The sum of three consecutive integers is -96. what is the larger integer

Let the first integer be x. Since the three integers are consecutive, the next two integers will be x + 1 and x + 2, respectively. Collectively the integers add to equal -96. That means you can set our expressions for the integers to -96, like so:

-96 = x + (x + 1) + (x + 2)

Now we can solve algebraically for x.

-96 = x + (x + 1) + (x + 2)
-96 = x + x + 1 + x + 2
-96 = 3x + 3
-99 = 3x
-33 = x

The first integer is x, so the first integer is -33.
The second integer is x + 1, so the second integer is -32.
The third integer is x + 2, so the third integer is -31.

Let's quickly add these integers to see if their sum is -96.

-33 + (-32) + (-31)
-33 - 32 - 31 = -96

These three consecutive integers add to equal -96. The question asks for the larger integer. Recall that the higher the number in front of a negative sign, the smaller the number. -2 is smaller than -1. That means the order of the integers from largest to smallest is -31, -32, -33.

The largest integer is -31.

4+2=-2. 1+2=3. y=-2/3x+b. 2=4/3+b. 2/3=b. the answer is y= -2/3x+2/3

This does not make sense