Question

Find three consecutive odd integers such that the sum of the smaller number and middle number is 27 less than 3 times the largest number. Please show how you did this.

Answer 1

$\bf \stackrel{\textit{smaller odd integer}}{2a+1}~\hspace{5em}\stackrel{\textit{middle odd integer}}{2a+3}~\hspace{5em}\stackrel{\textit{largest odd integer}}{2a+5} \\\\[-0.35em] \rule{34em}{0.25pt}$

$\bf \stackrel{\textit{sum of smaller and middle}}{(2a+1)+(2a+3)}~~\stackrel{is}{=}~~\stackrel{\textit{three times the largest, and less 27}}{3(2a+5)-27} \\\\\\ 4a+4=6a+15-27\implies 4a+4=6a-12\implies 12+4=6a-4a \\\\\\ 16=2a\implies \cfrac{16}{2}=a\implies 8=a \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \stackrel{2a+1}{17}~\hspace{7em}\stackrel{2a+3}{19}~\hspace{7em}\stackrel{2a+5}{21}~\hfill$