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1/4(3x+1)=1/2(x+5)
We move all terms to the left:
1/4(3x+1)-(1/2(x+5))=0
Domain of the equation: 4(3x+1)!=0
x∈R
Domain of the equation: 2(x+5))!=0We calculate fractions
x∈R
(2xx/(4(3x+1)*2(x+5)))+(-4x3/(4(3x+1)*2(x+5)))=0
We calculate terms in parentheses: +(2xx/(4(3x+1)*2(x+5))), so:
2xx/(4(3x+1)*2(x+5))
We multiply all the terms by the denominator
2xx
Back to the equation:
+(2xx)
We calculate terms in parentheses: +(-4x3/(4(3x+1)*2(x+5))), so:
-4x3/(4(3x+1)*2(x+5))
We multiply all the terms by the denominator
-4x3
We add all the numbers together, and all the variables
-4x^3
We do not support expression: x^3
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