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

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



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

We move all terms to the left:

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

We multiply parentheses

8z-2z-(3(z-4)+2(z-3))+24+8=0



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

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

We multiply parentheses

3z+2z-12-6

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

6z-(5z-18)+32=0

We get rid of parentheses

6z-5z+18+32=0

We add all the numbers together, and all the variables

z+50=0

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

z=-50

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