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

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Solution for 10(z+3)-2(z-3)=3(z-4)+4(z-2) equation: 



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

We move all terms to the left:

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

We multiply parentheses

10z-2z-(3(z-4)+4(z-2))+30+6=0



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

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

We multiply parentheses

3z+4z-12-8

We add all the numbers together, and all the variables

7z-20

Back to the equation:

-(7z-20)



We add all the numbers together, and all the variables

8z-(7z-20)+36=0

We get rid of parentheses

8z-7z+20+36=0

We add all the numbers together, and all the variables

z+56=0

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

z=-56

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