6(z-(10z+39)+3)=4(z+4)

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Solution for 6(z-(10z+39)+3)=4(z+4) equation: 



6(z-(10z+39)+3)=4(z+4)

We move all terms to the left:

6(z-(10z+39)+3)-(4(z+4))=0



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

4(z+4)

We multiply parentheses

4z+16

Back to the equation:

-(4z+16)



We get rid of parentheses

6(z-(10z+39)+3)-4z-16=0

We add all the numbers together, and all the variables

-4z+6(z-(10z+39)+3)-16=0

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

-4z+6(z-(10z+39)+3)=16

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