Fill in the blanks of a 4 by 4 table so that every row, every column and both diagonals add to the same total. Row 1 is -5, 9, blank, blank. Row 2 is blank, 0, 1, blank. Row 3 is 2, blank, 5, -1. Row 4 is 7, blank, blank, 10.

Step-by-step solution

1. 1

   Hunt for a line that is already complete. Nothing is gained by introducing an unknown magic constant if one line can be summed outright. The main diagonal $-5,\,0,\,5,\,10$ has no blanks, so

   $S=-5+0+5+10=10$

   Every row, column and diagonal must therefore total $10$.

2. 2

   Fill any line with exactly one blank. Row 3 and column 1 each have a single unknown:

   $2+e+5+(-1)=10 \Rightarrow e=4,\qquad -5+c+2+7=10 \Rightarrow c=6$

   Each newly found entry opens up further lines, so the puzzle unravels one cell at a time rather than needing a system of equations.

3. 3

   Use the new entries to unlock more lines. With $e=4$, column 2 has one blank left:

   $9+0+4+f=10 \Rightarrow f=-3$

   Row 4 then gives $7+(-3)+g+10=10 \Rightarrow g=-4$, and row 2 gives $6+0+1+d=10 \Rightarrow d=3$.

4. 4

   Finish the top row through column 3. Column 3 now reads $a,1,5,-4$:

   $a+1+5-4=10 \Rightarrow a=8$

   and row 1 gives $-5+9+8+b=10 \Rightarrow b=-2$.

5. 5

   Verify every line, including the anti-diagonal. The completed square is

   $\begin{array}{|c|c|c|c|}\hline -5&9&8&-2\\\hline 6&0&1&3\\\hline 2&4&5&-1\\\hline 7&-3&-4&10\\\hline\end{array}$

   Rows: $10,10,10,10$. Columns: $10,10,10,10$. Main diagonal $-5+0+5+10=10$; anti-diagonal $-2+1+4+7=10$. The anti-diagonal was never used to derive anything, so it is a genuine independent check.