((b-1)/4)+(2b-3)/4=b-1/2

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



((b-1)/4)+(2b-3)/4=b-1/2

We move all terms to the left:

((b-1)/4)+(2b-3)/4-(b-1/2)=0

We add all the numbers together, and all the variables

((b-1)/4)+(2b-3)/4-(+b-1/2)=0

We get rid of parentheses

((b-1)/4)+(2b-3)/4-b+1/2=0

We calculate fractions


-b+(((b-1)*2)/()+(4b-6)/()+()/()=0



We calculate terms in parentheses: +(((b-1)*2)/()+(4b-6)/()+()/(), so:

((b-1)*2)/()+(4b-6)/()+()/(

We add all the numbers together, and all the variables

((b-1)*2)/()+(4b-6)/()+1

We multiply all the terms by the denominator


((b-1)*2)+(4b-6)+1*()



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

(b-1)*2

We multiply parentheses

2b-2

Back to the equation:

+(2b-2)



We add all the numbers together, and all the variables

(2b-2)+(4b-6)

We get rid of parentheses

2b+4b-2-6

We add all the numbers together, and all the variables

6b-8

Back to the equation:

+(6b-8)



We add all the numbers together, and all the variables

-1b+(6b-8)=0

We get rid of parentheses

-1b+6b-8=0

We add all the numbers together, and all the variables

5b-8=0

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

5b=8

b=8/5

b=1+3/5

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