-15},y \in \left(-\infty,-3\right)\cup \left(-3,3\right)\cup \left(3,+\infty\right)\) - step2: Multiply both sides of the equation by LCD: \(\left(\frac{2}{y+3}-\frac{1}{y^{2}-9}\right)\times 5\left(y-3\right)\left(y+3\right)=\frac{3}{5y-15}\times 5\left(y-3\right)\left(y+3\right)\) - step3: Simplify the equation: \(10y-35=3y+9\) - step4: Move the variable to the left side: \(10y-35-3y=9\) - step5: Subtract the terms: \(7y-35=9\) - step6: Move the constant to the right side: \(7y=9+35\) - step7: Add the numbers: \(7y=44\) - step8: Divide both sides: \(\frac{7y}{7}=\frac{44}{7}\) - step9: Divide the numbers: \(y=\frac{44}{7}\) - step10: Check if the solution is in the defined range: \(y=\frac{44}{7},y \in \left(-\infty,-3\right)\cup \left(-3,3\right)\cup \left(3,+\infty\right)\) - step11: Find the intersection: \(y=\frac{44}{7}\) A equação dada é \( \frac{2}{y+3}-\frac{1}{y^{2}-9}=\frac{3}{5 y-15} \) para \( y \neq-3 \) e para \( y \neq 3 \). A solução para essa equação é \( y = \frac{44}{7} \) ou \( y = 6.\dot{2}8571\dot{4} \).