There are three consecutive odd integers such that the sum of twice the second and the smallest is seven more than twice the largest. Find the sum of the integers.

11 +13 + 15  = 39

Step-by-step explanation:

Let the consecutive odd integer be X , Y , Z .

X < Y < Z

Y = X + 2

Z = X + 4

Given ,

2 Y + X = 2 Z + 7

2 ( X + 2 ) + X = 2 ( X + 4 ) + 7

2 X + 4 + X = 2 X + 8 + 7

X = 15 - 4 = 11

So the integers are

11 , 13 , 15 .

Sum of integers = 39.

did you get the answer yet?

step-by-step explanation:

the simplified form is

step-by-step explanation:

we want to simplify the expression,

let us change the middle bar to a normal division sign to obtain,

we multiply the first fraction by the reciprocal of the second fraction to obtain,

we factor to obtain,

we cancel the common factors to get,

we simplify to get,

the correct answer is b