If they had been all unicycles there would have been only 9 wheels, So the
extra 13-9=4 wheels had to be on the bicycles, so there were 4 bicycles and
the other 9-4=5 were unicycles.  

Checking: 4 bicycles have 8 wheels and 5 unicycles have 5 wheels, so that
makes 4+5=9 things to ride with 8+5=13 wheels. 

No algebra needed. lol

Now square the number of bicycles and get 16.

Edwin

Here I am again under my alias AnlytcPhil.  But I'm Edwin.  It occurred 
to me that you're probably studying algebra and not basic math, so your teacher probably wants you to use algebra.

Let b = the number of bicycles
Let u = the number of unicycles

There are a total of 9 bicycles and unicycles in a path. 

So 

b + u = 9

Each bicycle has 2 wheels and each unicycle has 1 wheel, so

b bicycles has 2b wheels.
u unicycles has u wheels.

So 2b wheels + u wheels equals 13 wheels or

2b + u = 13.

So the system is



Solve the first equation for u, u = 9-b

Substitute (9-b) for u in

    2b+u = 13

2b+(9-b) = 13
  2b+9-b = 13
     b+9 = 13
       b = 4   <-- (4 bicycles)

Substitute (4) for b in

     b+u = 9
   (4)+u = 9 
       u = 5   <-- (5 unicycles)

This problem should be a problem in basic math, not in algebra,
don't you think?  lol

Edwin