15(z+2)-3(z-1)=7(z-1)+4(z-2)

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Solution for 15(z+2)-3(z-1)=7(z-1)+4(z-2) equation: 



15(z+2)-3(z-1)=7(z-1)+4(z-2)

We move all terms to the left:

15(z+2)-3(z-1)-(7(z-1)+4(z-2))=0

We multiply parentheses

15z-3z-(7(z-1)+4(z-2))+30+3=0



We calculate terms in parentheses: -(7(z-1)+4(z-2)), so:

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

We multiply parentheses

7z+4z-7-8

We add all the numbers together, and all the variables

11z-15

Back to the equation:

-(11z-15)



We add all the numbers together, and all the variables

12z-(11z-15)+33=0

We get rid of parentheses

12z-11z+15+33=0

We add all the numbers together, and all the variables

z+48=0

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

z=-48

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