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

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



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

We move all terms to the left:

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

We multiply parentheses

10z-2z-(4(z-4)+3(z-4))+10+8=0



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

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

We multiply parentheses

4z+3z-16-12

We add all the numbers together, and all the variables

7z-28

Back to the equation:

-(7z-28)



We add all the numbers together, and all the variables

8z-(7z-28)+18=0

We get rid of parentheses

8z-7z+28+18=0

We add all the numbers together, and all the variables

z+46=0

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

z=-46

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