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To do these problems the algebra way, make n the middle integer and call the 3 integers n-1, n, and n+1.

(n-1) + n + (n+1) = -93

simplifying the left side,

3n = -93

divide both sides by 3 to get n by itself ('isolate' n)

n = -31

So the 3 consecutive integers are -32, -31, and -30.

-32, -31, -30

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find three consecutive integers whose sum is - 93

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31, 32, 33

Q: What three consecutive integers have a sum of 96?

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If three consecutive integers have the sum of 96, then the problem can be expressed with the equation... N + (N+1) + (N+2) = 96 ...Simplify that and solve and you get... 3N + 3 = 96 3N = 93 N = 31 ... so the three integers are 31, 32, and 33.

31 32 33

The numbers are 31, 32 and 33.

31, 32 and 33

-31, -32, -33

Average is 96/3 ie 32 so integers are 30, 32 and 34.

The integers are 47 and 49.

They are odd consecutive integers: 21, 23, 25 and 27.

To figure this out, divide 96 by three, ignoring the remainder for now. This gives the number 30. The remainder is 6. Three consecutive numbers that add up to six are 1, 2 and 3. So the three consecutive integers are 31, 32, and 33.

The integers are -98 and -96.

31, 32, 33

Divide 96 by 3= 32 your three numbers are now: 32 -1 = 31 32 = 32 32 + 1=33 31 + 32 +33 = 96