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

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



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

We move all terms to the left:

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

We multiply parentheses

10z-3z-(2(z-3)+4(z-1))+30+6=0



We calculate terms in parentheses: -(2(z-3)+4(z-1)), so:

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

We multiply parentheses

2z+4z-6-4

We add all the numbers together, and all the variables

6z-10

Back to the equation:

-(6z-10)



We add all the numbers together, and all the variables

7z-(6z-10)+36=0

We get rid of parentheses

7z-6z+10+36=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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