8(z+2)-2(2-4)=2(z-3)+3(z-4)

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Solution for 8(z+2)-2(2-4)=2(z-3)+3(z-4) equation: 



8(z+2)-2(2-4)=2(z-3)+3(z-4)

We move all terms to the left:

8(z+2)-2(2-4)-(2(z-3)+3(z-4))=0

We add all the numbers together, and all the variables

8(z+2)-(2(z-3)+3(z-4))-2(-2)=0

We add all the numbers together, and all the variables

8(z+2)-(2(z-3)+3(z-4))+4=0

We multiply parentheses

8z-(2(z-3)+3(z-4))+16+4=0



We calculate terms in parentheses: -(2(z-3)+3(z-4)), so:

2(z-3)+3(z-4)

We multiply parentheses

2z+3z-6-12

We add all the numbers together, and all the variables

5z-18

Back to the equation:

-(5z-18)



We add all the numbers together, and all the variables

8z-(5z-18)+20=0

We get rid of parentheses

8z-5z+18+20=0

We add all the numbers together, and all the variables

3z+38=0

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

3z=-38

z=-38/3

z=-12+2/3

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