((3x-2)/(4x-4))=((x-4)/(x-7))

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


D( x )

x-7 = 0

4*x-4 = 0

x-7 = 0

x-7 = 0

x-7 = 0 // + 7

x = 7

4*x-4 = 0

4*x-4 = 0

4*x-4 = 0 // + 4

4*x = 4 // : 4

x = 4/4

x = 1

x in (-oo:1) U (1:7) U (7:+oo)

(3*x-2)/(4*x-4) = (x-4)/(x-7) // - (x-4)/(x-7)

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

(3*x-2)/(4*x-4)+(-1*(x-4))/(x-7) = 0

((3*x-2)*(x-7))/((4*x-4)*(x-7))+(-1*(x-4)*(4*x-4))/((4*x-4)*(x-7)) = 0

(3*x-2)*(x-7)-1*(x-4)*(4*x-4) = 0

-x^2-3*x-2 = 0

-x^2-3*x-2 = 0

-1*(x^2+3*x+2) = 0

x^2+3*x+2 = 0

DELTA = 3^2-(1*2*4)

DELTA = 1

DELTA > 0

x = (1^(1/2)-3)/(1*2) or x = (-1^(1/2)-3)/(1*2)

x = -1 or x = -2

-1*(x+2)*(x+1) = 0

(-1*(x+2)*(x+1))/((4*x-4)*(x-7)) = 0

(-1*(x+2)*(x+1))/((4*x-4)*(x-7)) = 0 // * (4*x-4)*(x-7)

-1*(x+2)*(x+1) = 0

( -1 )

-1 = 0

x belongs to the empty set

( x+1 )

x+1 = 0 // - 1

x = -1

( x+2 )

x+2 = 0 // - 2

x = -2

x in { -1, -2 }

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