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

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

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

We move all terms to the left:

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

We multiply parentheses

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


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

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

We multiply parentheses

4z+2z-12-4

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)+16=0

We get rid of parentheses

7z-6z+16+16=0

We add all the numbers together, and all the variables

z+32=0

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

z=-32