General proof that the minimum value of $x+y$ from $\frac{4}{11} < \frac{x}{y} < \frac{3}{8}$ is $4+11+3+8=26$ I found this problem and the solution on Twitter (translated).

$x$ and $y$ is a (positive) integer that satisfies $\frac{4}{11} < \frac{x}{y} < \frac{3}{8}$. If $y$ is the smallest number, what is $x+y$?

The given solution is $\frac{x}{y} = \frac{4+3}{11+8} = \frac{7}{19}$. Therefore, $x+y=7+19=26$.
Is the solution correct? If it is, why is the solution correct? What is the proof that it is correct? And if possible, what is the general formula?
Generally, the solution will be, given $a,b,c,d \in \mathbb{N}$ where $\frac{a}{b} < \frac {c}{d}$, $a$ and $b$ are coprimes, $c$ and $d$ are coprimes, and $a \neq b \neq c \neq d$, the minimum value of $x+y$ where $x,y \in \mathbb{N}$ and $\frac{a}{b} < \frac{x}{y} < \frac{c}{d}$ is $a+b+c+d$.
Edit: it seems the bounds are also unclear. What are the correct bounds for the general statement?
What is the proof that it's right or wrong? Thank you.
Edit: If it's possible, please explain it with only 10th-grade math or lower.
 A: If I were starting out from scratch here, I'd note that:

*

*$\frac{a}{b}<\frac{x}{y}$ means $bx-ay>0$. Since we are working with integers, $bx-ay\geq1$.

*$\frac{x}{y}<\frac{c}{d}$ means $-dx+cy>0$. Since we are working with integers, $-dx+cy\geq1$.

So we have the system
$$\left\{
\begin{aligned}
bx-ay&\geq1\\
-dx+cy&\geq1
\end{aligned}
\right.$$
So it is a linear programming question. It means
$$\begin{bmatrix}b&-a\\-d&c\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}\in R$$
where $R$ is the closed infinite rectangular region that has $(1,1)$ as a lower left corner.
So applying the matrix inverse,$$\begin{bmatrix}x\\y\end{bmatrix}\in \frac{1}{bc-ad}\begin{bmatrix}c&a\\d&b\end{bmatrix}R$$
Now $R=\{(u,v)|u,v\geq1\}$. So $$\begin{bmatrix}x\\y\end{bmatrix}\in \left\{\frac{1}{bc-ad}\begin{bmatrix}cu+av\\du+bv\end{bmatrix}\middle| u,v\geq1\right\}$$
Each of $a,b,c,d,u,v,bc-ad$ is positive. So the smallest that the second coordinate of the above could be is when $u$ and $v$ are at their smallest, $1$. So you will have the smallest possible value for $y$ when
$$\begin{bmatrix}x\\y\end{bmatrix}=\frac{1}{bc-ad}\begin{bmatrix}c+a\\d+b\end{bmatrix}$$
BUT we still require that $x$ and $y$ be integers. One way to guarantee this is to require that $bc-ad=1$. Then $\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}c+a\\d+b\end{bmatrix}$. And the smallest $y$ is acieved when $y=d+b$ and $x=c+a$.
So it is sufficient for $bc-ad=1$ to make it such that the minimal solution to the inequality $\frac{a}{b}<\frac{x}{y}<\frac{c}{d}$ is $y=d+b$, $x=c+a$.
For example, with $\frac{4}{11}<\frac{x}{y}<\frac{3}{8}$, the minimal solution is $y=19$, $x=7$.

Furthermore if $bc-ad\neq1$ but it happens to divide each of $d+b$ and $c+a$, that would make the minimal solution be $y=\frac{d+b}{bc-ad}$, $x=\frac{c+a}{bc-ad}$.
For example, with $\frac{3}{7}<\frac{x}{y}<\frac{5}{11}$, the minimal solution is $y=\frac{11+7}{2}=9$, $x=\frac{5+3}{2}=4$. (The fraction is $\frac{4}{9}$.)

If $bc-ad$ does not divide either $d+b$ or $c+a$, then $y=\frac{d+b}{bc-ad}$, $x=\frac{c+a}{bc-ad}$ give the minimal real solutions to the inequality, but at least one of them is not an integer. Instead, we would have to look for values of $(u,v)$ not both equal to $1$ such that $bc-ad$ divides both $du+bv$ and $cu+av$. And it's not immediately clear which values of $u$ and $v$ would lead to the minimal solution.
