7-5(z-6)+4=3-2(z-5)-3z+28

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Solution for 7-5(z-6)+4=3-2(z-5)-3z+28 equation: 



7-5(z-6)+4=3-2(z-5)-3z+28

We move all terms to the left:

7-5(z-6)+4-(3-2(z-5)-3z+28)=0

We add all the numbers together, and all the variables

-5(z-6)-(3-2(z-5)-3z+28)+11=0

We multiply parentheses

-5z-(3-2(z-5)-3z+28)+30+11=0



We calculate terms in parentheses: -(3-2(z-5)-3z+28), so:

3-2(z-5)-3z+28

determiningTheFunctionDomain
-2(z-5)-3z+3+28

We add all the numbers together, and all the variables

-3z-2(z-5)+31

We multiply parentheses

-3z-2z+10+31

We add all the numbers together, and all the variables

-5z+41

Back to the equation:

-(-5z+41)



We add all the numbers together, and all the variables

-5z-(-5z+41)+41=0

We get rid of parentheses

-5z+5z-41+41=0

We add all the numbers together, and all the variables

=0

z=0/1

z=0

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