14(z+3)-4(z-2)=4(z-2)+5(z-2)

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Solution for 14(z+3)-4(z-2)=4(z-2)+5(z-2) equation:



14(z+3)-4(z-2)=4(z-2)+5(z-2)
We move all terms to the left:
14(z+3)-4(z-2)-(4(z-2)+5(z-2))=0
We multiply parentheses
14z-4z-(4(z-2)+5(z-2))+42+8=0
We calculate terms in parentheses: -(4(z-2)+5(z-2)), so:
4(z-2)+5(z-2)
We multiply parentheses
4z+5z-8-10
We add all the numbers together, and all the variables
9z-18
Back to the equation:
-(9z-18)
We add all the numbers together, and all the variables
10z-(9z-18)+50=0
We get rid of parentheses
10z-9z+18+50=0
We add all the numbers together, and all the variables
z+68=0
We move all terms containing z to the left, all other terms to the right
z=-68

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