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

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

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

We move all terms to the left:

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

We multiply parentheses

10z-2z-(4(z-3)+3(z-4))+10+6=0


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

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

We multiply parentheses

4z+3z-12-12

We add all the numbers together, and all the variables

7z-24

Back to the equation:

-(7z-24)


We add all the numbers together, and all the variables

8z-(7z-24)+16=0

We get rid of parentheses

8z-7z+24+16=0

We add all the numbers together, and all the variables

z+40=0

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

z=-40