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

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

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

We move all terms to the left:

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

We multiply parentheses

10z-4z-(2(z-1)+3(z-2))+40+4=0


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

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

We multiply parentheses

2z+3z-2-6

We add all the numbers together, and all the variables

5z-8

Back to the equation:

-(5z-8)


We add all the numbers together, and all the variables

6z-(5z-8)+44=0

We get rid of parentheses

6z-5z+8+44=0

We add all the numbers together, and all the variables

z+52=0

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

z=-52