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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.

Respuesta :

Answer:

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.

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