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2/3(4x+1)=3/5(5-x)
We move all terms to the left:
2/3(4x+1)-(3/5(5-x))=0
Domain of the equation: 3(4x+1)!=0
x∈R
Domain of the equation: 5(5-x))!=0We add all the numbers together, and all the variables
x∈R
2/3(4x+1)-(3/5(-1x+5))=0
We calculate fractions
(10x(-)/(3(4x+1)*5(-1x+5)))+(-9x4/(3(4x+1)*5(-1x+5)))=0
We calculate terms in parentheses: +(10x(-)/(3(4x+1)*5(-1x+5))), so:
10x(-)/(3(4x+1)*5(-1x+5))
We add all the numbers together, and all the variables
10x0/(3(4x+1)*5(-1x+5))
We multiply all the terms by the denominator
10x0
We add all the numbers together, and all the variables
10x
Back to the equation:
+(10x)
We calculate terms in parentheses: +(-9x4/(3(4x+1)*5(-1x+5))), so:
-9x4/(3(4x+1)*5(-1x+5))
We multiply all the terms by the denominator
-9x4
We add all the numbers together, and all the variables
-9x^4
We do not support expression: x^4
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