((2x+1)/3)+((x-1)/4)=13/2

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Solution for ((2x+1)/3)+((x-1)/4)=13/2 equation: 



((2x+1)/3)+((x-1)/4)=13/2

We move all terms to the left:

((2x+1)/3)+((x-1)/4)-(13/2)=0

We add all the numbers together, and all the variables

((2x+1)/3)+((x-1)/4)-(+13/2)=0

We get rid of parentheses

((2x+1)/3)+((x-1)/4)-13/2=0

We calculate fractions


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



We calculate terms in parentheses: +(((2x+1)*4)*2)/(), so:

((2x+1)*4)*2)/(

We multiply all the terms by the denominator


((2x+1)*4)*2)



We calculate terms in parentheses: +((2x+1)*4)*2), so:

(2x+1)*4)*2

We multiply parentheses

4x+

We add all the numbers together, and all the variables

4x

Back to the equation:

+(4x)



Back to the equation:

+(4x)





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

((x-1)*3)*2)/(

We multiply all the terms by the denominator


((x-1)*3)*2)



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

(x-1)*3)*2

We multiply parentheses

2x+

We add all the numbers together, and all the variables

2x

Back to the equation:

+(2x)



Back to the equation:

+(2x)



We add all the numbers together, and all the variables

6x+1=0

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

6x=-1

x=-1/6

x=-1/6

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