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

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



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

We move all terms to the left:

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

We multiply parentheses

10z-2z-(3(z-4)+4(z-1))+20+4=0



We calculate terms in parentheses: -(3(z-4)+4(z-1)), so:

3(z-4)+4(z-1)

We multiply parentheses

3z+4z-12-4

We add all the numbers together, and all the variables

7z-16

Back to the equation:

-(7z-16)



We add all the numbers together, and all the variables

8z-(7z-16)+24=0

We get rid of parentheses

8z-7z+16+24=0

We add all the numbers together, and all the variables

z+40=0

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

z=-40

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