Consecutive Odd Integers

[KV]

Find two consecutive odd integers such that 5 times the first integer is 12 more than 3 times the second.
5(x+1)=3(x+1)+12

 

[Denis]

C'mon, KV: you can't have both integers = x + 1;
try again: use x and x+2

 

[KV]

Odd Integers

5x=3(x+2)+12
5x=3x+6+12
5x-3x=6+12
2x=18
x=9

That does not seem to wrok out right to me. FYI. Algebra is all new to me. :roll:

 

[Mrspi]

I'm not sure why you say "but that does not seem to work out right for me."

So, you got x = 9. The first of the consecutive odd integers, then, is 9.
And the next one is x + 2.....so what would that be?

Now, check.

Is 5 times the first of those integers the same thing as 12 more than three times the second?

You can always check an answer that "doesn't seem right." If it checks by satisfying the conditions of the problem, it's right!

 

[Denis]

Re: Odd Integers

5x=3(x+2)+12
5x=3x+6+12
5x-3x=6+12
2x=18
x=9

Well done! So integers are 9 and 9+2 = 11.

You can check that this is correct by substituting x=9 in the original equation:
5x = 3(x + 2) + 12
5*9 = 3(9 + 2) + 12
45 = 33 + 12
45 = 45

 

[KV]

Thank You

I appreciate all your help and it is a bit clearer for me. Denis Do you do tutoring:?:

 

[Denis]

Re: Thank You

I appreciate all your help and it is a bit clearer for me. Denis Do you do tutoring:?:

Nope. I'm just a retired guy who went back to maths 4 years ago,
and help when I can.

 

[KV]

OK thank you and enjoy our warmer weather.