7(z+1)-3z=5+4(2z-1)

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Solution for 7(z+1)-3z=5+4(2z-1) equation: 



7(z+1)-3z=5+4(2z-1)

We move all terms to the left:

7(z+1)-3z-(5+4(2z-1))=0

We add all the numbers together, and all the variables

-3z+7(z+1)-(5+4(2z-1))=0

We multiply parentheses

-3z+7z-(5+4(2z-1))+7=0



We calculate terms in parentheses: -(5+4(2z-1)), so:

5+4(2z-1)

determiningTheFunctionDomain
4(2z-1)+5

We multiply parentheses

8z-4+5

We add all the numbers together, and all the variables

8z+1

Back to the equation:

-(8z+1)



We add all the numbers together, and all the variables

4z-(8z+1)+7=0

We get rid of parentheses

4z-8z-1+7=0

We add all the numbers together, and all the variables

-4z+6=0

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

-4z=-6

z=-6/-4

z=1+1/2

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