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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