x = first integer
x+1 = integer right after x
For example, if x = 10, then the next one after that is x+1 = 10+1 = 11. So 10 and 11 are consecutive integers (one follows right after the other)
The sum of those two integers x and x+1 add to x+x+1 = 2x+1
This sum is no more than 209, therefore,
[tex]2x+1 \le 209[/tex]
[tex]2x+1-1 \le 209-1[/tex]
[tex]2x \le 208[/tex]
[tex]\frac{2x}{2} \le \frac{208}{2}[/tex]
[tex]x \le 104[/tex]
Therefore the largest x can be is 104
If x = 104, then x+1 = 104+1 = 105
Note how 104+105 = 209, which helps confirm the answer
