11-1/5(j-10)=2/5(25+j)

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Solution for 11-1/5(j-10)=2/5(25+j) equation: 



11-1/5(j-10)=2/5(25+j)

We move all terms to the left:

11-1/5(j-10)-(2/5(25+j))=0



Domain of the equation: 5(j-10)!=0

j∈R





Domain of the equation: 5(25+j))!=0

j∈R



We add all the numbers together, and all the variables

-1/5(j-10)-(2/5(j+25))+11=0

We calculate fractions


(-5jj/(5(j-10)*5(j+25)))+(-10jj/(5(j-10)*5(j+25)))+11=0



We calculate terms in parentheses: +(-5jj/(5(j-10)*5(j+25))), so:

-5jj/(5(j-10)*5(j+25))

We multiply all the terms by the denominator


-5jj

Back to the equation:

+(-5jj)





We calculate terms in parentheses: +(-10jj/(5(j-10)*5(j+25))), so:

-10jj/(5(j-10)*5(j+25))

We multiply all the terms by the denominator


-10jj

Back to the equation:

+(-10jj)



We get rid of parentheses

-5jj-10jj+11=0

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

-5jj-10jj=-11

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