-3/(y-2)=7/(2y-4)+2

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


D( y )

2*y-4 = 0

y-2 = 0

2*y-4 = 0

2*y-4 = 0

2*y-4 = 0 // + 4

2*y = 4 // : 2

y = 4/2

y = 2

y-2 = 0

y-2 = 0

y-2 = 0 // + 2

y = 2

y in (-oo:2) U (2:+oo)

-3/(y-2) = 7/(2*y-4)+2 // - 7/(2*y-4)+2

-3/(y-2)-(7/(2*y-4))-2 = 0

-3/(y-2)-7*(2*y-4)^-1-2 = 0

-3/(y-2)-7/(2*y-4)-2 = 0

(-3*(2*y-4))/((y-2)*(2*y-4))+(-7*(y-2))/((y-2)*(2*y-4))+(-2*(y-2)*(2*y-4))/((y-2)*(2*y-4)) = 0

-3*(2*y-4)-7*(y-2)-2*(y-2)*(2*y-4) = 0

16*y-4*y^2-13*y-16+26 = 0

3*y-4*y^2+10 = 0

3*y-4*y^2+10 = 0

3*y-4*y^2+10 = 0

DELTA = 3^2-(-4*4*10)

DELTA = 169

DELTA > 0

y = (169^(1/2)-3)/(-4*2) or y = (-169^(1/2)-3)/(-4*2)

y = -5/4 or y = 2

(y+5/4)*(y-2) = 0

((y+5/4)*(y-2))/((y-2)*(2*y-4)) = 0

((y+5/4)*(y-2))/((y-2)*(2*y-4)) = 0 // * (y-2)*(2*y-4)

(y+5/4)*(y-2) = 0

( y+5/4 )

y+5/4 = 0 // - 5/4

y = -5/4

( y-2 )

y-2 = 0 // + 2

y = 2

y in { 2} 

y = -5/4

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