3(7-z)-29=21z-4(2+6z)

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Solution for 3(7-z)-29=21z-4(2+6z) equation: 



3(7-z)-29=21z-4(2+6z)

We move all terms to the left:

3(7-z)-29-(21z-4(2+6z))=0

We add all the numbers together, and all the variables

3(-1z+7)-(21z-4(6z+2))-29=0

We multiply parentheses

-3z-(21z-4(6z+2))+21-29=0



We calculate terms in parentheses: -(21z-4(6z+2)), so:

21z-4(6z+2)

We multiply parentheses

21z-24z-8

We add all the numbers together, and all the variables

-3z-8

Back to the equation:

-(-3z-8)



We add all the numbers together, and all the variables

-3z-(-3z-8)-8=0

We get rid of parentheses

-3z+3z+8-8=0

We add all the numbers together, and all the variables

=0

z=0/1

z=0

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