1/(3x+1)^2+5=6/3x+1

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Solution for 1/(3x+1)^2+5=6/3x+1 equation: 



1/(3x+1)^2+5=6/3x+1

We move all terms to the left:

1/(3x+1)^2+5-(6/3x+1)=0



Domain of the equation: (3x+1)^2!=0

x∈R





Domain of the equation: 3x+1)!=0

x∈R



We get rid of parentheses

1/(3x+1)^2-6/3x-1+5=0

We calculate fractions


3x/((3x+1)^2*3x)+(-6*(3x+1)^2)/((3x+1)^2*3x)-1+5=0



We calculate terms in parentheses: +(-6*(3x+1)^2)/((3x+1)^2*3x), so:

-6*(3x+1)^2)/((3x+1)^2*3x

We multiply all the terms by the denominator


-6*(3x+1)^2)

We multiply parentheses

-18x-

We add all the numbers together, and all the variables

-18x

Back to the equation:

+(-18x)



We add all the numbers together, and all the variables

3x/((3x+1)^2*3x)+(-18x)+4=0

We get rid of parentheses

3x/((3x+1)^2*3x)-18x+4=0

We multiply all the terms by the denominator


3x-18x*((3x+1)^2*3x)+4*((3x+1)^2*3x)=0

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