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I need help with a math problem, please help?
I've been working on this for a while and i was hoping if you guys can show me a way of solving similar problems.
The question is:
The sum of four consecutive odd integers is three more than five times the least of the integers. Find the integers.
Is there a really easy way of doing this?
I know i learned it somewhere but i totally blanked out! Thanks
4 Answers
- 1 decade agoFavorite Answer
Let x = be the ist odd integer
remember that 2 is always the difference between odd and even numbers. (1,3,5,7,9,11,13 and so on)
2nd odd number= x+2
3rd = x+4
4th= x+ 6
sum= 5x+3
Lets solve:
x +(x+2)+(x+4)+(x+6)= 5x+3
x+x+2+x+4+x+6= 5x+3
4x+ 12= 5x+3
12-3= 5x-4x
9= x
1. x=9-----ist odd integer
Let's substitute the value of x:
2. x+ 2
9+2
= 11
3. x+4
9+4
= 13
4. x+6
9+6
= 15
Lets check our answer:
x +(x+2)+(x+4)+(x+6)= 5x+3
9+ (9+2)+(9+4)+(9+6)= 5(9) +3
9+11+13+15= 45 +3
48= 48
I hope this helps.
=)
- 1 decade ago
9+11+13+15=48
(sum of 4 consecutive odd integers)
5*9+3=48
(3 more than five times the least integer)
- piano1Lv 41 decade ago
w+x+y+z=3+5w
this would be the basic set-up for the problem I think where w is the smallest integer
