Why can't change of variables be done using the density distribution? Consider a random variable $X$. We want to compute $Z = 5X$.
Now, you could say that 
$$f_Z(z)= P(Z=z) = P(5X=z) = P(X=\frac{z}{5})= f_X(z/5)$$
All of those operations seem legal to me, but obviously the result is wrong. In the proper answer the density distribution should be replaced by the cumulative distribution. So which operation here is now allowed and why?
 A: The mechanics of doing a transformation differ, depending on whether the
distribution is discrete or continuous.
Continuous Distribution. Suppose $X \sim \mathsf{Exp}(rate = \lambda = 2),$ so that the PDF is
$f_X(x) = \lambda e^{-\lambda x},$ for $x > 0;$ and the CDF is
$F_X(x) = 1 - e^{\lambda x},$ for $x > 0.$
Method 1: The CDF of $Z = 5X$ can be found as 
$$F_Z(z) = P(Z \leq z) = P(5X \leq z) = P(X \leq .2z)
= 1 - e^{\lambda(.2x)} = 1 - e^{-.2\lambda z},$$
for $z > 0.$ 
Thus $Z \sim \mathsf{Exp}(rate = .2\lambda = \lambda/5).$
Notice that $E(X) = 1/2 = 1/\lambda$ and $E(Z) = E(5X) = 5E(X) = 5/\lambda = 2.5.$
Method 2: Denote the transformation as $z = h(x) = 5x,$ so that the
inverse transformation is $x = h^{-1}(z) = 0.2z$ and $dx/dz = 0.2.$ 
Then for single-valued
increasing transformations such as this one, there is a theorem that
states:
$$f_Z(z) = f_X(h^{-1}(z))\times \frac{dh^{-1}}{dz} = f_X(0.2z)\times 0.2
= \lambda e^{-\lambda(0.2z)}\times 0.2 = .4 e^{0.4 z},$$
for $z > 0.$ This is the density function of $\mathsf{Exp}(2/5).$ I suppose you can find this theorem in your text.
Intuitively, the reason for the factor $dh^{-1}/dz = .2$ is that the
PDF of $Z$ is 'five times as wide' as the PDF of $X,$ so it needs to be
'one-fifth as tall' in order to maintain the mandatory unit area under
the PDF curve. (Under each curve, 95% of the area is to the left of the
vertical red line.)

Discrete Distribution. Suppose that $Y$ is the number of spots showing
when a fair die is rolled, and that the payout in a game is \$5 per spot.
Then the payout from a roll of the die is $W = 5Y$ dollars.
The possible values of $Y$ are the integers from 1 through 6.
The possible values of $W$ are 5, 10, 15, 20, 25, 30.
For $i = 1, 2, 3, 4, 5,$ we have $p_Y(i) = P(Y = i) = 1/6.$
And $p_W(5i) = P(W = 5i) = 1/6.$
Here, $E(Y) = 3.5$ and $E(W) = E(5Y) = 5E(Y) = 5(3.5) = 17.5$ dollars.
