3(2-c)/2=-2(2c+3)/5

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



3(2-c)/2=-2(2c+3)/5
We move all terms to the left:
3(2-c)/2-(-2(2c+3)/5)=0
We add all the numbers together, and all the variables
3(-1c+2)/2-(-2(2c+3)/5)=0
We calculate fractions
(-3c)/()+(-(-2(2c+3)*2)/()=0
We calculate terms in parentheses: +(-(-2(2c+3)*2)/(), so:
-(-2(2c+3)*2)/(
We multiply all the terms by the denominator
-(-2(2c+3)*2)
We calculate terms in parentheses: -(-2(2c+3)*2), so:
-2(2c+3)*2
We multiply parentheses
-8c-12
Back to the equation:
-(-8c-12)
We get rid of parentheses
8c+12
Back to the equation:
+(8c+12)
We get rid of parentheses
(-3c)/()+8c+12=0
We multiply all the terms by the denominator
(-3c)+8c*()+12*()=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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