9(z+2)-4(z-4)=2(z-4)+2(z-4)

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Solution for 9(z+2)-4(z-4)=2(z-4)+2(z-4) equation: 



9(z+2)-4(z-4)=2(z-4)+2(z-4)

We move all terms to the left:

9(z+2)-4(z-4)-(2(z-4)+2(z-4))=0

We multiply parentheses

9z-4z-(2(z-4)+2(z-4))+18+16=0



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

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

We multiply parentheses

2z+2z-8-8

We add all the numbers together, and all the variables

4z-16

Back to the equation:

-(4z-16)



We add all the numbers together, and all the variables

5z-(4z-16)+34=0

We get rid of parentheses

5z-4z+16+34=0

We add all the numbers together, and all the variables

z+50=0

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

z=-50

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