Let the consecutive odd integers be #x, x+2, x+4, and x+6#
Even numbers are added to #x# because of the assumption that #x# is an odd integer and we need to add #2# to get the next odd integer. for example, if #x=3#, then we need to add #2# to #3# to get the next odd integer which is #5# and so on. Adding #1# will make the next number even.
Now, make and equation to match the condition given in the question,
#x+ (x+4)# = #1/2 {(x+2)+(x+6)}+7#
or, #2x +4# = #1/2(2x+8)+7#
or, #2x+4# =#(x+4)+7#
On the right side when #1/2# is multiplied to #2x#, both the twos get cancelled and only #x# remains. Similarly, #4# remains after #1/2# is multiplied to #8#.
So, now the equation becomes:
#2x+4# = #x+11# (Note: #4+7# = #11#)
Solving for #x#,
#2x-x# = #11=4#
or, #x# = #7#
Therefore, other consecutive odd numbers are:
#x+2# = #7+2=9#
#x+4# = #7+4 = 11#
and,
#x+6# = #7+6 = 13#
