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

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

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

We move all terms to the left:

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

We multiply parentheses

10z-4z-(3(z-3)+2(z-3))+20+12=0


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

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

We multiply parentheses

3z+2z-9-6

We add all the numbers together, and all the variables

5z-15

Back to the equation:

-(5z-15)


We add all the numbers together, and all the variables

6z-(5z-15)+32=0

We get rid of parentheses

6z-5z+15+32=0

We add all the numbers together, and all the variables

z+47=0

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

z=-47