(40/x)-(8/7)=20/(x-3)

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Solution for (40/x)-(8/7)=20/(x-3) equation:


D( x )

x = 0

x-3 = 0

x = 0

x = 0

x-3 = 0

x-3 = 0

x-3 = 0 // + 3

x = 3

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

40/x-(8/7) = 20/(x-3) // - 20/(x-3)

40/x-(20/(x-3))-(8/7) = 0

40/x-20*(x-3)^-1-8/7 = 0

40/x-20/(x-3)-8/7 = 0

(-20*7*x)/(7*x*(x-3))+(7*40*(x-3))/(7*x*(x-3))+(-8*x*(x-3))/(7*x*(x-3)) = 0

7*40*(x-3)-8*x*(x-3)-20*7*x = 0

140*x-8*x^2+24*x-840 = 0

164*x-8*x^2-840 = 0

164*x-8*x^2-840 = 0

4*(41*x-2*x^2-210) = 0

41*x-2*x^2-210 = 0

DELTA = 41^2-(-210*(-2)*4)

DELTA = 1

DELTA > 0

x = (1^(1/2)-41)/(-2*2) or x = (-1^(1/2)-41)/(-2*2)

x = 10 or x = 21/2

4*(x-10)*(x-21/2) = 0

(4*(x-10)*(x-21/2))/(7*x*(x-3)) = 0

(4*(x-10)*(x-21/2))/(7*x*(x-3)) = 0 // * 7*x*(x-3)

4*(x-10)*(x-21/2) = 0

( 4 )

4 = 0

x belongs to the empty set

( x-10 )

x-10 = 0 // + 10

x = 10

( x-21/2 )

x-21/2 = 0 // + 21/2

x = 21/2

x in { 10, 21/2 }

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