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

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

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

We move all terms to the left:

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

We multiply parentheses

10z-4z-(3(z-1)+2(z-1))+10+16=0


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

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

We multiply parentheses

3z+2z-3-2

We add all the numbers together, and all the variables

5z-5

Back to the equation:

-(5z-5)


We add all the numbers together, and all the variables

6z-(5z-5)+26=0

We get rid of parentheses

6z-5z+5+26=0

We add all the numbers together, and all the variables

z+31=0

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

z=-31