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

Solution for 9(z+4)-4(z-4)=2(z-3)+2(z-3) equation:

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

We move all terms to the left:

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

We multiply parentheses

9z-4z-(2(z-3)+2(z-3))+36+16=0


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

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

We multiply parentheses

2z+2z-6-6

We add all the numbers together, and all the variables

4z-12

Back to the equation:

-(4z-12)


We add all the numbers together, and all the variables

5z-(4z-12)+52=0

We get rid of parentheses

5z-4z+12+52=0

We add all the numbers together, and all the variables

z+64=0

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

z=-64