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We know that odd integers are of the form \[2n + 1,2n + 3,2n + 5....\]

The difference between two integers is

\[(2n + 3) - (2n + 1) = 2n + 3 - 2n - 1 = 2\]

So if we assume the first odd positive integer as \[x\]

Then consecutive odd positive integer will be \[x + 2\]

Now we know the sum of two consecutive odd integers is 80.

\[

\Rightarrow x + x + 2 = 80 \\

\Rightarrow 2x + 2 = 80 \\

\]

Shift all the constant values to one side

\[

\Rightarrow 2x = 80 - 2 \\

\Rightarrow 2x = 78 \\

\]

Dividing both sides by 2

\[

\Rightarrow \dfrac{{2x}}{2} = \dfrac{{78}}{2} \\

\Rightarrow x = 39 \\

\]

For second consecutive integer \[ \Rightarrow x + 2 = 39 + 2 = 41\]

So the first number is 39 and the second number is 41.

Students also make the mistake of considering \[(2n - 1)\] as the odd number which is wrong because if we put the value of as 0 we will get a negative odd integer and in this question we are taking values of positive odd integers.

Alternate method:

We know that odd integers are of the form \[2n + 1,2n + 3,2n + 5....\]

Considering two consecutive odd integers \[2n + 1,2n + 3\]

Then forming equation of sum of two numbers

\[

\Rightarrow 2n + 1 + 2n + 3 = 80 \\

\Rightarrow 4n + 4 = 80 \\

\]

Shift all constants to one side of the equation

\[

\Rightarrow 4n = 80 - 4 \\

\Rightarrow 4n = 76 \\

\]

Divide both sides of the equation by 4

\[

\Rightarrow \dfrac{{4n}}{4} = \dfrac{{76}}{4} \\

\Rightarrow n = 18 \\

\]

So substituting the value of n to the two numbers we get

\[

\Rightarrow 2n + 1 = 2(19) + 1 = 38 + 1 = 39 \\

\Rightarrow 2n + 3 = 2(19) + 3 = 38 + 3 = 41 \\

\]

So two consecutive odd positive integers are 39 and 41.