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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We calculate fractions


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


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

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

We multiply all the terms by the denominator


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


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

(2x-3)*3)*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+2)*6)*2)/(), so:

((x+2)*6)*2)/(

We multiply all the terms by the denominator


((x+2)*6)*2)


We calculate terms in parentheses: +((x+2)*6)*2), so:

(x+2)*6)*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