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

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Solution for 7(6-c)/2=-2(2c+1)/3 equation: 



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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We calculate fractions


(-7c)/()+(-(-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

(-7c)/()+8c+4=0

We multiply all the terms by the denominator


(-7c)+8c*()+4*()=0

We add all the numbers together, and all the variables

(-7c)+8c*()=0

We get rid of parentheses

-7c+8c*()=0

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