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1/9(a+5)=1/3(2-a)
We move all terms to the left:
1/9(a+5)-(1/3(2-a))=0
Domain of the equation: 9(a+5)!=0
a∈R
Domain of the equation: 3(2-a))!=0We add all the numbers together, and all the variables
a∈R
1/9(a+5)-(1/3(-1a+2))=0
We calculate fractions
(3a(-)/(9(a+5)*3(-1a+2)))+(-9aa/(9(a+5)*3(-1a+2)))=0
We calculate terms in parentheses: +(3a(-)/(9(a+5)*3(-1a+2))), so:
3a(-)/(9(a+5)*3(-1a+2))
We add all the numbers together, and all the variables
3a0/(9(a+5)*3(-1a+2))
We multiply all the terms by the denominator
3a0
We add all the numbers together, and all the variables
3a
Back to the equation:
+(3a)
We calculate terms in parentheses: +(-9aa/(9(a+5)*3(-1a+2))), so:We get rid of parentheses
-9aa/(9(a+5)*3(-1a+2))
We multiply all the terms by the denominator
-9aa
Back to the equation:
+(-9aa)
3a-9aa=0
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