3c+2(1c-6)+2=40

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Solution for 3c+2(1c-6)+2=40 equation: 



3c+2(1c-6)+2=40

We move all terms to the left:

3c+2(1c-6)+2-(40)=0

We add all the numbers together, and all the variables

3c+2(c-6)+2-40=0

We add all the numbers together, and all the variables

3c+2(c-6)-38=0

We multiply parentheses

3c+2c-12-38=0

We add all the numbers together, and all the variables

5c-50=0

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

5c=50

c=50/5

c=10

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