Find the unknowns from LCM HCF relation. Consider $2$ numbers $x,y$ such that $\frac{x+y}{lcm(x,y)}=\frac{7}{12}$, and it is given that hcf(x,y) is $4$.
How to find $\mathbf{x,y}$.

I have tried the question like this.
\begin{align*}
  \frac{x + y}{(x\cdot y)/4} \Rightarrow \frac{4(x+y)}{x\cdot y} &= \frac{7}{12}\\
48x+ 48y &= 7xy\\
48x &= 7xy-48y\\
48x &= y\cdot(7x-48)\\
y &= \frac{48\cdot x}{7x-48}\\
\end{align*}
As $y$ is a positive number the denominator have to be $<0$, so $x \ge 7$. Now If I put $x =7$
I get $y = \frac{48\cdot 7}{49-48} = 336$. But
lcm(7,336) is 336
and $\frac{7+336}{336} \neq \frac{7}{12}$. Where am I making mistake? the hcf is not 4 for (7,336)

second method I Tried was,
$x = 4a, y =4b.$
now ,
\begin{align*}
  \frac{4a + 4b}{4\cdot a\cdot b} &= \frac{7}{12} \\ 
  \frac{a+b}{a\cdot b} &= \frac{7}{12} \\ 
    \frac{a+b}{a\cdot b} &= \frac{7}{12} \\ 
   \frac{1}{a}+\frac{1}{b} &= \frac{7}{12}\\
   \frac{1}{a} &= \frac{7}{12} - \frac{1}{b}\\
   \frac{1}{a} &= \frac{7b-12}{12b}\\
or\\
a &= \frac{12b}{7b-12}
\end{align*}
so $b\ge 2$ to get an integer, now for $b=2, a= 12. \Rightarrow x=48, y=4$, also if I choose b=2, the hcf is not 4, so if I put $b=3, a = 4 \Rightarrow (12, 16)$ , here the hcf and lcm are,
$4$ and $48$ respectively.and $\frac{12+16}{48} = \frac{7}{12}$.
What is the mistake I am making. ?

Cancelling $\,4\,$ from our gcd $\,(x,y)=4\,\Rightarrow\, \color{#c00}{(a,b)=1}$ thus  $\,\color{#0a0}{(a\!+\!b,\,ab) = 1}\,$ by here.
So $\ \dfrac{a\!+\!b}{\color{#90f}{ab}}\! =\! \dfrac{7}{\color{#90f}{12}}\,$  are $\rm\color{#0a0}{reduced}$, thus $\,\color{#90f}{ab = 12},\,$ so  $\,a,b\,$ are $\rm\color{#c00}{coprime}$ factors of $12$, so $\,a,b = \begin{cases}\!3,4\\\! 4,3\end{cases}$
So the "mistake" was not invoking enough number theory (viz. the $3$ linked well-known theorems).
The first method boils down to the second, but scaled by the gcd $4$ so their is no need to treat it separately. Generally, as explained here homogeneous problems like this are usually simplified by reducing to the coprime case by cancelling the gcd throughout (using gcd & lcm distributive laws).
