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14-1/5(j-10)=2/5(j+25)
We move all terms to the left:
14-1/5(j-10)-(2/5(j+25))=0
Domain of the equation: 5(j-10)!=0
j∈R
Domain of the equation: 5(j+25))!=0We calculate fractions
j∈R
(-5jj/(5(j-10)*5(j+25)))+(-10jj/(5(j-10)*5(j+25)))+14=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:We get rid of parentheses
-10jj/(5(j-10)*5(j+25))
We multiply all the terms by the denominator
-10jj
Back to the equation:
+(-10jj)
-5jj-10jj+14=0
We move all terms containing j to the left, all other terms to the right
-5jj-10jj=-14
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