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

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



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

We move all terms to the left:

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

We multiply parentheses

14z-5z-(4(z-2)+4(z-1))+28+15=0



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

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

We multiply parentheses

4z+4z-8-4

We add all the numbers together, and all the variables

8z-12

Back to the equation:

-(8z-12)



We add all the numbers together, and all the variables

9z-(8z-12)+43=0

We get rid of parentheses

9z-8z+12+43=0

We add all the numbers together, and all the variables

z+55=0

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

z=-55

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