Impasta

I think there's an easier way. If you add the two equations together and divide by 7 you'll get the answer.

My knowledge of logarithms is a bit rusty but I think you can say that

\(b^{2/3}=81\).

Now to solve for \(b\) we need to take the square root of both sides and then cube it, giving you 

\(b^{2/3\cdot3/2}=81^{3/2}\)

The exponent of b will be just 1, so now we have

\(b=729.\)

What I meant to say is, in AoPS, they have a sneaky April Fools prank that I totally did not fall for.

First put both equations in slope-intercept form to express the slope in terms of a.

Then multiply them together and set that product equal to -1 (The product of the slopes of two perpendicular lines is, for the most part, -1)

Then you solve for a using the quadratic formula or factoring.

You're reposting a problem. Again. Did you not get it last time? I'm pretty sure you said you did.

First room: \(\binom{20}{2}\)

Second room: \(\binom{18}{2}\)

And so on. Use multiplication I think. Someone check my work because knowing these types of problems there's usually a catch or my method is completely wrong.