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

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



10(z+3)-4(z-2)=2(z-4)+3(z-4)
We move all terms to the left:
10(z+3)-4(z-2)-(2(z-4)+3(z-4))=0
We multiply parentheses
10z-4z-(2(z-4)+3(z-4))+30+8=0
We calculate terms in parentheses: -(2(z-4)+3(z-4)), so:
2(z-4)+3(z-4)
We multiply parentheses
2z+3z-8-12
We add all the numbers together, and all the variables
5z-20
Back to the equation:
-(5z-20)
We add all the numbers together, and all the variables
6z-(5z-20)+38=0
We get rid of parentheses
6z-5z+20+38=0
We add all the numbers together, and all the variables
z+58=0
We move all terms containing z to the left, all other terms to the right
z=-58

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