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

Step-by-step explanation:

let the numbers be n,n+2,n+4

2n+3(n+2)=2(n+4)+55

2n+3n+6=2n+8+55

2n+3n-2n=63-6

3n=57

n=57/3=19

so numbers are 19,21,23

-19.333…

-18.333…

-17.333…

Step-by-step explanation:

First is x

Second is x+1

Third is x+2

2x+3(x+1)=2(x+2)-55

2x+3x+3=2x+4-55

5x+3=2x-51

5x-2x=-51-3

3x=-58

3x/3=-58/3

x=-19.333…

x+1=-19.333+1

=-18.333...

x+2=-19.333+2

=-17.333...

its 60

4 x 12 = 48

3 x 4 = 12

48 + 12 = 60

yeeeet