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-4/5(x+2)=1/2(x+2)
We move all terms to the left:
-4/5(x+2)-(1/2(x+2))=0
Domain of the equation: 5(x+2)!=0
x∈R
Domain of the equation: 2(x+2))!=0We calculate fractions
x∈R
(-8xx/(5(x+2)*2(x+2)))+(-5xx/(5(x+2)*2(x+2)))=0
We calculate terms in parentheses: +(-8xx/(5(x+2)*2(x+2))), so:
-8xx/(5(x+2)*2(x+2))
We multiply all the terms by the denominator
-8xx
Back to the equation:
+(-8xx)
We calculate terms in parentheses: +(-5xx/(5(x+2)*2(x+2))), so:We get rid of parentheses
-5xx/(5(x+2)*2(x+2))
We multiply all the terms by the denominator
-5xx
Back to the equation:
+(-5xx)
-8xx-5xx=0
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