11(z+1)+6(3z-7)=6(z-3)+14

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Solution for 11(z+1)+6(3z-7)=6(z-3)+14 equation: 



11(z+1)+6(3z-7)=6(z-3)+14

We move all terms to the left:

11(z+1)+6(3z-7)-(6(z-3)+14)=0

We multiply parentheses

11z+18z-(6(z-3)+14)+11-42=0



We calculate terms in parentheses: -(6(z-3)+14), so:

6(z-3)+14

We multiply parentheses

6z-18+14

We add all the numbers together, and all the variables

6z-4

Back to the equation:

-(6z-4)



We add all the numbers together, and all the variables

29z-(6z-4)-31=0

We get rid of parentheses

29z-6z+4-31=0

We add all the numbers together, and all the variables

23z-27=0

We move all terms containing z to the left, all other terms to the right

23z=27

z=27/23

z=1+4/23

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