# four consecutive even integers sum to -92

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### 2 Answers

Let the first integer be n.

==> Then the next consecutive even integers are: n+2, n+4, and n+6.

Given that the sum of the four integers is -92.

==> n + (n+2) + (n+4) + (n+6)= -92

Now we will combine like terms.

==> (n+n+n+n) + (2+4 + 6)= -92

==> 4n + 12 = -92

Now we will solve for n.

We will subtract 12 from both sides.

==> 4n= -92 -12

==> 4n= -104

Now we will divide by 4.

==> n = -104/4 = -26

Then the numbers are:

-26 , (-26+2) , (-26+4), and (-26+6)

**==> -26, -24, -22, and -20.**

To check to will find the sum.

==> -26 + -24+ -22 + -20 = -92.

Let: x = 1st #

x+2 = 2nd #

x+4 = 3rd #

x+6 = 4th #

Equation: (x) + (x+2) + (x+4) + (x+6) = -92

Solutuion: (x) + (x+2) + (x+4) + (x+6) = -92

4x + 12 = -92

4x = -92 - 12

4x = -104

4x/4 = -104/4

x = -26

substitute:

x = -26

x+2 = -24

x+4 = -22

x+6 = -20

Therefore : 4 consecutive integers are -26, -24, -22, and -20...

Check: (x) + (x+2) + (x+4) + (x+6) = -92

(-26) + (-24) + (-22) + (-20) = -92