Question

The sum of three consecutive integers is 16 more than 2 times the smallest integer. find the integers

Answer 1

X, x + 1, x + 2

sum of three consecutive integers is 1 more than 2 times the smallest integer
(x) + (x + 1) + (x + 2) - 16 = 2x
3x + 3 - 16 = 2x
3x -13 = 2x
-13 = -x
13 = x
The smallest integer is 13, the second integer is 14, the third integer is 15

Answer 2

When we refer to integers as consecutive, we're referring to the fact that they follow immediately after one another on the number line. For instance, 1, 2, and 3 are consecutive since they come right after one another sequentially. For here, we don't know what our integers are, but if we call the first one "a", we can make some statements about the other two. We know that the next integer will be one more than $a$, or $a+1$, in other words. We also know that the third integer will be one more than the second, or $(a+1)+1=a+2$. With that in mind, let's approach the rest of the problem:

"The sum of three consecutive integers..." From this, we know we'll be adding all three of our integers together, so let's collect those:

$a+(a+1)+(a+2)=3a+3$

"...is 16 more than 2 times the smallest integer." Translating that into math: "is" is shorthand for "equals," "16 more than" means we'll be adding 16 to some value, and "2 times" obviously implies multiplication plus two. Finally, the smallest integer we have is $a$, so let's compose these facts into a full expression and combine them with our sum above to form a complete equation:

$3a+3=2a+16$

From here, all that's left is to solve for $a$. Once you have that value, $a$, $a+1$, and $a+2$ will be your integers.