The sum of three consecutive integers is three more than twice the middle integer. create an equation and solve to find the integers. can someone me out?

Three consecutive integers...
Let call the first integer as ''x''.
They are consecutive integers. So, we have: x, x+1, x+2.
The sum is (x+(x+1)+(x+2)).
The sum is three more than twice the middle integer (x+1). So, we get the following equation:

3*(x+(x+1)+(x+2)) = 2*(x+1)

Let's solve the equation:
1. Multiplying...
3x + 3x + 3 + 3x + 6 = 2x + 2

2. Adding the terms...
9x + 9 = 2x + 2

3. Let's subtract 2x from both sides...
9x - 2x + 9 = 2x - 2x+ 2

7x + 9 = 2

Now, let's subtract 9...
7x + 9 - 9 = 2 - 9
7x = -7

Now, divide both sides by 7...
7x/7 = -7/7

So, we have:
x = - 1

First integer: x = -1
Second integer: x + 1 => -1 + 1 = 0
Third integer: x + 2 => -1 + 2 = 1

Ican’t answer if there is no diagram