In a class of 45 students, the number of girls is 5. If 5 more boys and 3 more girls join the class, the number of boys will be twice the number of girls. Find the initial number of boys and girls.

Let $B$ be the initial number of boys and $G$ be the initial number of girls.
From the problem statement:
The total number of students is 45:

$$B + G = 45 \quad (\text{Equation 1})$$

The number of boys is three times the number of girls:

$$B = 3G \quad (\text{Equation 2})$$

Substitute Equation 2 into Equation 1:

$$3G + G = 45$$ $$4G = 45$$ $$G = \frac{45}{4} = 11.25$$

Now find $B$:

$$B = 3G = 3 \times 11.25 = 33.75$$

Since the number of students must be whole numbers, these non-integer values indicate that the initial conditions as stated are problematic. We will proceed to check consistency with the second set of conditions.

5 more boys join, so the new number of boys is $B' = B + 5$.
3 more girls join, so the new number of girls is $G' = G + 3$.
The new number of boys will be twice the new number of girls:

$$B' = 2G'$$

Substitute the expressions for $B'$ and $G'$:

$$B + 5 = 2(G + 3) \quad (\text{Equation 3})$$

Substitute $B = 33.75$ and $G = 11.25$ into Equation 3:

$$33.75 + 5 = 2(11.25 + 3)$$ $$38.75 = 2(14.25)$$ $$38.75 = 28.5$$

The statement $38.75 = 28.5$ is false. This means that the conditions given in the problem cannot be simultaneously satisfied. Therefore, there is no solution for the initial number of boys and girls that fits all the criteria.

$$\text{No solution}$$