Question

Find three consecutive odd numbers when three times of their sum is 5 more than
8 times the middle number.​

Answer 1

$\huge \colorbox {pink}{\red{★Solution࿐}}$

Let the numbers be, x, (x + 2), (x + 4)

According to the question,

$3(x + x + 2 + x + 4) = 5 + 8(x + 2)$

$3(3x + 6) = 5 + 8x + 16$

$9x + 18 = 21 + 8x$

$9x - 8x = 21 - 18$

$\large {\boxed {x = 3}}$

So, the numbers are 3, 5 and 7.

Answer 2

Answer:

The three consecutive numbers are 3,5,7

Step-by-step explanation:

Let the three consecutive numbers be

x, x+2, x+4

3(x+x+2+x+4) = 5+8(x+2)

3(3x+6) = 5+(8x+16)

9x+18 = 5+8x+16

9x + 18 = 8x + 21

9x - 8x = 21-18

x = 3

Substitute the value of x in the consecutive numbers

x, x+2 , x+4

= 3, (3+2) , (3+4)

= 3, 5, 7