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

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

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

We move all terms to the left:

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

We multiply parentheses

10z-2z-(3(z-3)+4(z-1))+10+8=0


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

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

We multiply parentheses

3z+4z-9-4

We add all the numbers together, and all the variables

7z-13

Back to the equation:

-(7z-13)


We add all the numbers together, and all the variables

8z-(7z-13)+18=0

We get rid of parentheses

8z-7z+13+18=0

We add all the numbers together, and all the variables

z+31=0

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

z=-31