4(2z+3)=33+7z+15

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Solution for 4(2z+3)=33+7z+15 equation: 



4(2z+3)=33+7z+15

We move all terms to the left:

4(2z+3)-(33+7z+15)=0

We add all the numbers together, and all the variables

4(2z+3)-(7z+48)=0

We multiply parentheses

8z-(7z+48)+12=0

We get rid of parentheses

8z-7z-48+12=0

We add all the numbers together, and all the variables

z-36=0

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

z=36

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