6(2j-4)+4(4j-3)=20

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Solution for 6(2j-4)+4(4j-3)=20 equation: 



6(2j-4)+4(4j-3)=20

We move all terms to the left:

6(2j-4)+4(4j-3)-(20)=0

We multiply parentheses

12j+16j-24-12-20=0

We add all the numbers together, and all the variables

28j-56=0

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

28j=56

j=56/28

j=2

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