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2/5(3k-2)=5/2(2k-3)
We move all terms to the left:
2/5(3k-2)-(5/2(2k-3))=0
Domain of the equation: 5(3k-2)!=0
k∈R
Domain of the equation: 2(2k-3))!=0We calculate fractions
k∈R
(4k2/(5(3k-2)*2(2k-3)))+(-25k3/(5(3k-2)*2(2k-3)))=0
We calculate terms in parentheses: +(4k2/(5(3k-2)*2(2k-3))), so:
4k2/(5(3k-2)*2(2k-3))
We multiply all the terms by the denominator
4k2
We add all the numbers together, and all the variables
4k^2
Back to the equation:
+(4k^2)
We calculate terms in parentheses: +(-25k3/(5(3k-2)*2(2k-3))), so:
-25k3/(5(3k-2)*2(2k-3))
We multiply all the terms by the denominator
-25k3
We add all the numbers together, and all the variables
-25k^3
We do not support ekpression: k^3
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