The sum of four consecutive integers is −46. What is the smallest of these integers?

The integers are -13, -12 and -11.

Step-by-step explanation:

Let the four numbers be n, n+1, n +2, and n+3

Therefore (n )+ (n+1 )+(n+2) + (n +3) = -46

4n + (1 +2+3) = -46

4n + 6= -46

Subtract 6 from both sides to get the value of n

4n +6 - 6= -46 -6

4n = -52

Divide both sides by 4

n = -52/4

n = -13

Therefore, n + 1 = -13 + 1= -12

n + 2= -13 + 2 = -11

The integers are -13, -12 and -11.

I hope this helps, please mark as brainliest.

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