6(z+1)=5(z-4)+z

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Solution for 6(z+1)=5(z-4)+z equation: 



6(z+1)=5(z-4)+z

We move all terms to the left:

6(z+1)-(5(z-4)+z)=0

We multiply parentheses

6z-(5(z-4)+z)+6=0



We calculate terms in parentheses: -(5(z-4)+z), so:

5(z-4)+z

We add all the numbers together, and all the variables

z+5(z-4)

We multiply parentheses

z+5z-20

We add all the numbers together, and all the variables

6z-20

Back to the equation:

-(6z-20)



We get rid of parentheses

6z-6z+20+6=0

We add all the numbers together, and all the variables

26!=0

There is no solution for this equation

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