Si la mitad del reciproco de un numero se suma al duplo del reciproco de otro, la suma es uno mas que el reciproco del primer numero. la diferencia de los reciprocos de los numeros es una decima.​

Answer:

Ambos números son [tex]-\frac{25}{8}[/tex] y [tex]\frac{50}{21}[/tex]

Step-by-step explanation:

La mitad de una cantidad es el resultado de partir en dos partes iguales dicha cantidad, siendo cada una de esas partes es la mitad de la cantidad inicial. Por lo que la mitad de un número se calcula dividiendo el número por 2.

El recíproco o inverso multiplicativo de un número x es el número que, cuando es multiplicado por x , da 1.  Es decir, el producto de un número y su recíproco es 1. Para obtener el recíproco de un número, sólo divide 1 por el número. Entonces el reciproco de un numero cualesquiera x es [tex]\frac{1}{x}[/tex].

El doble o duplo de un número se calcula multiplicando el número por 2.

La diferencia es el resultado de la resta o sustracción.

Entonces:

Entonces:

[tex]\frac{1}{x}[/tex]÷2 + [tex]\frac{1}{y}[/tex]*2 = [tex]\frac{1}{x}+1[/tex] Ecuación (A)

Siendo la diferencia de los recíprocos de los números igual a una decima, entonces

[tex]\frac{1}{x}+\frac{1}{y}=\frac{1}{10}[/tex] Ecuación (B)

Entonces, de la ecuación (B) se obtiene:

[tex]\frac{1}{y}=\frac{1}{10}-\frac{1}{x}[/tex] Ecuación (C)

Reemplazando en la ecuación (A):

[tex]\frac{1}{x}[/tex]÷2 +( [tex]\frac{1}{10}-\frac{1}{x}[/tex])*2 = [tex]\frac{1}{x}+1[/tex]

Resolviendo:

[tex]\frac{1}{x}[/tex]÷2 + [tex]\frac{1}{10}*2-\frac{1}{x}*2[/tex] = [tex]\frac{1}{x}+1[/tex]

[tex]\frac{1}{x}[/tex]*[tex]\frac{1}{2}[/tex] - [tex]\frac{1}{x}*2[/tex] - [tex]\frac{1}{x}[/tex]= 1- [tex]\frac{1}{10} *2[/tex]

[tex]-\frac{5}{2}*\frac{1}{x}=\frac{4}{5}[/tex]

[tex]\frac{1}{x} =\frac{4}{5}[/tex] ÷ [tex](-\frac{5}{2})[/tex]

[tex]\frac{1}{x} =-\frac{8}{25}[/tex]

[tex]x=-\frac{25}{8}[/tex]

Entonces, reemplazando en ecuación (C)

[tex]\frac{1}{y}=\frac{1}{10}-\frac{1}{-\frac{25}{8} }[/tex]

[tex]\frac{1}{y}=\frac{1}{10}+\frac{8}{25}[/tex]

[tex]\frac{1}{y}=\frac{21}{50}[/tex]

[tex]y=\frac{50}{21}[/tex]

Entonces, ambos números son [tex]-\frac{25}{8}[/tex] y [tex]\frac{50}{21}[/tex]

Usando un sistema de ecuaciones, se encuentra que  los números son: [tex]-\frac{1}{4}, -2[/tex]

Si la mitad del reciproco de un numero se suma al duplo del reciproco de otro, la suma es uno mas que el reciproco del primer numero, o sea:

[tex]\frac{1}{2x} + \frac{2}{y} = 1 + \frac{1}{x}[/tex]

La diferencia de los reciprocos de los numeros es una decima, o sea:

[tex]\frac{1}{x} - \frac{1}{y} = 10[/tex]

[tex]\frac{y - x}{xy} = 10[/tex]

[tex]y - x = 10xy[/tex]

[tex]y = 10xy + x[/tex]

[tex]y = x(10 + y)[/tex]

[tex]x = \frac{y}{10 + y}[/tex]

Reemplazando en la primera ecuación:

[tex]\frac{1}{2x} + \frac{2}{y} = 1 + \frac{1}{x}[/tex]

[tex]\frac{10 + y}{2y} + \frac{2}{y} = 1 + \frac{10 + y}{y}[/tex]

[tex]\frac{10 + y + 4}{2y} = \frac{y + 10 + y}{y}[/tex]

[tex]\frac{14 + y}{2y} = \frac{2y + 10}{y}[/tex]

[tex]14y + y^2 = 4y^2 + 20y[/tex]

[tex]3y^2 + 6y = 0[/tex]

[tex]3y(y + 2) = 0[/tex]

[tex]y + 2 = 0[/tex]

[tex]y = -2[/tex]

[tex]x = \frac{y}{10 + y} = \frac{-2}{10 - 2} = -\frac{1}{4}[/tex]

Los números son: [tex]-\frac{1}{4}, -2[/tex]

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