3(c+5)-2+4c=7(c+2)

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Solution for 3(c+5)-2+4c=7(c+2) equation: 



3(c+5)-2+4c=7(c+2)

We move all terms to the left:

3(c+5)-2+4c-(7(c+2))=0

We add all the numbers together, and all the variables

4c+3(c+5)-(7(c+2))-2=0

We multiply parentheses

4c+3c-(7(c+2))+15-2=0



We calculate terms in parentheses: -(7(c+2)), so:

7(c+2)

We multiply parentheses

7c+14

Back to the equation:

-(7c+14)



We add all the numbers together, and all the variables

7c-(7c+14)+13=0

We get rid of parentheses

7c-7c-14+13=0

We add all the numbers together, and all the variables

-1!=0

There is no solution for this equation

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