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3(c+5)=12+3(c+4)
We move all terms to the left:
3(c+5)-(12+3(c+4))=0
We multiply parentheses
3c-(12+3(c+4))+15=0
We calculate terms in parentheses: -(12+3(c+4)), so:We get rid of parentheses
12+3(c+4)
determiningTheFunctionDomain 3(c+4)+12
We multiply parentheses
3c+12+12
We add all the numbers together, and all the variables
3c+24
Back to the equation:
-(3c+24)
3c-3c-24+15=0
We add all the numbers together, and all the variables
-9!=0
There is no solution for this equation
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