(z-20)+z=2(-z+56)

Solution for (z-20)+z=2(-z+56) equation:

(z-20)+z=2(-z+56)

We move all terms to the left:

(z-20)+z-(2(-z+56))=0

We add all the numbers together, and all the variables

(z-20)+z-(2(-1z+56))=0

We add all the numbers together, and all the variables

z+(z-20)-(2(-1z+56))=0

We get rid of parentheses

z+z-(2(-1z+56))-20=0


We calculate terms in parentheses: -(2(-1z+56)), so:

2(-1z+56)

We multiply parentheses

-2z+112

Back to the equation:

-(-2z+112)


We add all the numbers together, and all the variables

2z-(-2z+112)-20=0

We get rid of parentheses

2z+2z-112-20=0

We add all the numbers together, and all the variables

4z-132=0

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

4z=132

z=132/4

z=33