Question

Three consecutive even integers have a sum of
48. Find the integers.

Answer 1

what do u think it is?

Answer 2

I got 6,7,8

Answer 3

@likeabossssssss thanks for the help

Answer 4

it says it's incorrect

Answer 5

it said it would be 14,16, and 18

Answer 6

We know three consecutive numbers equal 48, due to the fact that the number are even we would go up by multiples of 2.

\(\huge\bf{(n)+(n+2)+(n+4)=48}\)

Combine Like terms.

\(\huge\bf{3n+6=48}\)

Subtract 6 from both sides.

\(\huge\bf{3n=42}\)

Divide by 3.

\(\huge\bf{n=14}\)

Now input 14.

\(\huge\bf{(14)=14}\)

\(\huge\bf{14+2=16}\)

\(\huge\bf{14+4=18}\)

\(\huge\bf\color{red}{14,16,18}\)

That should be your answer @Ebonyserrano

Answer 7

At least I think thats how you do it...do you agree @KJSaif ?