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

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



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

We move all terms to the left:

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



Domain of the equation: 3x!=0

x!=0/3

x!=0

x∈R





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

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

x!=-1

x∈R



We add all the numbers together, and all the variables

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

We add all the numbers together, and all the variables

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

We calculate fractions


((x+3)*(x+1))/(3x^2+3x)+(3x^2+6x)/(3x^2+3x)-4=0



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

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

We add all the numbers together, and all the variables

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

We multiply all the terms by the denominator


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

Back to the equation:

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



We multiply all the terms by the denominator


((3x*(3x^2+(x+3)*(x+1))))*(3x^2+3x)+(3x^2+6x)-4*(3x^2+3x)=0



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

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

We add all the numbers together, and all the variables

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

Back to the equation:

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



We multiply parentheses

-12x^2+(3x+(3x*(3x^2+(x+3)*(x+1))))*3x^2+(3x^2+6x)-12x=0

We get rid of parentheses

-12x^2+3x^2+(3x+(3x*(3x^2+(x+3)*(x+1))))*3x^2+6x-12x=0

We add all the numbers together, and all the variables

-9x^2-6x+(3x+(3x*(3x^2+(x+3)*(x+1))))*3x^2=0

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