6(32-z)+4(z-5)=78

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Solution for 6(32-z)+4(z-5)=78 equation: 



6(32-z)+4(z-5)=78

We move all terms to the left:

6(32-z)+4(z-5)-(78)=0

We add all the numbers together, and all the variables

6(-1z+32)+4(z-5)-78=0

We multiply parentheses

-6z+4z+192-20-78=0

We add all the numbers together, and all the variables

-2z+94=0

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

-2z=-94

z=-94/-2

z=+47

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