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

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

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

We move all terms to the left:

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

We multiply parentheses

10z-4z-(3(z-4)+2(z-2))+40+16=0


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

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

We multiply parentheses

3z+2z-12-4

We add all the numbers together, and all the variables

5z-16

Back to the equation:

-(5z-16)


We add all the numbers together, and all the variables

6z-(5z-16)+56=0

We get rid of parentheses

6z-5z+16+56=0

We add all the numbers together, and all the variables

z+72=0

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

z=-72