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

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Solution for 10(z+1)-3(z-3)=2(z-4)+4(z-4) equation: 



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

We move all terms to the left:

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

We multiply parentheses

10z-3z-(2(z-4)+4(z-4))+10+9=0



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

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

We multiply parentheses

2z+4z-8-16

We add all the numbers together, and all the variables

6z-24

Back to the equation:

-(6z-24)



We add all the numbers together, and all the variables

7z-(6z-24)+19=0

We get rid of parentheses

7z-6z+24+19=0

We add all the numbers together, and all the variables

z+43=0

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

z=-43

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