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

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Solution for 5(c-2)+c=6(c+3)-4c equation: 



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

We move all terms to the left:

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

We add all the numbers together, and all the variables

c+5(c-2)-(6(c+3)-4c)=0

We multiply parentheses

c+5c-(6(c+3)-4c)-10=0



We calculate terms in parentheses: -(6(c+3)-4c), so:

6(c+3)-4c

We add all the numbers together, and all the variables

-4c+6(c+3)

We multiply parentheses

-4c+6c+18

We add all the numbers together, and all the variables

2c+18

Back to the equation:

-(2c+18)



We add all the numbers together, and all the variables

6c-(2c+18)-10=0

We get rid of parentheses

6c-2c-18-10=0

We add all the numbers together, and all the variables

4c-28=0

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

4c=28

c=28/4

c=7

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