14-2/5(j-10)=2/5(25+j)

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Solution for 14-2/5(j-10)=2/5(25+j) equation: 



14-2/5(j-10)=2/5(25+j)

We move all terms to the left:

14-2/5(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

-2/5(j-10)-(2/5(j+25))+14=0

We calculate fractions


(-10jj/(5(j-10)*5(j+25)))+(-10jj/(5(j-10)*5(j+25)))+14=0



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 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

-10jj-10jj+14=0

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

-10jj-10jj=-14

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