3(6-c)/2=-4(2c+3)/7

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Solution for 3(6-c)/2=-4(2c+3)/7 equation: 



3(6-c)/2=-4(2c+3)/7

We move all terms to the left:

3(6-c)/2-(-4(2c+3)/7)=0

We add all the numbers together, and all the variables

3(-1c+6)/2-(-4(2c+3)/7)=0

We calculate fractions


(-3c)/()+(-(-4(2c+3)*2)/()=0



We calculate terms in parentheses: +(-(-4(2c+3)*2)/(), so:

-(-4(2c+3)*2)/(

We multiply all the terms by the denominator


-(-4(2c+3)*2)



We calculate terms in parentheses: -(-4(2c+3)*2), so:

-4(2c+3)*2

We multiply parentheses

-16c-24

Back to the equation:

-(-16c-24)



We get rid of parentheses

16c+24

Back to the equation:

+(16c+24)



We get rid of parentheses

(-3c)/()+16c+24=0

We multiply all the terms by the denominator


(-3c)+16c*()+24*()=0

We add all the numbers together, and all the variables

(-3c)+16c*()=0

We get rid of parentheses

-3c+16c*()=0

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