4(2z-3)+2(7z+1)+4=3(5z-1)+4

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Solution for 4(2z-3)+2(7z+1)+4=3(5z-1)+4 equation: 



4(2z-3)+2(7z+1)+4=3(5z-1)+4

We move all terms to the left:

4(2z-3)+2(7z+1)+4-(3(5z-1)+4)=0

We multiply parentheses

8z+14z-(3(5z-1)+4)-12+2+4=0



We calculate terms in parentheses: -(3(5z-1)+4), so:

3(5z-1)+4

We multiply parentheses

15z-3+4

We add all the numbers together, and all the variables

15z+1

Back to the equation:

-(15z+1)



We add all the numbers together, and all the variables

22z-(15z+1)-6=0

We get rid of parentheses

22z-15z-1-6=0

We add all the numbers together, and all the variables

7z-7=0

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

7z=7

z=7/7

z=1

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