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

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



11(z+4)-3(z-3)=3(z-2)+4(z-1)
We move all terms to the left:
11(z+4)-3(z-3)-(3(z-2)+4(z-1))=0
We multiply parentheses
11z-3z-(3(z-2)+4(z-1))+44+9=0
We calculate terms in parentheses: -(3(z-2)+4(z-1)), so:
3(z-2)+4(z-1)
We multiply parentheses
3z+4z-6-4
We add all the numbers together, and all the variables
7z-10
Back to the equation:
-(7z-10)
We add all the numbers together, and all the variables
8z-(7z-10)+53=0
We get rid of parentheses
8z-7z+10+53=0
We add all the numbers together, and all the variables
z+63=0
We move all terms containing z to the left, all other terms to the right
z=-63

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