Question

3 consecutive odd integers such that the sum of twice the first and three times the second is 55 more than twice the third

Answer 1

Answer:

Consecutive odd integers are 19 , 21 & 23

Step-by-step explanation:

Let the first 3 consecutive odd intergers be x , (x + 2) and (x + 4).

According to the question,

$2x + 3(x + 2) = 2(x + 4) + 55$

$= > 2x + 3x + 6 = 2x + 8 + 55$

Eliminating 2x from both the sides,

$= > 3x + 6 = 63$

$= > 3x = 63 - 6 = 57$

$= > x = \frac{57}{3} = 19$

So, the consecutive odd integers are = 19 , 21 & 23.