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

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



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

We move all terms to the left:

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

We multiply parentheses

10z-3z-(4(z-2)+2(z-4))+30+3=0



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

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

We multiply parentheses

4z+2z-8-8

We add all the numbers together, and all the variables

6z-16

Back to the equation:

-(6z-16)



We add all the numbers together, and all the variables

7z-(6z-16)+33=0

We get rid of parentheses

7z-6z+16+33=0

We add all the numbers together, and all the variables

z+49=0

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

z=-49

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