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1/5(7j+9)=7/5j-2
We move all terms to the left:
1/5(7j+9)-(7/5j-2)=0
Domain of the equation: 5(7j+9)!=0
j∈R
Domain of the equation: 5j-2)!=0We get rid of parentheses
j∈R
1/5(7j+9)-7/5j+2=0
We calculate fractions
5j/(175j^2+225j)+(-35j7/(175j^2+225j)+2=0
We calculate fractions
We do not support ejpression: j^3
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