15(z+3)-4(z-3)=4(z-2)+6(z-3)

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Solution for 15(z+3)-4(z-3)=4(z-2)+6(z-3) equation: 



15(z+3)-4(z-3)=4(z-2)+6(z-3)

We move all terms to the left:

15(z+3)-4(z-3)-(4(z-2)+6(z-3))=0

We multiply parentheses

15z-4z-(4(z-2)+6(z-3))+45+12=0



We calculate terms in parentheses: -(4(z-2)+6(z-3)), so:

4(z-2)+6(z-3)

We multiply parentheses

4z+6z-8-18

We add all the numbers together, and all the variables

10z-26

Back to the equation:

-(10z-26)



We add all the numbers together, and all the variables

11z-(10z-26)+57=0

We get rid of parentheses

11z-10z+26+57=0

We add all the numbers together, and all the variables

z+83=0

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

z=-83

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