9(1z+42)=3(6z-18)

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Solution for 9(1z+42)=3(6z-18) equation: 



9(1z+42)=3(6z-18)

We move all terms to the left:

9(1z+42)-(3(6z-18))=0

We add all the numbers together, and all the variables

9(z+42)-(3(6z-18))=0

We multiply parentheses

9z-(3(6z-18))+378=0



We calculate terms in parentheses: -(3(6z-18)), so:

3(6z-18)

We multiply parentheses

18z-54

Back to the equation:

-(18z-54)



We get rid of parentheses

9z-18z+54+378=0

We add all the numbers together, and all the variables

-9z+432=0

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

-9z=-432

z=-432/-9

z=+48

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