((19-2(3x-1))/5=x+2

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Solution for ((19-2(3x-1))/5=x+2 equation: 



((19-2(3x-1))/5=x+2

We move all terms to the left:

((19-2(3x-1))/5-(x+2)=0



Domain of the equation: 5-(x+2)!=0

We move all terms containing x to the left, all other terms to the right

-(x+2)!=-5

x∈R



We multiply all the terms by the denominator


((19-2(3x-1))=0



We calculate terms in parentheses: +((19-2(3x-1)), so:

(19-2(3x-1)



We calculate terms in parentheses: +(19-2(3x-1), so:

19-2(3x-1

determiningTheFunctionDomain
-2(3x-1+19

Back to the equation:

+(-2(3x-1+19)



We add all the numbers together, and all the variables

(-2(3x+18)



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-2(3x+18





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-2(3x+18





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-2(3x+18





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-2(3x+18





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-2(3x+18





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-2(3x+18





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-2(3x+18





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-2(3x+18





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-2(3x+18





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-2(3x+18





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