Finding the equation to tangent line to a function given certain info - Multiple answers? 
The problem states: find the equation of the tangent line to the graph of continuous function $y = f(x)$ at $x=3$ given the following information.
$\lim_{x \to 3} {\dfrac{f(x)+7}{2x-6}}=8$      and      $f(3) = -7$
How you are supposed to solve it
$\lim_{x \to 3} {{\dfrac{1}{2}}\dfrac{f(x)+7}{x-3}}=8$ Take out 1/2
${\dfrac{1}{2}}\lim_{x \to 3} {\dfrac{f(x)+7}{x-3}}=8$ Take 1/2 out of limit
${\dfrac{1}{2}}\lim_{x \to 3} {\dfrac{f(x)-f(3)}{x-3}}=8$ As we know $f(3)=-7$, plug that in
${\dfrac{1}{2}}f'(3)=8$ By definition of limit, change $\lim_{x \to 3} {\dfrac{f(x)-f(3)}{x-3}}$ to $f'(3)$
$f'(3)=16$
Therefore, the tangent line at $f(3)=-7$ is $y+7={16}(x-3)$
How I solved it
As $\lim_{x \to 3} {\dfrac{f(x)+7}{2x-6}}=8$ and this is a continuous function where it exists (as $f(x)$ is continuous and $\dfrac{f(x)+7}{2x-6}$ is a rational function so $\lim_{x \to a}{\dfrac{f(x)-f(a)}{x-a}}=f(a)$)
I know that the ratio between the numerator and the denominator of $ {\dfrac{f(x)+7}{2x-6}}$ approaches 8 as $\lim_{x \to 3}$.
Therefore, at $x=3$ (we know the function $f(x)$ exists at $f(3)$ as stated in the directions)
$8(f(x)+7)=2x-6$
$8(y+7)=2x-6$ Directions stated $f(x)=y$
$8y+56=2x-6$ Factoring in the 8
$8dy=2dx$ Via implicit differentiation
$dy={\dfrac{1}{4}}dx$
${\dfrac{dy}{dx}}={\dfrac{1}{4}}$
Therefore, the tangent line at $f(3)=-7$ is
$y+7={\dfrac{1}{4}}(x-3)$
Questions
Could someone find an example where either the 'correct' answer or 'my answer' gives a correct answer?
Or perhaps explain where I went wrong?
Or is the problem itself fundamentally flawed?
Thanks for the help.

As they have given the : $f(x=3) = -7$
And they have also given the value of $$\lim_{x\to3}\frac{f(x)+7}{2x-6} = 8$$
Now, this must be the undermined form($\frac{0}{0}$) $\implies$ you can use L'Hôpital's rule 
Now,
$$\begin{align*} \frac{f'(x =3)}{2} = 
8 \implies f'(x=3) = 8×2 = 16 
\end{align*}$$
Which is the slope of the tangent at $x=3$
Equation of tangent at point: $P=(3, -7)$ As they have already given $f(3) = -7$
$$\begin{align*} \frac{y+7}{x-3} = 16
\end{align*}$$
