6(z-2)-5z=4(z+4)-4(z-1)

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Solution for 6(z-2)-5z=4(z+4)-4(z-1) equation: 



6(z-2)-5z=4(z+4)-4(z-1)

We move all terms to the left:

6(z-2)-5z-(4(z+4)-4(z-1))=0

We add all the numbers together, and all the variables

-5z+6(z-2)-(4(z+4)-4(z-1))=0

We multiply parentheses

-5z+6z-(4(z+4)-4(z-1))-12=0



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

4(z+4)-4(z-1)

We multiply parentheses

4z-4z+16+4

We add all the numbers together, and all the variables

20

Back to the equation:

-(20)



We add all the numbers together, and all the variables

z-32=0

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

z=32

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