13-2(z-5)=(5+z)+2z-1

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Solution for 13-2(z-5)=(5+z)+2z-1 equation: 



13-2(z-5)=(5+z)+2z-1

We move all terms to the left:

13-2(z-5)-((5+z)+2z-1)=0

We add all the numbers together, and all the variables

-2(z-5)-((z+5)+2z-1)+13=0

We multiply parentheses

-2z-((z+5)+2z-1)+10+13=0



We calculate terms in parentheses: -((z+5)+2z-1), so:

(z+5)+2z-1

We add all the numbers together, and all the variables

2z+(z+5)-1

We get rid of parentheses

2z+z+5-1

We add all the numbers together, and all the variables

3z+4

Back to the equation:

-(3z+4)



We add all the numbers together, and all the variables

-2z-(3z+4)+23=0

We get rid of parentheses

-2z-3z-4+23=0

We add all the numbers together, and all the variables

-5z+19=0

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

-5z=-19

z=-19/-5

z=3+4/5

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