(3-k)/4=(k+3)/2

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Solution for (3-k)/4=(k+3)/2 equation: 



(3-k)/4=(k+3)/2

We move all terms to the left:

(3-k)/4-((k+3)/2)=0

We add all the numbers together, and all the variables

(-1k+3)/4-((k+3)/2)=0

We calculate fractions


(-k)/()+(-((k+3)*4)/()=0



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

-((k+3)*4)/(

We multiply all the terms by the denominator


-((k+3)*4)



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

(k+3)*4

We multiply parentheses

4k+12

Back to the equation:

-(4k+12)



We get rid of parentheses

-4k-12

Back to the equation:

+(-4k-12)



We add all the numbers together, and all the variables

(-1k)/()+(-4k-12)=0

We get rid of parentheses

(-1k)/()-4k-12=0

We multiply all the terms by the denominator


(-1k)-4k*()-12*()=0

We add all the numbers together, and all the variables

(-1k)-4k*()=0

We get rid of parentheses

-1k-4k*()=0

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