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14-2/5(x+10)=2/5(25+x)
We move all terms to the left:
14-2/5(x+10)-(2/5(25+x))=0
Domain of the equation: 5(x+10)!=0
x∈R
Domain of the equation: 5(25+x))!=0We add all the numbers together, and all the variables
x∈R
-2/5(x+10)-(2/5(x+25))+14=0
We calculate fractions
(-10xx/(5(x+10)*5(x+25)))+(-10xx/(5(x+10)*5(x+25)))+14=0
We calculate terms in parentheses: +(-10xx/(5(x+10)*5(x+25))), so:
-10xx/(5(x+10)*5(x+25))
We multiply all the terms by the denominator
-10xx
Back to the equation:
+(-10xx)
We calculate terms in parentheses: +(-10xx/(5(x+10)*5(x+25))), so:We get rid of parentheses
-10xx/(5(x+10)*5(x+25))
We multiply all the terms by the denominator
-10xx
Back to the equation:
+(-10xx)
-10xx-10xx+14=0
We move all terms containing x to the left, all other terms to the right
-10xx-10xx=-14
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