Question

Find three consecutive integers such that the sum of the first two integers is equal to 32 more than the third.

Answer 1

34, 35, 36

Step-by-step explanation:

Let the first consecutive integer =x

The next two consecutive integers are: x+1 and x+2

The sum of the first two integers =x+(x+1)

32 more than the third integer = (x+3)+32.

Therefore, we have the equation

`x+(x+1)=(x+3)+32`

We then solve for x.

`\begin{gathered} 2x+1=x+35 \\ \text{Collect like terms} \\ 2x-x=35-1 \\ x=34 \end{gathered}`

The three consecutive integers are: 34, 35 and 36.