A student solves the following equation and
determines that the solution is -2 . Is the student
correct? Explain.

Ask by Powers Blake. in the United States

Jan 23,2025

Question

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

UpStudy AI · Accepted Answer

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:
  $$\frac{3}{a+2}-\frac{6a}{a^{2}-4}=\frac{1}{a-2},a \in \left(-\infty,-2ight)\cup \left(-2,2ight)\cup \left(2,+\inftyight)$$
- step2: Multiply both sides of the equation by LCD:
  $$\left(\frac{3}{a+2}-\frac{6a}{a^{2}-4}ight)\left(a-2ight)\left(a+2ight)=\frac{1}{a-2}	imes \left(a-2ight)\left(a+2ight)$$
- 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:
  $$a=-2,a \in \left(-\infty,-2ight)\cup \left(-2,2ight)\cup \left(2,+\inftyight)$$
- 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.
The student is incorrect because there are no real solutions to the equation, so -2 is not a valid solution.

A student solves the following equation and
determines that the solution is -2 . Is the student
correct? Explain.

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