((n-5)/6)+((n+3)/7)=11/21

Solution for ((n-5)/6)+((n+3)/7)=11/21 equation:

((n-5)/6)+((n+3)/7)=11/21

We move all terms to the left:

((n-5)/6)+((n+3)/7)-(11/21)=0

We add all the numbers together, and all the variables

((n-5)/6)+((n+3)/7)-(+11/21)=0

We get rid of parentheses

((n-5)/6)+((n+3)/7)-11/21=0

We calculate fractions


(((n-5)*7)*21)/()+(((n+3)*6)*21)/()+()/()=0


We calculate terms in parentheses: +(((n-5)*7)*21)/(), so:

((n-5)*7)*21)/(

We multiply all the terms by the denominator


((n-5)*7)*21)


We calculate terms in parentheses: +((n-5)*7)*21), so:

(n-5)*7)*21

We multiply parentheses

21n+

We add all the numbers together, and all the variables

21n

Back to the equation:

+(21n)


Back to the equation:

+(21n)



We calculate terms in parentheses: +(((n+3)*6)*21)/(), so:

((n+3)*6)*21)/(

We multiply all the terms by the denominator


((n+3)*6)*21)


We calculate terms in parentheses: +((n+3)*6)*21), so:

(n+3)*6)*21

We multiply parentheses

21n+

We add all the numbers together, and all the variables

21n

Back to the equation:

+(21n)


Back to the equation:

+(21n)


We add all the numbers together, and all the variables

42n+1=0

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

42n=-1

n=-1/42

n=-1/42