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

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



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

We move all terms to the left:

3(3-c)/2-(-2(2c+5)/3)=0

We add all the numbers together, and all the variables

3(-1c+3)/2-(-2(2c+5)/3)=0

We calculate fractions


(-3c)/()+(-(-2(2c+5)*2)/()=0



We calculate terms in parentheses: +(-(-2(2c+5)*2)/(), so:

-(-2(2c+5)*2)/(

We multiply all the terms by the denominator


-(-2(2c+5)*2)



We calculate terms in parentheses: -(-2(2c+5)*2), so:

-2(2c+5)*2

We multiply parentheses

-8c-20

Back to the equation:

-(-8c-20)



We get rid of parentheses

8c+20

Back to the equation:

+(8c+20)



We get rid of parentheses

(-3c)/()+8c+20=0

We multiply all the terms by the denominator


(-3c)+8c*()+20*()=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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