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

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



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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

3z-(3-(z+4))-15=0



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

3-(z+4)

determiningTheFunctionDomain
-(z+4)+3

We get rid of parentheses

-z-4+3

We add all the numbers together, and all the variables

-1z-1

Back to the equation:

-(-1z-1)



We get rid of parentheses

3z+1z+1-15=0

We add all the numbers together, and all the variables

4z-14=0

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

4z=14

z=14/4

z=3+1/2

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