# What are three consecutive integers whose sum is 96?

Mar 22, 2018

I got $31 , 32 \mathmr{and} 33$

#### Explanation:

$n$
$n + 1$
$n + 2$

you get:

$n + n + 1 + n + 2 = 96$

rearrange:

$3 n = 93$
and so:
$n = \frac{93}{3} = 31$

so our integers are:

$n = 31$
$n + 1 = 32$
$n + 2 = 33$

Mar 22, 2018

You must symbolize the first integer with $x$.

#### Explanation:

Lets pretend the first number was $5$. What would you do to get to the immediate next integer? (Integers are whole numbers like $1 , 2 , 3$ ) You would add $1$. So the next number is symbolized as "$x + 1$".

How would you get from $5$ to $7$? You would add $2$ to the $x$. So the next number is written in symbols as "$x + 2$."

Now add them all up like this: $x + x + 1 + x + 2 = 96$
Combine like terms: $3 x + 3 = 96$
Subtract the 3 from both sides $3 x = 93$
Divide both sides by $3$: $x = 32$
Answer: $x = 32$.

BTW, "consecutive" means to come right after. In my pretend answer, $6$ came right after $5$, and $7$ came right after $6$.

Mar 22, 2018

31, 32, 33

#### Explanation:

If you represent the first integer with the letter $x$, then:
$x + \left(x + 1\right) + \left(x + 2\right) = 96$

This simplifies to:
$x + x + 1 + x + 2 = 96$
$x + x + 1 + x + 2 = 96$
$3 x + 3 = 96$
$3 x = 93$
$x = 31$

The first integer is 31. The next two consecutive integers are 32 $\left(x + 1\right)$ and 33 $\left(x + 2\right)$.