5z-(29-(4z-3)-(6-11))=29-(4z-(3z+2))

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Solution for 5z-(29-(4z-3)-(6-11))=29-(4z-(3z+2)) equation: 



5z-(29-(4z-3)-(6-11))=29-(4z-(3z+2))

We move all terms to the left:

5z-(29-(4z-3)-(6-11))-(29-(4z-(3z+2)))=0

We add all the numbers together, and all the variables

5z-(29-(4z-3)-(-5))-(29-(4z-(3z+2)))=0



We calculate terms in parentheses: -(29-(4z-3)-(-5)), so:

29-(4z-3)-(-5)

determiningTheFunctionDomain
-(4z-3)+29-(-5)

We add all the numbers together, and all the variables

-(4z-3)+34

We get rid of parentheses

-4z+3+34

We add all the numbers together, and all the variables

-4z+37

Back to the equation:

-(-4z+37)





We calculate terms in parentheses: -(29-(4z-(3z+2))), so:

29-(4z-(3z+2))

determiningTheFunctionDomain
-(4z-(3z+2))+29



We calculate terms in parentheses: -(4z-(3z+2)), so:

4z-(3z+2)

We get rid of parentheses

4z-3z-2

We add all the numbers together, and all the variables

z-2

Back to the equation:

-(z-2)



We get rid of parentheses

-z+2+29

We add all the numbers together, and all the variables

-1z+31

Back to the equation:

-(-1z+31)



We get rid of parentheses

5z+4z+1z-37-31=0

We add all the numbers together, and all the variables

10z-68=0

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

10z=68

z=68/10

z=6+4/5

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