A student solves the following equation and determines that the solution is -2 . Is the student correct? Explain. \( \frac{3}{a+2}-\frac{6 a}{a^{2}-4}=\frac{1}{a-2} \)

Solve the equation \( \frac{3}{a+2}-\frac{6a}{a^{2}-4}=\frac{1}{a-2} \). Solve the equation by following steps: - step0: Solve for \(a\): \(\frac{3}{a+2}-\frac{6a}{a^{2}-4}=\frac{1}{a-2}\) - step1: Find the domain: \(\frac{3}{a+2}-\frac{6a}{a^{2}-4}=\frac{1}{a-2},a \in \left(-\infty,-2\right)\cup \left(-2,2\right)\cup \left(2,+\infty\right)\) - step2: Multiply both sides of the equation by LCD: \(\left(\frac{3}{a+2}-\frac{6a}{a^{2}-4}\right)\left(a-2\right)\left(a+2\right)=\frac{1}{a-2}\times \left(a-2\right)\left(a+2\right)\) - step3: Simplify the equation: \(-3a-6=a+2\) - step4: Move the expression to the left side: \(-3a-a=2+6\) - step5: Add and subtract: \(-4a=2+6\) - step6: Add and subtract: \(-4a=8\) - step7: Change the signs: \(4a=-8\) - step8: Divide both sides: \(\frac{4a}{4}=\frac{-8}{4}\) - step9: Divide the numbers: \(a=-2\) - step10: Check if the solution is in the defined range: \(a=-2,a \in \left(-\infty,-2\right)\cup \left(-2,2\right)\cup \left(2,+\infty\right)\) - step11: Find the intersection: \(a \in \varnothing \) The solution to the equation \( \frac{3}{a+2}-\frac{6a}{a^{2}-4}=\frac{1}{a-2} \) is that there are no real solutions for \( a \). Therefore, the student is incorrect in stating that the solution is -2.