Question

find 3 consecutive odd intergers with a sum of 63. show work.

Answer 1

Solve

\[n+(n+2)+(n+4)=63\]

for n

Answer 2

Let 2x+1 = first odd integer
then 2x+3 = next
and 2x + 5 = last one
6x+9 = 63
6x=54
x=9
2x+1=19
2x+3=21
2x+5 =23 The answer is 19, 21, and 23

Answer 3

\[n+(n+2)+(n+4)=63\]

\[3n+6=63\]

\[3n=57\]

\[n=19\]

so 19,21,23