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

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



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

We move all terms to the left:

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

We multiply parentheses

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



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

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

We multiply parentheses

3z+3z-12-9

We add all the numbers together, and all the variables

6z-21

Back to the equation:

-(6z-21)



We add all the numbers together, and all the variables

7z-(6z-21)+36=0

We get rid of parentheses

7z-6z+21+36=0

We add all the numbers together, and all the variables

z+57=0

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

z=-57

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