what is the solution to the equation $\frac{3}{m + 3}-\frac{m}{3 - m}=\frac{m^{2}+9}{m^{2}-9}$?\n$m = 3$\n$m…

Explanation:

Step1: Factor the denominator

Note that $m^{2}-9=(m + 3)(m - 3)$. The given equation $\frac{3}{m + 3}-\frac{m}{3 - m}=\frac{m^{2}+9}{m^{2}-9}$ can be rewritten as $\frac{3}{m + 3}+\frac{m}{m - 3}=\frac{m^{2}+9}{(m + 3)(m - 3)}$.

Step2: Find the common - denominator

The common denominator of the left - hand side is $(m + 3)(m - 3)$. So we have $\frac{3(m - 3)+m(m + 3)}{(m + 3)(m - 3)}=\frac{m^{2}+9}{(m + 3)(m - 3)}$.

Step3: Expand the numerator

Expand $3(m - 3)+m(m + 3)$: $3(m - 3)+m(m + 3)=3m-9+m^{2}+3m=m^{2}+6m - 9$. The equation becomes $\frac{m^{2}+6m - 9}{(m + 3)(m - 3)}=\frac{m^{2}+9}{(m + 3)(m - 3)}$. Since the denominators are the same, we can set the numerators equal: $m^{2}+6m - 9=m^{2}+9$.

Step4: Solve for m

Subtract $m^{2}$ from both sides of the equation: $m^{2}+6m - 9-m^{2}=m^{2}+9-m^{2}$, which simplifies to $6m-9 = 9$. Add 9 to both sides: $6m-9 + 9=9 + 9$, so $6m=18$. Divide both sides by 6: $m = 3$.

Step5: Check for extraneous solutions

When $m = 3$, the original denominators $m + 3=6$ and $m^{2}-9=0$ and $3 - m=0$. Since division by zero is undefined, $m = 3$ is an extraneous solution.

Answer:

no solution