Could someone please explain this parametric surface?

Question

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

Answer 1

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.

Answer 2

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.