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[tex]\huge \bf༆ Answer ༄[/tex]

Let's solve ~

  • [tex] \sf \dfrac{3}{x} - \dfrac{3}{x + 4} = 1[/tex]

  • [tex] \sf \dfrac{3(x + 4) -3x }{x(x + 4)} = 1[/tex]

  • [tex] \sf \dfrac{3x + 12-3x }{x(x + 4)} = 1[/tex]

  • [tex] \sf 12 = {x(x + 4)} [/tex]

  • [tex] \sf {x}^{2} + 4x = 12[/tex]

  • [tex] \sf{ {x}^{2} + 4x - 12 = 0 }[/tex]

  • [tex] \sf{ {x}^{2} + 6x - 2x - 12 = 0}[/tex]

  • [tex] \sf{x(x + 6) - 2(x + 6) = 0}[/tex]

  • [tex] \sf(x - 2)(x + 6) = 0[/tex]

Hence,

  • [tex] \sf x = 2 \: \: or \: \: x = - 6[/tex]

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