5/2(j+3)=19+6/7j

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Solution for 5/2(j+3)=19+6/7j equation: 



5/2(j+3)=19+6/7j

We move all terms to the left:

5/2(j+3)-(19+6/7j)=0



Domain of the equation: 2(j+3)!=0

j∈R





Domain of the equation: 7j)!=0

j!=0/1

j!=0

j∈R



We add all the numbers together, and all the variables

5/2(j+3)-(6/7j+19)=0

We get rid of parentheses

5/2(j+3)-6/7j-19=0

We calculate fractions


35j/(14j^2+42j)+(-12jj/(14j^2+42j)-19=0

We calculate fractions

We do not support ejpression: j^3

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