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

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Solution for 13(z+4)-3(z-3)=5(z-2)+4(z-1) equation: 



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

We move all terms to the left:

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

We multiply parentheses

13z-3z-(5(z-2)+4(z-1))+52+9=0



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

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

We multiply parentheses

5z+4z-10-4

We add all the numbers together, and all the variables

9z-14

Back to the equation:

-(9z-14)



We add all the numbers together, and all the variables

10z-(9z-14)+61=0

We get rid of parentheses

10z-9z+14+61=0

We add all the numbers together, and all the variables

z+75=0

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

z=-75

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