Question

what is the solution to the equation \\(\frac{3}{m + 3} - \frac{m}{3 - m} = \frac{m^2 + 9}{m^2 - 9}\\)?
\\(\bigcirc\\) \\(m = 3\\)
\\(\bigcirc\\) \\(m = 6\\)
\\(\bigcirc\\) all real numbers
\\(\bigcirc\\) no solution

Explanation:

Step1: Factor denominator

Note that $m^2 - 9=(m+3)(m-3)$, and rewrite $3-m=-(m-3)$ to get a common denominator:
$$\frac{3}{m+3} + \frac{m}{m-3} = \frac{m^2+9}{(m+3)(m-3)}$$

Step2: Multiply by common denominator

Multiply all terms by $(m+3)(m-3)$ (where $m
eq\pm3$ to avoid division by 0):
$$3(m-3) + m(m+3) = m^2+9$$

Step3: Expand left-hand side

$$3m - 9 + m^2 + 3m = m^2+9$$

Step4: Simplify left-hand side

Combine like terms:
$$m^2 + 6m - 9 = m^2+9$$

Step5: Isolate variable terms

Subtract $m^2$ from both sides:
$$6m - 9 = 9$$

Step6: Solve for m

Add 9 to both sides, then divide by 6:
$$6m=18 \implies m=3$$

Step7: Check validity

$m=3$ makes the original equation's denominators $m-3$ and $m^2-9$ equal to 0, so it is an extraneous solution.

Answer:

no solution