1/(2x+3)=7/(x+4)

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Solution for 1/(2x+3)=7/(x+4) equation: 


D( x )

x+4 = 0

2*x+3 = 0

x+4 = 0

x+4 = 0

x+4 = 0 // - 4

x = -4

2*x+3 = 0

2*x+3 = 0

2*x+3 = 0 // - 3

2*x = -3 // : 2

x = -3/2

x in (-oo:-4) U (-4:-3/2) U (-3/2:+oo)

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

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

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

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

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

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

-13*x-17 = 0

(-13*x-17)/((2*x+3)*(x+4)) = 0

(-13*x-17)/((2*x+3)*(x+4)) = 0 // * (2*x+3)*(x+4)

-13*x-17 = 0

-13*x-17 = 0 // + 17

-13*x = 17 // : -13

x = 17/(-13)

x = -17/13

x = -17/13

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