41-1(2c+3)=4(1c+6)+1c

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Solution for 41-1(2c+3)=4(1c+6)+1c equation: 



41-1(2c+3)=4(1c+6)+1c

We move all terms to the left:

41-1(2c+3)-(4(1c+6)+1c)=0

We add all the numbers together, and all the variables

-1(2c+3)-(4(c+6)+1c)+41=0



We calculate terms in parentheses: -(4(c+6)+1c), so:

4(c+6)+1c

We add all the numbers together, and all the variables

c+4(c+6)

We multiply parentheses

c+4c+24

We add all the numbers together, and all the variables

5c+24

Back to the equation:

-(5c+24)



We get rid of parentheses

-1(2c+3)-5c-24+41=0

We add all the numbers together, and all the variables

-5c-1(2c+3)+17=0

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

-5c-1(2c+3)=-17

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