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

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



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

We move all terms to the left:

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

We add all the numbers together, and all the variables

5(z+4)-(+5(-1z+2))=0

We multiply parentheses

5z-(+5(-1z+2))+20=0



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

+5(-1z+2)

determiningTheFunctionDomain
5(-1z+2)

We multiply parentheses

-5z+10

Back to the equation:

-(-5z+10)



We get rid of parentheses

5z+5z-10+20=0

We add all the numbers together, and all the variables

10z+10=0

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

10z=-10

z=-10/10

z=-1

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