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
