-5+4(2z+3)=23+2(3z-3)

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Solution for -5+4(2z+3)=23+2(3z-3) equation: 



-5+4(2z+3)=23+2(3z-3)

We move all terms to the left:

-5+4(2z+3)-(23+2(3z-3))=0

We multiply parentheses

8z-(23+2(3z-3))+12-5=0



We calculate terms in parentheses: -(23+2(3z-3)), so:

23+2(3z-3)

determiningTheFunctionDomain
2(3z-3)+23

We multiply parentheses

6z-6+23

We add all the numbers together, and all the variables

6z+17

Back to the equation:

-(6z+17)



We add all the numbers together, and all the variables

8z-(6z+17)+7=0

We get rid of parentheses

8z-6z-17+7=0

We add all the numbers together, and all the variables

2z-10=0

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

2z=10

z=10/2

z=5

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