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

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



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

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