Hello Meh,
Let's break this down into the upstream portion of the trip and the downstream portion, using the relation distance equals speed times time. We will let $0<c$ be the speed of the current and $t_1$ be the time going upstream and $t_2$ be the time going downstream.
$$30=(15-c)t_1$$
$$30=(15+c)t_2$$
Now, we also know the total time is 4.5 hours, so we may state:
$$t_1+t_2=4.5\implies t_2=4.5-t_1$$
And so our second equation becomes:
$$30=(15+c)\left(4.5-t_1 \right)$$
Solving the first equation for $t_1$, we find:
$$t_1=\frac{30}{15-c}$$
Now, substituting this into our new second equation, we obtain an equation in one variable $c$, which is what we are trying to find:
$$30=(15+c)\left(4.5-\frac{30}{15-c} \right)$$
Multiplying through by $15-c$, we get:
$$30(15-c)=(15+c)\left(4.5(15-c)-30 \right)$$
Distribute on the right:
$$30(15-c)=(15+c)\left(67.5-4.5c-30 \right)$$
Combine like terms:
$$30(15-c)=(15+c)\left(37.5-4.5c \right)$$
Multiply through by $2$:
$$30(30-2c)=(15+c)\left(75-9c \right)$$
Distribute on both sides:
$$900-60c=1125-60c-9c^2$$
Combine like terms and arrange as:
$$9c^2=225$$
Divide through by $9$:
$$c^2=25$$
Take the positive root:
$$c=5$$
Thus, we conclude that the current in the river is $$5\,\frac{\text{km}}{\text{hr}}$$.