(3u+1)/8=(7u-9)/4+(1)

Solution for (3u+1)/8=(7u-9)/4+(1) equation:

(3u+1)/8=(7u-9)/4+(1)

We move all terms to the left:

(3u+1)/8-((7u-9)/4+(1))=0

We calculate fractions


(12u+4)/()+(-((7u-9)*8)/()=0


We calculate terms in parentheses: +(-((7u-9)*8)/(), so:

-((7u-9)*8)/(

We multiply all the terms by the denominator


-((7u-9)*8)


We calculate terms in parentheses: -((7u-9)*8), so:

(7u-9)*8

We multiply parentheses

56u-72

Back to the equation:

-(56u-72)


We get rid of parentheses

-56u+72

Back to the equation:

+(-56u+72)


We get rid of parentheses

(12u+4)/()-56u+72=0

We multiply all the terms by the denominator


(12u+4)-56u*()+72*()=0

We add all the numbers together, and all the variables

(12u+4)-56u*()=0

We get rid of parentheses

12u-56u*()+4=0

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

12u-56u*()=-4