4z+9+3(22)=129

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Solution for 4z+9+3(22)=129 equation:



4z+9+3(22)=129
We move all terms to the left:
4z+9+3(22)-(129)=0
We add all the numbers together, and all the variables
4z+202=0
We move all terms containing z to the left, all other terms to the right
4z=-202
z=-202/4
z=-50+1/2

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