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

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



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

We move all terms to the left:

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

We multiply parentheses

11z-4z-(2(z+4)+4(z+1))+11+16=0



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

2(z+4)+4(z+1)

We multiply parentheses

2z+4z+8+4

We add all the numbers together, and all the variables

6z+12

Back to the equation:

-(6z+12)



We add all the numbers together, and all the variables

7z-(6z+12)+27=0

We get rid of parentheses

7z-6z-12+27=0

We add all the numbers together, and all the variables

z+15=0

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

z=-15

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