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

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



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

We move all terms to the left:

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



Domain of the equation: (6-5x))!=0

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

-5x)!=-6

x!=-6/1

x!=-6

x∈R



We add all the numbers together, and all the variables

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

We multiply all the terms by the denominator


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



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

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

We multiply parentheses

(3x+1)+10x-12

We get rid of parentheses

3x+10x+1-12

We add all the numbers together, and all the variables

13x-11

Back to the equation:

+(13x-11)



We get rid of parentheses

13x-11=0

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

13x=11

x=11/13

x=11/13

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