3(z-4)=2(z-5)+z

Solution for 3(z-4)=2(z-5)+z equation:

3(z-4)=2(z-5)+z

We move all terms to the left:

3(z-4)-(2(z-5)+z)=0

We multiply parentheses

3z-(2(z-5)+z)-12=0


We calculate terms in parentheses: -(2(z-5)+z), so:

2(z-5)+z

We add all the numbers together, and all the variables

z+2(z-5)

We multiply parentheses

z+2z-10

We add all the numbers together, and all the variables

3z-10

Back to the equation:

-(3z-10)


We get rid of parentheses

3z-3z+10-12=0

We add all the numbers together, and all the variables

-2!=0

There is no solution for this equation