Find three consecutive integers whose sum is -33

Question

Math questiabout for my high school class in chiolllll 

Answer 1

Hi omar,

First, lets make sure we know exactly what we need to find. Three consecutive integers, so there are three unknowns. We could start out by saying that each of these integers needs a label: x₁, x₂, and x₃. But the key piece of information here is the word consecutive, meaning following one after the other without interruption. 10 and 11 are consecutive numbers, 11 and 12 are consecutive numbers, and they differ by 1. So, I'll call one integer x and the other two integers x + 1 and x + 2.

The second piece of information is "whose sum is -33". Add them up and set it to -33

x + (x + 1) + (x + 2) = -33

In the last equation, x is the first integer, x + 1 is the next one, and x + 2 is the third integer. Now, combine like terms.

3x + 3 = -33

Subtract 3 from both sides

3x = -36

Divide by 3

x = -12

Adding 1 and 2 gives -11 and -10. 

Now check your answer.

-12 + -11 + -10 = -33.

Answer 2

Hey Omar -- here's a "kitchen table" approach ... the -33 sum has 3 elements -- what's the average? ... -11 serves as the middle number, with -10 and -12 on each side ... -10,-11,-12.  Best wishes, sir :)

Answer 3

The way to setup this problem is to identify one number, or integer, (the lowest) as x. For consecutive numbers, you add 1, then 2 (and so on). The word "sum" relates to addition. Your three numbers are x, x+1, and x+2.

x + (x+1) + (x+2) = -33

3x + 3 = -33

3x = -36

x = -12; x+1 = -11; x+2 = -10

Check your work: (-12) + (-11) + (-10) = -33