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

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



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

We move all terms to the left:

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

We multiply parentheses

10z-3z-(3(z-2)+3(z-1))+20+9=0



We calculate terms in parentheses: -(3(z-2)+3(z-1)), so:

3(z-2)+3(z-1)

We multiply parentheses

3z+3z-6-3

We add all the numbers together, and all the variables

6z-9

Back to the equation:

-(6z-9)



We add all the numbers together, and all the variables

7z-(6z-9)+29=0

We get rid of parentheses

7z-6z+9+29=0

We add all the numbers together, and all the variables

z+38=0

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

z=-38

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