Could someone please explain this parametric surface? I have an example problem in my textbook that I cannot understand. Could someone please explain how the answer the textbook gives of [ x = u, y = v, and z = 3u + 2v] is correct? I am just beginning to learn this so a very basic overview is much appreciated. thank you for the help. 
Problem:
Find a parametric representation of the surface? z = 3x + 4y
 A: In the most simple  parametrizing case you can take $ (u,v)  = (x,y)$ itself directly when defining a plane surface, using what is known as the Monge form $ z = f(x,y). $ 
When each of $(x,y,z)$ are defined as functions of $(u,v)$ then it is next stage of surface definition through proper parametrization.
For example in $( x = u + v , y = u-v , z = 4  u v,\,) $ we have 
$$ 4\, z =(x^2+y^2)$$ in Monge form representing a paraboloid of unit focal length.
But the case of your plane is much simpler than that.
A: Since you are asking for a very basic overview, then let me expand the answer and comments
you already got.
In 3D, a generic point is individuated by the $3$ variables $x,y,z$, and they are "free", meaning that they can get 
any value independently from the others: so (in physics and engineering) we say that the point has "$3$ degrees of freedom (DOF)".
A function $f(x,y,z)=0$ has the meaning that not all of the three variables can assume whichever value: if you fix the value of two 
of them, the other is forced to have the value that satisfies the condition $f(x,y,z)=0$.
Thereby the DOF are reduced to 2: the constraint
given by the function to be null deletes $1$ DOF.
With $2$ DOF remaining, only two variables are independent and, in general, you can choose two parameters, e.g. $u$ and $v$, and put
$$
\left\{ \begin{gathered}
  x = x(u,v) \hfill \\
  y = y(u,v) \hfill \\
  z = z(u,v) \hfill \\
  f(x,y,z) = \varphi (u,v) = 0 \hfill \\ 
\end{gathered}  \right.
$$
If the choice of the three functions tying $x,y,z$ to $u,v$ is done in such a way that the $\phi (u,v)=0$ is automatically
satisfied, then they alone represent the $2$DOF object.
But caution, what is difficult in many instances is to ensure that the range of variation
of $u,v$  will cover , through the three functions, the range of variation of the original variables, 
as it is allowed by the constraint of $f(x,y,z)=0$: no less (holes), no more (repetitions).  
Since you are at the beginning, you are recommended to always do such a check, and explicit
the range of $u,v$ under the brace bracket above.
So in the example you give, you can choose
$$
\left\{ \begin{gathered}
  x,y,z \in \;\;\mathbb{R}\,\quad \;u,v \in \;\;\mathbb{R}\, \hfill \\
  x = u \hfill \\
  y = v \hfill \\
  z = 3x + 4y = 3u + 4v \hfill \\
  z - 3x - 4y = \varphi (u,v) \equiv 0 = 0 \hfill \\ 
\end{gathered}  \right.
$$
But if the given equation was, for instance, that of a (half) cone (probably you will study little ahead)
$$
\left\{ \begin{gathered}
  x,y,z \in \;\;\mathbb{R} \hfill \\
  z = \sqrt {x^{\,2}  + y^{\,2} }  \hfill \\ 
\end{gathered}  \right.
$$
then probably you will choose
$$
\left\{ \begin{gathered}
  x,y,z \in \;\;\mathbb{R}\; \hfill \\
  r,\theta  \in \;\;\mathbb{R}\quad \;0 \leqslant r\quad  - \pi  < \theta  \leqslant \pi  \hfill \\
  x = r\cos \theta  \hfill \\
  y = r\sin \theta  \hfill \\
  z = r \hfill \\
  0 = z - \sqrt {x^{\,2}  + y^{\,2} }  = 0(r,\theta ) \hfill \\ 
\end{gathered}  \right.
$$
Exercise yourself in changing the bounds above and see what happens,
to acquaint yourself with this subject.
