Any other way to calculate range of transformed variable? I have been doing questions related to transformations of 2d random variables. 
In a question I have to find a range for 
$$u = (x-y)/2
$$
and
$$v = y 
$$
where $x,y > 0$
So is it necessary that I plot a graph for $2u+v>0$ and $v>0$ to find the range for pdf or is there any other way?
 A: Support of a Bivariate Distribution.
As @whuber points out this is a mathematical problem just about the boundaries of a region after transformation. I anticipate that
the next step for you may be to use such transformations for
probability distributions, which you mention in your question. I hope you will look at the boundaries in this answer for now---and for later, take a look at the probability distributions.
Looking at two univariate distributions. Let's look at a particular example, using simulation in R.
Then I hope a general solution will make more sense. We begin by looking
at the two univariate distributions, for $U$ and $V,$ separately.
Let $X$ and $Y$ be independently distributed as $\mathsf{Exp}(\lambda = 1/\mu = 0.1).$
Then $E(X) = E(Y) = 10,$ so $E(X-Y) = E(X) - E(Y) = 0,$ and 
$E(U) = E\left(\frac{X-Y}{2}\right) = 0.$ 
Also, $Var(X) = Var(Y) = 10^2,$ so $Var(X-Y) = Var(X) + Var(Y) = 200,$ and
$Var(U)  = \frac 1 4 (200) = 50.$
Random variables $X$ and $Y$ both have support $(, \infty).$ The support of the random variable $U$ is $(-\infty, \infty).$ For example, if $X$ is huge and $Y$ is near $0,$ and $U$ could be huge. Also, extremely negative values
of $U$ are easy to imagine. The support of random variable $V$ is $(0, \infty),$ the same as for $Y.$
Now for a simulation. With only $10\,000$ sampled values from each distribution we can
only expect to approximate means and standard deviations to about 1 or 2 significant digits, but that is good enough to see our theoretical computations above aren't terribly wrong.
set.seed(2020)
x = rexp(10000, .1);  y = rexp(10000, .1)
u = (x-y)/2;  v = y
mean(u);  sd(u)
[1] 0.1129585    # aprx E(U) = 0
[1] 7.034309     # aprx SD(U) = sqrt(50) =  7.0711
mean(v);  sd(v)
[1] 9.919974     # aprx E(V) = 10
[1] 9.968185     # aprx SD(V) = sqrt(100) = 10

The shape of the distribution of $U$ is suggested by a histogram of
our simulated sample of size $10\,000. This is a Laplace distribution.
hist(u, br=40, prob=T, col="skyblue2")


Looking at their joint distribution. However, for the full story about your transformation, we need to investigate
the joint distribution of $(U,V).$ We can get an intuitive view of it
by looking at a scatterplot of our simulated values. Perhaps you were surprised
that the left boundary of the support of the bivariate distribution is
given by a diagonal line. Maybe you can try some values of $x$ and $y$ to determine
why some points are 'missing' from the support. For example why is $u =-20, v = 20$
impossible? 

Notes: (1) Better estimates of means and standard deviations of $U$ and $V$ could
have been attained by using more than $10\,000$ values of each. But with more
points the scatterplot becomes more difficult to interpret. So $10\,000$ is a compromise.
(2) From the scatterplot it is clear that $U$ and $V$ are not independent. For example,
$P(-21 < U < -19) > 0$ and $P(19 < V < 21) > 0,$ but $P(-21 < U < -19,\,19 < V < 21) = 0.$
Addendum: Implementing suggestion by @whuber just for boundary, no distribution (color added); see Comment:
x = c(rep(0,50),1:50, 0, 10)
y = c(1:50, rep(0,50),0, 10)
u = (x-y)/2; v=y
frb=hsv((0:102)/102, 1, .7)
par(mfrow=c(1,2))
 plot(x,y, pch=20, col=frb)
 plot(u,v, pch=20, col=frb)
par(mfrow=c(1,2))


