How do you find two consecutive even integers whose sum is 66?

Hint: An integer is a real number that does not have a fractional component. Even integers are the integers that are divisible by 2. Consecutive integers are the integers that follow a sequence such that each subsequent number is one more than its preceding number.

Complete step by step answer:
As per the given question, we need to find the pair of even consecutive integers. Given that, the sum of the two even consecutive integers is 66. Even consecutive integers are the integers that follow a sequence such that each subsequent integer is two more than its preceding integer. Let, n be an integer divisible by 2. Then, the even consecutive integer of n is (n+2).

As it is given that the sum of the two consecutive even integers is 66, we can write an equation showing their relation as
\[\Rightarrow n+(n+2)=66\]

Removing the brackets, we can rewrite the above equation as
\[\Rightarrow n+n+2=66\]

We know that the summation of two \[n\] is \[2n\]. So, we can rewrite the equation as
\[\Rightarrow 2n+2=66\]

We can remove 2 from left hand side of the equation by subtracting 2 on both sides. Then the equation becomes
\[\begin{align}
  & \Rightarrow 2n+2-2=66-2 \\
 & \Rightarrow 2n=64 \\
\end{align}\]

On the left-hand side of the equation, we have \[2n\]. So, we have to divide by 2 into both sides. Then, we get
\[\Rightarrow \dfrac{2n}{2}=\dfrac{64}{2}\]

We know that \[2n\] by 2 gives \[n\] and 64 by 2 is equal to 32. On substituting these values into the equation, we get
\[\Rightarrow n=32\]

For \[n=32\], the consecutive even integer is \[n+2=32+2=34\].

\[\therefore\] The two consecutive even integers which add up to 66 are 32 and 34.

Note:
In order to such types of problems, we should know the concept of consecutive numbers. While solving such problems, a common mistake is made with the calculation part. So, we should avoid mistakes while solving the problem.