14(z+1)-4(z-3)=4(z-2)+5(z-1)

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Solution for 14(z+1)-4(z-3)=4(z-2)+5(z-1) equation: 



14(z+1)-4(z-3)=4(z-2)+5(z-1)

We move all terms to the left:

14(z+1)-4(z-3)-(4(z-2)+5(z-1))=0

We multiply parentheses

14z-4z-(4(z-2)+5(z-1))+14+12=0



We calculate terms in parentheses: -(4(z-2)+5(z-1)), so:

4(z-2)+5(z-1)

We multiply parentheses

4z+5z-8-5

We add all the numbers together, and all the variables

9z-13

Back to the equation:

-(9z-13)



We add all the numbers together, and all the variables

10z-(9z-13)+26=0

We get rid of parentheses

10z-9z+13+26=0

We add all the numbers together, and all the variables

z+39=0

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

z=-39

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