<s>[INST] 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. ?
 [/INST] For your first method since hcf(x,y)=4 then $4|x$ and $4|y$ so clearly $x=7$ is not  a possibility since $4\not|7.$ So you can only choose $x$ of the form $4k$ for some $k\in\mathbb N.$
In fact the possible integer solutions are given by $(x,y)=(-48,6),(6,-48),(7,336),(7,48)$ and $(12,16).$ Then checking, the correct one is $(12,16).$
For your second method you have $x=4a$ and $y=4b$ then $$\frac{a+b}{ab}=\frac{7}{12}=\frac{7}{3\cdot 4}=\frac{3+4}{3\cdot 4}$$
so clearly you can have $(a,b)=(3,4),(4,3)$ (and the other solutions are $(2,12)$ and $(12,2)$ which are not valid since they have a common divisor of $2$).
So setting $(a,b)=(3,4)$ we have $(x,y)=(12,16)$ and setting $(a,b)=(4,3)$ you have $(x,y)=(16,12).$
</s>