Let's denote the three whole numbers as $x$, $y$, and $z$. According to the problem, when added two at a time, their sums are as follows:

$x + y = 39$ $y + z = 48$ $z + x = 51$

Now, we add all three equations together:

$(x + y) + (y + z) + (z + x) = 39 + 48 + 51$ $2(x + y + z) = 138$ $x + y + z = \frac{138}{2} = 69$

Now, to find each number individually, we subtract each of the original equations from $x + y + z = 69$:

1. $z = 69 - (x + y) = 69 - 39 = 30$
2. $x = 69 - (y + z) = 69 - 48 = 21$
3. $y = 69 - (z + x) = 69 - 51 = 18$

Thus, the three numbers are $x = 21$, $y = 18$, and $z = 30$.

The largest of the three numbers is 30.

Would you like further details on this solution or have any additional questions?

Here are 5 related questions you could explore:

1. What if the sums were different? How would the solution change?
2. Can you generalize this method for any number of equations like this?
3. What if the numbers were not whole numbers? How would you handle fractions?
4. How can this method be applied to solving systems of linear equations in general?
5. What if you only knew two of the sums, could you still solve for the numbers?

Tip: Adding all the given equations and then subtracting them individually is a powerful strategy to isolate unknowns when dealing with sums of pairs of numbers.