#[Solving by elimination in redundant or inconsistent systems][Resolución por eliminación en sistemas redundantes o inconsistentes]#
#[To
eliminate an unknown means to obtain an equation in which that unknown does not occur. To eliminate an unknown that occurs in both equations, multiply one or both equations by suitable nonzero constant (if necessary) so that adding the resulting equations cancels out ("eliminates") that unknown.
When the system has a unique solution, eliminating one of the variables will result in an equation in the other variable, for which you can then solve. But sometimes, eliminating one of the variables causes both of them to be eliminated, indicating that the system is either redundant or inconsistent.][
Eliminar un desconocido significa obtener una ecuación en la que no ocurre ese desconocido. Para eliminar un desconocido que ocurre en ambas ecuaciones, multiplica una o ambas ecuaciones por constantes adecuadas distintas de cero (si sea necesario) tal que la adición de las ecuaciones resultantes anula ("elimina") aquella incógnita.
Cuando el sistema tiene una solución única, eliminar una de las variables resultará en una ecuación en la otra variable, que luego se puede resolver. Sin embargo, a veces, eliminar una de las variables provoca la eliminación de ambas, lo que indica que el sistema es redundante o inconsistente.]#
#[Example][Ejemplo]#
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#[System][Sistema]#: |
\t !r!$x - 2y$ \t $=$ \t $-8$
\\ \t !r!$-2x + 4y$ \t $=$ \t $16$
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#[We eliminate $x$. Notice that just adding the equations does not eliminate the $x,$ but results in $-x + 2y = 8.$ However, if we multiply the first equation by $2$ we get an equivalent system:][Eliminamos la $x.$ Observa que simplemente sumando las ecuaciones elimina los $x,$ pero resulta en $-x + 2y = 8.$ Sin embargo, si multiplicamos la primera ecuación por $2$ obtenemos un sistema equivalente:]# |
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\t !r!$2x - 4y$ \t $=$ \t $-16$ \t
\\ \t !r!$-2x + 4y$ \t $=$ \t $5$
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#[Now if we add them, the $x$s and also the $y$s cancel, and we get the following rather uninteresting equation:][Ahora si las sumamos, las $x$ y tambi&ecute;n las $y$ se cancelan y obtenemos la siguiente , bastante poco interesante ecuación:]# |
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$0 = 0$
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The reason we get $0 = $ is that the two equations say the same thing! (one is just the negative of the other; graphing them resuts in the same line). In other words, the system is
redundant, and we really have only one equation's worth of information. We also know that a system consisting of a single equation has infinitely many solutions: we can, for instance, choose $x$ to be any number we like and solve either equation for $y$, getting $y = (x+8)/2$.
This gives us a
general solution of the form $(x, (x+8)/2)$; $x$ arbitrary. (Open a practice session and, under
Practice Options go to "Finding particular and general solutions of a linear equation").
Or, we could choose $y$ to be arbitrary, and solve for $x$ getting $x = 2y - 8$, and hence another form of the general solution: $(2y-8, y)$; $y$ arbitrary.
#[However, it sometimes also happens that we don\'t get $0 = 0$. For instance, suppose our original system was][Sin embargo, a veces también sucede que no obtenemos $0 = 0$. Por ejemplo, supongamos que nuestro sistema original era]#\
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#[System][Sistema]#: |
\t !r!$2x - 4y$ \t $=$ \t $-2$
\\ \t !r!$-3x + 6y$ \t $=$ \t $-1$
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#[To eliminate $x,$ we can accomplish this by multiplying the first equation by $3$ and the second by $2$:][Para eliminar $x$, podemos multiplicar la primera ecuación por $3$ y la segunda por $2$:]# |
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\t !r!$6x - 12y$ \t $=$ \t $-6$
\\ \t !r!$-6x + 12y$ \t $=$ \t $-2$
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#[Now if we add them, the $y$s cancel, and we get:][Ahora si las sumamos, las $y$ se cancelan y obtenemos:]# |
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$0 = -8$
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#[But this is absurd! What this is telling us is that there can be no real numbers $x$ and $y$ making both equations true, as assuming there were would lead to the contradiction $0 = -8$. In otherwords, the given system has
no solution at all. Geometrically, if you plot both equations, the resuting two lines are parallel and hence do not intersect.][¡Pero esto es absurdo! Lo que nos dice es que no puede haber números reales $x$ e $y$ que hagan que ambas ecuaciones sean verdaderas, ya que asumir que hubieras verdaderas nos lleva a la contradicción $0 = -8$. En otras palabras, el sistema dado no tiene solución. Geométricamente, si se trazan ambas ecuaciones, las dos rectas resultantes son paralelas y, por lo tanto, no se intersecan.]#