11(z+3)-4(z-2)=4(z-3)+2(z-1)

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Solution for 11(z+3)-4(z-2)=4(z-3)+2(z-1) equation: 



11(z+3)-4(z-2)=4(z-3)+2(z-1)

We move all terms to the left:

11(z+3)-4(z-2)-(4(z-3)+2(z-1))=0

We multiply parentheses

11z-4z-(4(z-3)+2(z-1))+33+8=0



We calculate terms in parentheses: -(4(z-3)+2(z-1)), so:

4(z-3)+2(z-1)

We multiply parentheses

4z+2z-12-2

We add all the numbers together, and all the variables

6z-14

Back to the equation:

-(6z-14)



We add all the numbers together, and all the variables

7z-(6z-14)+41=0

We get rid of parentheses

7z-6z+14+41=0

We add all the numbers together, and all the variables

z+55=0

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

z=-55

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