20=(b+(b-3))/2

Solution for 20=(b+(b-3))/2 equation:

20=(b+(b-3))/2

We move all terms to the left:

20-((b+(b-3))/2)=0

We multiply all the terms by the denominator


-((b+(b-3))+20*2)=0


We calculate terms in parentheses: -((b+(b-3))+20*2), so:

(b+(b-3))+20*2

We add all the numbers together, and all the variables

(b+(b-3))+40


We calculate terms in parentheses: +(b+(b-3)), so:

b+(b-3)

We get rid of parentheses

b+b-3

We add all the numbers together, and all the variables

2b-3

Back to the equation:

+(2b-3)


We get rid of parentheses

2b-3+40

We add all the numbers together, and all the variables

2b+37

Back to the equation:

-(2b+37)


We get rid of parentheses

-2b-37=0

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

-2b=37

b=37/-2

b=-18+1/2