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

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Solution for (x-5)/3x+6/(2x+1)=0 equation: 



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



Domain of the equation: 3x!=0

x!=0/3

x!=0

x∈R





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

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

2x!=-1

x!=-1/2

x!=-1/2

x∈R



We calculate fractions


((x-5)*(2x+1))/(6x^2+3x)+18x/(6x^2+3x)=0



We calculate terms in parentheses: +((x-5)*(2x+1))/(6x^2+3x), so:

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

We add all the numbers together, and all the variables

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

We multiply all the terms by the denominator


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

Back to the equation:

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



We multiply all the terms by the denominator


((3x*(6x^2+(x-5)*(2x+1))))*(6x^2+3x)+18x=0



We calculate terms in parentheses: +((3x*(6x^2+(x-5)*(2x+1))))*(6x^2+3x), so:

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

We add all the numbers together, and all the variables

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

Back to the equation:

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



We add all the numbers together, and all the variables

18x+(3x+(3x*(6x^2+(x-5)*(2x+1))))*6x^2=0

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