There are 14 desks of 4 types; desks with 1 drawer, 2 drawers, 3 drawers and 4 drawers. There are 33 drawers altogether.The number of 2 drawer...

This problem can be solved using systems and
substitution.

Assign variables as
follows:

a:  1 drawer
desks

b:  2 drawer desks

c:  3
drawer desks

d:  4 drawer
desks

Here is what we know:

a
+ b + c + d = 14 (because there are 14 desks)

1a + 2b + 3c
+ 4d = 33 (because there are 33 drawers)

b + c = a (because
the combination of 2 and 3 drawer desks equals the number of 1 drawer
desks)

Substitute (b + c) in for a in both
equations.

(b + c) + b + c + d =
14

2b + 2c + d = 14

1(b + c) +
2b + 3c + 4d = 33

3b + 4c + 4d =
33

You now have the following system of
equations:

2b + 2c + d = 14

3b
+ 4c + 4d = 33

Multiplying the first equation by
-4.

-8d + -8c + -4d = -56

3b +
4c + 4d = 33

Now add the equations.  The variable d has
been eliminated.

-5b + -4c =
-23

To make this equation easier to work with, multiply
through by -1.

5b + 4c =
23

Solve for b.

b = (23 - 4c)
/ 5

Since b represents 2 drawer desks, it must be a whole
number.  The only value of c that will produce a whole number answer is
2.

b = (23 - 4 * 2) / 5

b =
(23 - 8) / 5

b = 15 / 5

b =
3

So now we know that b = 3 and c =
2. 

We already know that b + c =
a.

Therefore...

b + c =
a

3 + 2 = 5

a =
5

We also know that there are 14 desks. 
Therefore...

a + b + c + d =
14

5 + 3 + 2 + d = 14

10 + d =
14

d = 4

There
are 5 desks with one drawer, 3 desks with two drawers, 2 desks with three drawers, and 4
desks with four drawers.

You can check this
using the second equation.

1a + 2b + 3c + 4d =
33

1(5) + 2(3) + 3(2) + 4(4) =
33

5 + 6 + 6 + 16 = 33

33 =
33

This answer
works.

Answer:  There are 5 desks
with one drawer.