4(3z-7)+3(5z+1)=109

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Solution for 4(3z-7)+3(5z+1)=109 equation: 



4(3z-7)+3(5z+1)=109

We move all terms to the left:

4(3z-7)+3(5z+1)-(109)=0

We multiply parentheses

12z+15z-28+3-109=0

We add all the numbers together, and all the variables

27z-134=0

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

27z=134

z=134/27

z=4+26/27

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