Question:
Find all pairs of consecutive odd positive integers both of which are smaller than 10 such that their sum is more than 11.
Solution:
Let x be the smaller of the two consecutive odd positive integers. Then, the other integer is x + 2.
Since both the integers are smaller than 10,
$x+2<10$
$\Rightarrow x<10-2$
$\Rightarrow x<8 \ldots$ (i)
Also, the sum of the two integers is more than 11.
$\therefore x+(x+2)>11$
$\Rightarrow 2 x+2>11$
$\Rightarrow 2 x>11-2$
$\Rightarrow 2 x>9$
$\Rightarrow x>\frac{9}{2}$
$\Rightarrow x>4.5$ (ii)
From (i) and (ii), we obtain $4.5
Since $x$ is an odd number, $x$ can take the values, 5 and 7 .
Thus, the required possible pairs are $(5,7)$ and $(7,9)$.