Two consecutive integers have a sum of 73. find the integers.

Here's how to do it using algebra:

Let the smaller integer be 'x' .  Then the larger integer is (x+1).

Their sum is (x) + (x+1).

That's (2x + 1) .

The problem says their sum is 73, so we can write . . . 2x + 1 = 73 .

Subtract 1 from each side . . . 2x = 72

Divide each side by 2 . . . x = 36

The smaller integer is 36 .

The larger integer is (x+1) .  That would be 37.

The two consecutive integers are  36 and  37 .

Here's how to do it using your brain:

-- If they were both the same number, then their sum would be double it, and each integer would be half of  73 .

-- That's 36.5 .

-- But  36.5  is not even an integer, so neither integer can be 36.5.  Is there something we can do to a pair of 36.5s to make them whole numbers (integers), make them different, and make them consecutive, but not let their sum change ?

-- Well, if we slice off a piece from one of them and glue it onto the other one, then we'll wind up with two numbers that are different, and their sum won't change.

-- What happens if we take the  .5  off of one of them and glue it onto the other one ?   Then one of the 36.5s shrinks to become  36, and the other one grows to become  37 .

-- Well whaddaya know !  Now we have two numbers that are consecutive integers, and they add up to  73 .  Great move !

a c

step-by-step explanation: