11(z+1)-3(z-4)=3(z-4)+4(z-4)

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Solution for 11(z+1)-3(z-4)=3(z-4)+4(z-4) equation: 



11(z+1)-3(z-4)=3(z-4)+4(z-4)

We move all terms to the left:

11(z+1)-3(z-4)-(3(z-4)+4(z-4))=0

We multiply parentheses

11z-3z-(3(z-4)+4(z-4))+11+12=0



We calculate terms in parentheses: -(3(z-4)+4(z-4)), so:

3(z-4)+4(z-4)

We multiply parentheses

3z+4z-12-16

We add all the numbers together, and all the variables

7z-28

Back to the equation:

-(7z-28)



We add all the numbers together, and all the variables

8z-(7z-28)+23=0

We get rid of parentheses

8z-7z+28+23=0

We add all the numbers together, and all the variables

z+51=0

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

z=-51

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