Need Help W/ Finding 3 Consecutive Odd Integers

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loverose82

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Im begining solving integers and I am confused. I have reviewed in my book, watched videos, but I still don't understand. Here is a homework question to show you all. Please explain so I can possibly understand better in english terms. Im going crazy trying to figure this out. I havent done Algebra in 13 years, Im just returning back to school, and I need help.

1. Find three consecutive odd integers such that four times the sum of the first and third integer is five more than the poduct of three and the second number.
 
loverose82 said:
Im begining solving integers and I am confused. I have reviewed in my book, watched videos, but I still don't understand. Here is a homework question to show you all. Please explain so I can possibly understand better in english terms. Im going crazy trying to figure this out. I havent done Algebra in 13 years, Im just returning back to school, and I need help.

1. Find three consecutive odd integers such that four times the sum of the first and third integer is five more than the poduct of three and the second number.
Click to expand...
Call the first integer n. Then the next two "consecutive odd integers" are n+2 and n+ 4 "Four times the sum of the first and third integers" is 4(n+ (n+ 4))= 4(2n+ 4)= 8n+ 16. "Five more than the product of three and the second number" is 5+ 3(n+ 2)= 5+ 3n+ 6= 3n+ 11. Those are equal so 8n+ 16= 3n+ 11. Solve that for n.
 
loverose82 said:
Im begining solving integers and I am confused. I have reviewed in my book, watched videos, but I still don't understand. Here is a homework question to show you all. Please explain so I can possibly understand better in english terms. Im going crazy trying to figure this out. I havent done Algebra in 13 years, Im just returning back to school, and I need help.

1. Find three consecutive odd integers such that four times the sum of the first and third integer is five more than the poduct of three and the second number.
Click to expand...

Three consecutive odd integers can be written as:

(2*n-1), (2*n +1) & (2*n + 3)

sum of first and third: (2n-1) + (2n+3) = 2n + 2

four times the sum of the first and third integer = 4 * (2n + 2) = 8n +8

the product of three and the second number = 3 * (2n +1) = 6n + 3

four times the sum of the first and third integer is five more than the product of three and the second number.

8n +8 = 5 + 6n + 3

Can you solve for 'n' now?

Then calculate (2*n-1), (2*n +1) & (2*n + 3) to find the numbers.
 
Thank you

Thank you everyone for your response and help, Im still a little confused but I understand better now.
 
Hey:)

loverose82 said:
Thank you everyone for your response and help, Im still a little confused but I understand better now.
Click to expand...

I may be able to help:)
Okay so start what to call the integers.
The first one is always x.
Now since they are consecutive, odd integers instead of just adding one to x we have to add two.
So your integers would look like this:
x, x+2, x+4
Now, lets create the equation.
4[x+(x+4)]=5+(x+2)(3)
Three consecutive odd integers such that four times the sum of the first and third integer is five more than the poduct of three and the second number.
Now, we solve.
I got:
x = -1, 1, and 3.
What did you get?
 
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