the som of three consecutive even integers is -30
a = first number
b = secnod number = a + 1
c = third number = b + 1 = a + 1 + 1 = a + 2
a + b + c = 30
a + a + 1 + a + 2 = 30
3 a + 3 = 30 Subtract 3 to both sides
3 a + 3 - 3 = 30 - 3
3 a = 27 Divide both sides by 3
a = 9
first number = 9
secnod number = 9 + 1 = 10
third number = 9 + 2 = 11
Proof:
a + b + c = 30
9 + 10 + 11 = 30
30 = 30
I believe it was a sum of -30
no big deal ....
1st integer x-2
2nd integer x
3rd integer x+2
x-2 + x + x+2 = -30
3x = -30
x = -10
1st is -12
2nd is -10
3rd is -8
To solve this problem, we can use algebra. Let's assume that the first even integer is x.
Since we are looking for three consecutive even integers, the next two consecutive even integers would be x + 2 and x + 4, respectively.
According to the problem, the sum of these three consecutive even integers is -30. So, we can write the equation as follows:
x + (x + 2) + (x + 4) = -30
Now, let's simplify and solve the equation:
3x + 6 = -30
Subtracting 6 from both sides of the equation:
3x = -36
Finally, dividing both sides of the equation by 3:
x = -12
So, the first even integer is -12.
The next two consecutive even integers are -10 and -8, respectively.
Therefore, the three consecutive even integers that have a sum of -30 are -12, -10, and -8.