Find three consecutive integers such that the sum of the first and third is equal to five less than three times the second integer.

Answer 1

We are looking for the value of the second integer, so let's mark it as X.

Then three consecutive integers are : (x-1) , x , (x+1)

The sum of the first and the third integers (X-1) + (X+1)

Five less than three times the second integer 3X - 5

Now we can write an equation and solve it for X

1. (X-1)+(X+1)= 3X - 5 simplifying left part
2. 2X = 3X -5 subtract 2X from both sides
3. 0 = X-5
4. X=5

SO THREE CONSECUTIVE INTEGERS ARE 4, 5, 6

Answer 2

Let x = the smallest of the three consecutive integers. Since the integers are consecutive, the other two are therefore x+1 and x+2. From the problem statement, the sum of the first and the third,

x + (x+2)

is equal to five less than three times the second integer, 3(x+1) -5

The equation is therefore

X+ (x+2) = 3(x+1) - 5

Combine, distribute and simplify to get

2x+2 = 3x -2

Subtract 2x from both sides to get

2 = x-2

Add 2 to both sides to get

4=x. Since x is the smallest integer, the other two are 5 and 6

To check,

4+6 = 3(5) -5

10=10