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

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

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

We move all terms to the left:

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

We multiply parentheses

10z-4z-(3(z-1)+2(z-3))+20+4=0


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

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

We multiply parentheses

3z+2z-3-6

We add all the numbers together, and all the variables

5z-9

Back to the equation:

-(5z-9)


We add all the numbers together, and all the variables

6z-(5z-9)+24=0

We get rid of parentheses

6z-5z+9+24=0

We add all the numbers together, and all the variables

z+33=0

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

z=-33