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5(2c-3)=3(c-5)
We move all terms to the left:
5(2c-3)-(3(c-5))=0
We multiply parentheses
10c-(3(c-5))-15=0
We calculate terms in parentheses: -(3(c-5)), so:We get rid of parentheses
3(c-5)
We multiply parentheses
3c-15
Back to the equation:
-(3c-15)
10c-3c+15-15=0
We add all the numbers together, and all the variables
7c=0
c=0/7
c=0
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