Question 12097
there are 4 consecutive odd intergers. 


#1= x
#2= x+2
#3= x+4
#4= x+6


the sum of these 4 intergers equals 96
x+(x+2)+(x+4)+(x+6)=96
4x+12=96        <--subtract 12 from both sides
4x+12-12=96-12
4x=84   <--divide both sides by 4
{{{4x/4}}}={{{84/4}}}
x=21


since#1=x and x=21, the first number is 21
x+2=21+2=23
x+4=21+4=25
x+6=27


the four numbers are 21, 23, 25, 27