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

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



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

We move all terms to the left:

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

We multiply parentheses

13z-3z-(4(z-2)+5(z-2))+39+9=0



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

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

We multiply parentheses

4z+5z-8-10

We add all the numbers together, and all the variables

9z-18

Back to the equation:

-(9z-18)



We add all the numbers together, and all the variables

10z-(9z-18)+48=0

We get rid of parentheses

10z-9z+18+48=0

We add all the numbers together, and all the variables

z+66=0

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

z=-66

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