(8/x-3)-1/10=32/(4x-12)

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Solution for (8/x-3)-1/10=32/(4x-12) equation:


D( x )

4*x-12 = 0

x = 0

4*x-12 = 0

4*x-12 = 0

4*x-12 = 0 // + 12

4*x = 12 // : 4

x = 12/4

x = 3

x = 0

x = 0

x in (-oo:0) U (0:3) U (3:+oo)

8/x-(1/10)-3 = 32/(4*x-12) // - 32/(4*x-12)

8/x-(32/(4*x-12))-(1/10)-3 = 0

8/x-32*(4*x-12)^-1-1/10-3 = 0

8/x-32/(4*x-12)-1/10-3 = 0

(-32*10*x)/(10*x*(4*x-12))+(8*10*(4*x-12))/(10*x*(4*x-12))+(-1*x*(4*x-12))/(10*x*(4*x-12))+(-3*10*x*(4*x-12))/(10*x*(4*x-12)) = 0

8*10*(4*x-12)-1*x*(4*x-12)-3*10*x*(4*x-12)-32*10*x = 0

12*x-4*x^2-120*x^2+360*x-960 = 0

372*x-124*x^2-960 = 0

372*x-124*x^2-960 = 0

4*(93*x-31*x^2-240) = 0

93*x-31*x^2-240 = 0

DELTA = 93^2-(-240*(-31)*4)

DELTA = -21111

DELTA < 0

4 = 0

4/(10*x*(4*x-12)) = 0

4/(10*x*(4*x-12)) = 0 // * 10*x*(4*x-12)

4 = 0

x belongs to the empty set

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