10(z+3)-3(z-3)=4(z-1)+2(z-2)

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Solution for 10(z+3)-3(z-3)=4(z-1)+2(z-2) equation: 



10(z+3)-3(z-3)=4(z-1)+2(z-2)

We move all terms to the left:

10(z+3)-3(z-3)-(4(z-1)+2(z-2))=0

We multiply parentheses

10z-3z-(4(z-1)+2(z-2))+30+9=0



We calculate terms in parentheses: -(4(z-1)+2(z-2)), so:

4(z-1)+2(z-2)

We multiply parentheses

4z+2z-4-4

We add all the numbers together, and all the variables

6z-8

Back to the equation:

-(6z-8)



We add all the numbers together, and all the variables

7z-(6z-8)+39=0

We get rid of parentheses

7z-6z+8+39=0

We add all the numbers together, and all the variables

z+47=0

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

z=-47

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