6+2(3j-2)=4(j+1)

Solution for 6+2(3j-2)=4(j+1) equation:

6+2(3j-2)=4(j+1)

We move all terms to the left:

6+2(3j-2)-(4(j+1))=0

We multiply parentheses

6j-(4(j+1))-4+6=0


We calculate terms in parentheses: -(4(j+1)), so:

4(j+1)

We multiply parentheses

4j+4

Back to the equation:

-(4j+4)


We add all the numbers together, and all the variables

6j-(4j+4)+2=0

We get rid of parentheses

6j-4j-4+2=0

We add all the numbers together, and all the variables

2j-2=0

We move all terms containing j to the left, all other terms to the right

2j=2

j=2/2

j=1