Find three consecutive integers if the first is two-fifths of their sum?

Question

Having trouble. Please help!

Answer 1

Since three consecutive numbers could be represented by:

x, x + 1 and x +2, their sum would be 3x +3

Since the first is 2 fifths of the total, we can make the equation:

x = (3x + 3) 2/5, multiply both side by 5

5 x = (3x + 3) x 2

5x = 6x + 6 subtract 6x from both sides

-x = 6; x = -6

Check the answer: -6 x 5/2 = [-6 + -5 + -4] = -15

Answer 2

first number : x
second number : x + 1 (next integer)
third number : x + 2

sum = x + (x + 1) + (x + 2)
3x +3

first number (x) = 2/5 sum (3x +3)
x=2/5*(3x + 3)

that finds first integer then go back to the beginning to find the last two. Hope this helps :D

Answer 3

Let x,x+1,x+2 be the three consecutive integers.
sum=x+(x+1)+(x+2)=3x+3
x=2/5(3x+3)
5x=2(3x+3)
5x=6x+6
-x=6
x=-6
The numbers are -6,-5,-4 (Answer)
check: first one=-6
sum=-15
2/5(sum)=2/5(-15)=-6