(7z-5)+(3z+)=29+5z

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Solution for (7z-5)+(3z+)=29+5z equation: 



(7z-5)+(3z+)=29+5z

We move all terms to the left:

(7z-5)+(3z+)-(29+5z)=0

We add all the numbers together, and all the variables

(7z-5)+(+3z)-(5z+29)=0

We get rid of parentheses

7z+3z-5z-5-29=0

We add all the numbers together, and all the variables

5z-34=0

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

5z=34

z=34/5

z=6+4/5

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