I need help figuring out how to approach solving the attached problem and others like it. Single digit numbers are a piece of cake, no logic involved, but when the numbers become double digit you can spin your wheels filling in the squares. There has to be a trick to starting, any ideas????

2. Hello, Hunter385!

You said the first one was "a piece of cake".
How did you solve it? . . . Trial and error?

. . . $\begin{array}{|c|c|}\hline a & b \\ \hline c & d \\ \hline \end{array}\begin{array}{c}38 \\ 64\end{array}$
. $49\;\;45\;\;57\;\; 53$

We have six equations:

. . $\begin{array}{cc}a+b\:=\:38 & [1] \\ c+d\:=\:64 & [2] \\ a+c \:=\:45 & [3] \\ b+d\:=\:57 & [4] \\ a+d\:=\:53 & [5] \\ b+c\:=\:49 & [6] \end{array}$

$\begin{array}{cccccc}
\text{Subtract [1] - [3]:} & b-c &=&\text{-}7 \\
\text{Add [6]:} & b + c &=& 49 \end{array}$

And we have: . $2b \:= \:42\quad\Rightarrow\quad \boxed{b \:=\:21} \quad\Rightarrow\quad\boxed{ c \:=\:28}$

$\begin{array}{cccccc}
\text{Subtract [1] - [4]:} & a - d &=& \text{-}19 \\
\text{Add [5]:} & a + d &=& 53 \end{array}$

And we have: . $2a \:=\:34\quad\Rightarrow\quad \boxed{a \:=\:17} \quad\Rightarrow\quad \boxed{d \:=\:36}$

. . . Solution: . $\begin{array}{|c|c|}\hline 17 & 21 \\ \hline 28 & 36 \\ \hline \end{array}$