7(5-c)/2=-4(2c+5)/3

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Solution for 7(5-c)/2=-4(2c+5)/3 equation: 



7(5-c)/2=-4(2c+5)/3

We move all terms to the left:

7(5-c)/2-(-4(2c+5)/3)=0

We add all the numbers together, and all the variables

7(-1c+5)/2-(-4(2c+5)/3)=0

We calculate fractions


(-7c)/()+(-(-4(2c+5)*2)/()=0



We calculate terms in parentheses: +(-(-4(2c+5)*2)/(), so:

-(-4(2c+5)*2)/(

We multiply all the terms by the denominator


-(-4(2c+5)*2)



We calculate terms in parentheses: -(-4(2c+5)*2), so:

-4(2c+5)*2

We multiply parentheses

-16c-40

Back to the equation:

-(-16c-40)



We get rid of parentheses

16c+40

Back to the equation:

+(16c+40)



We get rid of parentheses

(-7c)/()+16c+40=0

We multiply all the terms by the denominator


(-7c)+16c*()+40*()=0

We add all the numbers together, and all the variables

(-7c)+16c*()=0

We get rid of parentheses

-7c+16c*()=0

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