To solve the present problem, we need to remember that, for every change made on one of the sides of the equality, it must be done the same to the other side, so you will maintain the equality.
The first step is to group the terms which have j together, and to perform the calculation with the numbers which are without j among themselves, as follows:
[tex]\begin{gathered} 4j+6j+3-9=j+9j+3-9 \\ \to(4j+6j)+(3-9)=(j+9j)+(3-9) \\ \text{Now we p}\operatorname{erf}orm\text{ the n}eeded\text{ calculation:} \\ \to10j-6=10j-6 \end{gathered}[/tex]What we are already able to see from this point is that both sides result in the same expression, which will lead to the trivial solution: 0 = 0
It means that, doesn't matter the value of j, equality will always be true.
From the above-developed solution, we are able to conclude that the answer is:
[tex]j\in\mathfrak{\Re }[/tex]