Is it possible for this puzzle to be solved without the use of computational software? This puzzle was found from the Hot Network Questions on the right. A repost of a question in a way (from Mathematica SE), but I was wondering if the following puzzle could be done:

A quick translation:

Can anyone solve this? It's so hard!
Fill in the numbers $1–9$ in the boxes, the numbers cannot repeat.

It has already been solved in the given link, the answers being (hidden for those who wish to try and solve it):

 $17\times4=68+25=93$

The answer to the Mathematica question stated there's a way to solve it without brute-forcing, but I didn't understand it too well. How would one—if possible—solve this without using too complex softwares?
You can reduce the effort to a level that can be handled without a computer.

Let's write the calculation as

\begin{array}{cc}
&amp;A&amp;B\\
\times&amp;&amp;C\\\hline
&amp;D&amp;E\\
+&amp;F&amp;G\\\hline
&amp;H&amp;I
\end{array}

There are strong restrictions on $A$ and $C$, since their product must be single-digit. We can't have $C=1$, since otherwise $DE=AB$. So either $A=1$, or $(A,C)$ must be one of $(2,3)$, $(2,4)$, $(3,2)$ or $(4,2)$. We can exclude $(2,4)$ and $(4,2)$: This implies $D=8$, $F=1$ and $H=9$, but then adding $E$ and $G$ mustn't result in a carry, which is impossible, since $I$ can't be $8$ or $9$ and $E+G$ must be at least $3+5$.

So unless $A=1$, $(A,C)$ must be either $(2,3)$ or $(3,2)$. Also $B\notin\{1,5\}$. That leaves $10$ possible triples for $(A,B,C)$, which each determine values for $D$ and $E$, and it shouldn't be too much effort to go through those $10$ cases, find the $4$ remaining options for $F$, $G$, $H$ and $I$ and show that they don't work out.

That leaves us with $A=1$. In that case the options for $(C,B)$ can be enumerated as $(2,7)$, $(2,8)$, $(2,9)$, $(3,4)$, $(3,6)$, $(3,8)$, $(3,9)$, $(4,3)$, $(4,7)$, $(4,8)$, $(4,9)$, $(6,3)$, $(6,7)$, $(6,9)$, $(7,2)$, $(7,4)$, $(7,6)$, $(7,8)$, $(7,9)$, $(8,2)$, $(8,3)$, $(8,4)$, $(8,7)$, $(8,9)$. That's $24$ more cases, for a total of $34$ cases to be checked.
$1 7* 4$ 

$=6 8
+2 5$ 

$9 3$

is the right answer! 


  $$17*4 = 68 + 25 = 93$$ 


There may be more combinations too!