SOLUTION: find three consecutive odd integers such that the sum of the first three exceeds twice the fourth by 5.

Question 12334: find three consecutive odd integers such that the sum of the first three exceeds twice the fourth by 5.
Answer by bonster(299) (Show Source): You can put this solution on YOUR website!
three consecutive odd integers
#1=x
#2=x+2
#3=x+4
#4=x+6
the sum of the first three exceeds twice the fourth by 5
keep in mind: if it says the sum exceeds another number by 5, add 5 to the sum
[x+(x+2)+(x+4)]+5=2(x+6)
3x+11=2x+12 <--subtract 2x from both sides
(3x+11)-2x=(2x+12)-2x
x+11=12 <--subtract 11 from both sides
(x+11)-11=12-11
x=1

if #1=x then #2 (x+2)=1+2=3, #3 (x+4)=1+4=5
your three numbers are 1, 3, 5

Check:
the sum of the first three exceeds twice the fourth by 5
(1+3+5)+5=2(7)
4+10=14
14=14

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