40-(3c+4)=2(c+3)+6

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Solution for 40-(3c+4)=2(c+3)+6 equation: 



40-(3c+4)=2(c+3)+6

We move all terms to the left:

40-(3c+4)-(2(c+3)+6)=0

We get rid of parentheses

-3c-(2(c+3)+6)-4+40=0



We calculate terms in parentheses: -(2(c+3)+6), so:

2(c+3)+6

We multiply parentheses

2c+6+6

We add all the numbers together, and all the variables

2c+12

Back to the equation:

-(2c+12)



We add all the numbers together, and all the variables

-3c-(2c+12)+36=0

We get rid of parentheses

-3c-2c-12+36=0

We add all the numbers together, and all the variables

-5c+24=0

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

-5c=-24

c=-24/-5

c=4+4/5

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