6(3c-2)+3(3-2c)=100

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Solution for 6(3c-2)+3(3-2c)=100 equation: 



6(3c-2)+3(3-2c)=100

We move all terms to the left:

6(3c-2)+3(3-2c)-(100)=0

We add all the numbers together, and all the variables

6(3c-2)+3(-2c+3)-100=0

We multiply parentheses

18c-6c-12+9-100=0

We add all the numbers together, and all the variables

12c-103=0

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

12c=103

c=103/12

c=8+7/12

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