abcdefghijklmnopqrst

I tried multiplying top and bottom by (x-a) and plugging in a for x in the factorization, but then you get lots of 0's in numerator and denominator, and it leads nowhere, but it seems like the right thing to do.

ok thanks

We can first set the smaller one to be $a$, and the larger one to be $a+2$. Now, the problem tells us that $a(a+2)=99$. Multiplying out and rearranging, we obtain the equation $a^2+2a-99=0$. We can factor this as $(a-9)(a+11)=0$, so the solutions are $a=-11$ or $a=9$. This means that the odd integers could either be $\boxed{-11,-9}$, or they could be $\boxed{9,11}$.

Hope this helped 

Sorry again...

Thanks Alan!

First divide everything by 3, and we get -5x-y=45. Now add y to both sides, and we have -5x=y+45. Finally, subtract 45 from both sides to get

$\boxed{y=-5x-45}$

Can you please show how you got this answer Guest?

I'm very sorry, I made a rookie mistake and forgot to save my changes on the desmos 

I've fixed it now if you want to see

Why are you trolling?

Everyone's time is valuable please do not post trivial questions. Not only does this take up time, it takes away space from problems from people that actually might need help.

If you really don't know (I doubt it), then sorry, and the answer is

1+1=2

I just played with the numbers until the circles overlapped, and I got a=5, b=3, c=15, and d=15

Then I actually did the problem 

Do you want to see my work?