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

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

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

We move all terms to the left:

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

We multiply parentheses

10z-3z-(4(z-3)+2(z-1))+40+6=0


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

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

We multiply parentheses

4z+2z-12-2

We add all the numbers together, and all the variables

6z-14

Back to the equation:

-(6z-14)


We add all the numbers together, and all the variables

7z-(6z-14)+46=0

We get rid of parentheses

7z-6z+14+46=0

We add all the numbers together, and all the variables

z+60=0

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

z=-60