6(b+1)-7(3b+2)=10+0,5(-20b+6)

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Solution for 6(b+1)-7(3b+2)=10+0,5(-20b+6) equation: 



6(b+1)-7(3b+2)=10+0.5(-20b+6)

We move all terms to the left:

6(b+1)-7(3b+2)-(10+0.5(-20b+6))=0

We multiply parentheses

6b-21b-(10+0.5(-20b+6))+6-14=0



We calculate terms in parentheses: -(10+0.5(-20b+6)), so:

10+0.5(-20b+6)

determiningTheFunctionDomain
0.5(-20b+6)+10

We multiply parentheses

-10b+3+10

We add all the numbers together, and all the variables

-10b+13

Back to the equation:

-(-10b+13)



We add all the numbers together, and all the variables

-15b-(-10b+13)-8=0

We get rid of parentheses

-15b+10b-13-8=0

We add all the numbers together, and all the variables

-5b-21=0

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

-5b=21

b=21/-5

b=-4+1/5

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