3(6-c)/2=-2(2c+1)/3

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Solution for 3(6-c)/2=-2(2c+1)/3 equation:



3(6-c)/2=-2(2c+1)/3
We move all terms to the left:
3(6-c)/2-(-2(2c+1)/3)=0
We add all the numbers together, and all the variables
3(-1c+6)/2-(-2(2c+1)/3)=0
We calculate fractions
(-3c)/()+(-(-2(2c+1)*2)/()=0
We calculate terms in parentheses: +(-(-2(2c+1)*2)/(), so:
-(-2(2c+1)*2)/(
We multiply all the terms by the denominator
-(-2(2c+1)*2)
We calculate terms in parentheses: -(-2(2c+1)*2), so:
-2(2c+1)*2
We multiply parentheses
-8c-4
Back to the equation:
-(-8c-4)
We get rid of parentheses
8c+4
Back to the equation:
+(8c+4)
We get rid of parentheses
(-3c)/()+8c+4=0
We multiply all the terms by the denominator
(-3c)+8c*()+4*()=0
We add all the numbers together, and all the variables
(-3c)+8c*()=0
We get rid of parentheses
-3c+8c*()=0

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