Using trial and improvement:

The first two numbers, 15 and 5, add to 20.

The last three numbers, 17, 7 and 9, add to 33.

This means that $x$ added onto 20 and 33 makes two different totals that have the same mean.

If you assume that the mean will be a whole number you can try numbers, that when added to 20 give a total that is divisible by 3 (e.g. 1, 4, 7, 10...) and when added to 33 give a total that is divisible by 4 (e.g. 3, 7, 11, 15, ...)

After some trials you may find that the number you are looking for is 19.

Noticing difference:

The first three numbers must add up to 3 times the mean, and the last four numbers must add up to 4 times the mean, so the difference between their sums will be the mean.

5 + 15 + x = 20 + x

7 + 9 + 17 + x = 33 + x

(33 + x) - (20 + x) = 13

Therefore 13 is the mean of both groups of numbers.

3 x 13 = 39

4 x 13 = 52

Therefore the first three numbers must add to 39, and the last four numbers to 52.

39 - 20 = 19

52 - 33 = 19

Therefore 19 is the missing number.

Using algebra:

The mean of the first three numbers is $\dfrac{20 + x}{3}$

and the mean of the last four numbers is $\dfrac{33 + x}{4}$

Both means are equal, so:

$$\frac{20 + x}{3}=\frac{33+ x}{4}$$

Multiplying both sides by $12$:

$$80 + 4x = 99 + 3x$$

Subtracting $80$ from both sides:

$$4x = 19 + 3x$$

Subtracting $3x$ from both sides:

$$x = 19$$

Therefore, the missing number is $19$.

When we plug it in into the original question, we find that the mean of the first three numbers is $13$ and so is the mean of the last four numbers.