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

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



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

We move all terms to the left:

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

We multiply parentheses

11z-4z-(4(z-3)+2(z-4))+44+12=0



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

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

We multiply parentheses

4z+2z-12-8

We add all the numbers together, and all the variables

6z-20

Back to the equation:

-(6z-20)



We add all the numbers together, and all the variables

7z-(6z-20)+56=0

We get rid of parentheses

7z-6z+20+56=0

We add all the numbers together, and all the variables

z+76=0

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

z=-76

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