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

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Solution for 11(z+3)-4(z-1)=3(z-1)+3(z-2) equation:



11(z+3)-4(z-1)=3(z-1)+3(z-2)
We move all terms to the left:
11(z+3)-4(z-1)-(3(z-1)+3(z-2))=0
We multiply parentheses
11z-4z-(3(z-1)+3(z-2))+33+4=0
We calculate terms in parentheses: -(3(z-1)+3(z-2)), so:
3(z-1)+3(z-2)
We multiply parentheses
3z+3z-3-6
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)+37=0
We get rid of parentheses
7z-6z+9+37=0
We add all the numbers together, and all the variables
z+46=0
We move all terms containing z to the left, all other terms to the right
z=-46

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