7z+35=5(3z-1)

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Solution for 7z+35=5(3z-1) equation: 



7z+35=5(3z-1)

We move all terms to the left:

7z+35-(5(3z-1))=0



We calculate terms in parentheses: -(5(3z-1)), so:

5(3z-1)

We multiply parentheses

15z-5

Back to the equation:

-(15z-5)



We get rid of parentheses

7z-15z+5+35=0

We add all the numbers together, and all the variables

-8z+40=0

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

-8z=-40

z=-40/-8

z=+5

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