13(z+3)-2(z-1)=4(z-1)+6(z-2)

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Solution for 13(z+3)-2(z-1)=4(z-1)+6(z-2) equation: 



13(z+3)-2(z-1)=4(z-1)+6(z-2)

We move all terms to the left:

13(z+3)-2(z-1)-(4(z-1)+6(z-2))=0

We multiply parentheses

13z-2z-(4(z-1)+6(z-2))+39+2=0



We calculate terms in parentheses: -(4(z-1)+6(z-2)), so:

4(z-1)+6(z-2)

We multiply parentheses

4z+6z-4-12

We add all the numbers together, and all the variables

10z-16

Back to the equation:

-(10z-16)



We add all the numbers together, and all the variables

11z-(10z-16)+41=0

We get rid of parentheses

11z-10z+16+41=0

We add all the numbers together, and all the variables

z+57=0

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

z=-57

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