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1/5(x+4)=1/2(x-2)
We move all terms to the left:
1/5(x+4)-(1/2(x-2))=0
Domain of the equation: 5(x+4)!=0
x∈R
Domain of the equation: 2(x-2))!=0We calculate fractions
x∈R
(2xx/(5(x+4)*2(x-2)))+(-5xx/(5(x+4)*2(x-2)))=0
We calculate terms in parentheses: +(2xx/(5(x+4)*2(x-2))), so:
2xx/(5(x+4)*2(x-2))
We multiply all the terms by the denominator
2xx
Back to the equation:
+(2xx)
We calculate terms in parentheses: +(-5xx/(5(x+4)*2(x-2))), so:We get rid of parentheses
-5xx/(5(x+4)*2(x-2))
We multiply all the terms by the denominator
-5xx
Back to the equation:
+(-5xx)
2xx-5xx=0
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