Let's assume that the speed of the boat in still water is x km/hr and speed of the stream is y km/hr.
So, the speed of the boat in upstream will be (x-y) km/hr.
Similarly, the speed of the boat downstream will be (x+y) km/hr.
We know $\text{time}=\left(\frac{\text{distance}}{\text{speed}}\right)$.
Using the above formula we can form the equations in two variables.
Taking the first case,
$\frac{30}{\text{x - y}}+\frac{44}{\text{x + y}}=14$.
Taking the second case,
$\frac{60}{\text{x - y}}+\frac{55}{\text{x + y}}=25$.
Now, we have the equations in two variables but the equations are not linear.
So, we will assume  $\frac{1}{\text{x - y}}=\text{u}$  and  $\frac{1}{\text{x + y}}=\text{v}$.
So on substituting $\text{u}$ and $\text{v}$ in the above two equations, we get
$30\text{u}+44\text{v}=14$   ...(1)
$60\text{u}+55\text{v}=25$   ...(2)
We can solve the above two equations using the elimination method.
$60\text{u}+88\text{v}=28$   ...(3)
(by multiplying equation (1) by 2)
On subtracting equation (2) from (3), we get $\text{v}=\frac{1}{11}$
On substituting $\text{v}$ in equation (2) we get $\text{u}=\frac{1}{3}$
Now as we have assumed 
$\frac{1}{\text{x - y}}=\text{u}$              and                $\frac{1}{\text{x + y}}=\text{v}$ 
On substituting the values of $\text{u}$ and $\text{v}$,
we get a pair of linear equations in $\text{x and y}$
$\text{x - y}=3$...(4)
$\text{x + y}=11$...(5)
On adding (5) from (4), we have 
$2x=14$
$x=7$ 
On subsituting the value of x in $x-y=3$,  we get y = 4.
So, the speed of the boat in still water is 7 km/hr and the speed of the stream is 4 km/hr.