what are 3 consecutive even integers that have a sum of -48?

Answer 1

The answer is -48 = (-17) + (-16) + (-15)

Here is how I did it:

Let a represent the first integer in the 3 integer series

The second integer will be one greater than a, represented as (a+1)

The third integer will be two greater than a, represented as (a+2)

Now, we can write the equation below:

-48 = a + (a+1) + (a+2)

Now group the a's and the numbers

-48 = 3a + 3

Subtract 3 from both sides

-51 = 3a

Divide both sides by 3

-17 = a

So the first integer in the series is -17

Add positive 1 to get the second integer: -17 + 1 = -16

Add positive 2 to get the third integer: -17 + 2 = -15

Hope that helps

Answer 2

Hi Sammy W

-14

-16

-18

Let your even intergers be

Let first even integer = -x

Let next consecutive even integer = -x - 2

Let final consecutive even integer = -x - 4

Then combine and solve for x

-x - x - 2 - x - 4 = -48

Give it a try

I hope this helps