# Three consecutive integers are such that the sum of the first and twice the second is 15 more than twice the third. Form a equation & solve?

This is for Algebra 1 honors (8th grade) and I need help asap. I am supposed to show all work but I have no idea how to.

Relevance
• :-

x + 2 ( x + 1 ) = 2 ( x + 2 ) + 15

3x + 2 = 2x + 19

x = 17

Integers are 17 , 18 , 19

• Three consecutive integers are such that

the sum of the first and twice the second is 15 more than twice the third.

Form a equation & solve?

(x - 1) + 2x = 2(x + 1) + 15

3x - 1 = 2x + 17

x = 18

The three consecutive integers are 17, 18, and 19.

• 3 consecutive integers = n, n+1, and n+2

first + twice the second = 15 more than twice the third

n + 2(n+1) = 2(n+2)+15

3n + 2 = 2n + 19

3n - 2n = 19 - 2

n = 17

Ans:

3 integers are 17, 18, and 19

• n = 1st,  n+1 2nd,  n+2 = 3rd

n + 2(n+1) = 15 + 2(n+2)

n + 2n + 2 = 15 + 2n + 4

n + 2n - 2n = 15 + 4 - 2

n = 17

n +1 = 18

n + 2 = 19

• Hint:

Three consecutive integers are x , x+1 , x+2

Form an equation then solve

• x - 1 , x , x + 1

(x - 1) + 2 * x = 15 + 2 * (x + 1)

Expand, combine like terms and solve for x

x + 2x - 1 = 15 + 2x + 2

3x - 1 = 17 + 2x

3x - 2x = 17 + 1

x = 18

18 - 1 , 18 , 18 + 1

17 , 18 , 19

Those are your numbers.  Test it out

17 + 2 * 18 = 17 + 36 = 53

19 * 2 + 15 = 38 + 15 = 53