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

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



10(z+1)-3(z-2)=4(z-3)+2(z-2)
We move all terms to the left:
10(z+1)-3(z-2)-(4(z-3)+2(z-2))=0
We multiply parentheses
10z-3z-(4(z-3)+2(z-2))+10+6=0
We calculate terms in parentheses: -(4(z-3)+2(z-2)), so:
4(z-3)+2(z-2)
We multiply parentheses
4z+2z-12-4
We add all the numbers together, and all the variables
6z-16
Back to the equation:
-(6z-16)
We add all the numbers together, and all the variables
7z-(6z-16)+16=0
We get rid of parentheses
7z-6z+16+16=0
We add all the numbers together, and all the variables
z+32=0
We move all terms containing z to the left, all other terms to the right
z=-32

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