3(c-10)+35=5(c+1)-2c

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Solution for 3(c-10)+35=5(c+1)-2c equation: 



3(c-10)+35=5(c+1)-2c

We move all terms to the left:

3(c-10)+35-(5(c+1)-2c)=0

We multiply parentheses

3c-(5(c+1)-2c)-30+35=0



We calculate terms in parentheses: -(5(c+1)-2c), so:

5(c+1)-2c

We add all the numbers together, and all the variables

-2c+5(c+1)

We multiply parentheses

-2c+5c+5

We add all the numbers together, and all the variables

3c+5

Back to the equation:

-(3c+5)



We add all the numbers together, and all the variables

3c-(3c+5)+5=0

We get rid of parentheses

3c-3c-5+5=0

We add all the numbers together, and all the variables

=0

c=0/1

c=0

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