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

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



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

We move all terms to the left:

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

We multiply parentheses

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



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

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

We multiply parentheses

4z+3z-8-3

We add all the numbers together, and all the variables

7z-11

Back to the equation:

-(7z-11)



We add all the numbers together, and all the variables

8z-(7z-11)+32=0

We get rid of parentheses

8z-7z+11+32=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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