Tutorial: Systems of two linear equations in two unknowns

This tutorial: Part B: Solving by elimination

(This topic is also in Section 3.1 in Finite Mathematics or Section 4.1 in Finite Mathematics and Applied Calculus)

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#[Solving systems with two unknowns by elimination][Solucionar sistemas con dos incógnitas por eliminación]#

%%Q Do we really need another method of solving a system of linear equations?
%%A A The problem with the graphical approach is that it only gives approximate solutions; locating the exact point of intersection of two lines would require perfect accuracy, which is impossible in practice.

#[The method of elimination is an algebraic procedure to obtain the exact solution(s) of a system of equations in two unknowns by manipulating the equations to eliminate one of the variables ($x$ or $y$).][El método de eliminación es un procedimiento algebraico para obtener la solución exacta de un sistema de ecuaciones con dos incógnitas por manipular las ecuaciones para eliminar una de las variables ($x$ o $y$).]#

We will focus mostly on cases where all the coefficients are rational numbers (which include finite or recurring decimals), as such systems reduce us to working with integer coefficients. At the end of this tutorial we will see how solution by elimination works in cases where not all the coefficients are rational.

Solution of systems with rational coefficients

#[Step 1: Get rid of fractions and decimals.][Paso 1: Deshágate de fracciones y decimales.]#
%%Example

\t !r!$\frac{1}{6}x - \frac{1}{3}y$ \t $=$ \t $-\frac{4}{3}$ \\ \t !r!$x + 1.5y$ \t $=$ \t $2.5$

#[Ugh!][¡Uf!]#

#[You can get rid of the fractions in the first equation by multiplying by a common multiple of the denominators like 6 in this case, and you can get rid of the decimals in the second equation by multiplying by 2 in this case:][Puedes deshacerte de las fracciones en la primera ecuación multiplicando por un múltiplo común de los denominadores como 6 en este caso, y puedes deshacerte de los decimales en la segunda ecuación multiplicando por 2 en este caso:]#

#[Solving by elimination][Solucionar por eliminación]#

#[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. Then solve that equation for the remaining variable. Repeat by eliminating the other variable.][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. Luego resuelve la ecuación para la variable restante. Repite por eliminar la otra incógnita.]#

%%Example
#[System][Sistema]#:

$x - 2y$ \t ${}= -8$ \\ $2x + 3y$ \t ${}=5$

#[Solving by elimination][Solucionar por eliminación]# #[We eliminate $x$. Notice that just adding the equations does not eliminate the $x,$ but results in $3x + y = -3.$ 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 $3x + y = -3.$ Sin embargo, si multiplicamos la primera ecuación por $-2$ obtenemos un sistema equivalente:]#

$-2x + 4y$ \t ${}= 16$ \t $\color{blue}{(-2)}(x - 2y) = \color{blue}{(-2)}(-8)$ \\ $2x + 3y$ \t ${}=5$

#[Now if we add them, the $x$s cancel, and we get:][Ahora si las sumamos, las $x$ se cancelan y obtenemos:]#

$7y = 21$

#[So][Por lo tanto]#

$y = \dfrac{21}{7} = 3.$ \t $\qquad$ #[We have solved for $y.$][Hemos despeja a $y.$]#

#[Now that we know the value of $y,$ we could obtain $x$ by substituting this value in one of the original equations, but instead, just for the fun of it, let's eliminate $y$ starting again with the original system:][Ahora que sabemos que el valor de $y,$ podríamos obtener $x$ sustituyendo este valor en una de las ecuaciones originales, pero en cambio, por el gusto de hacerlo, vamos a eliminar $y$ comenzando de nuevo con el sistema original:]#

#[Original system][Sistema original]#:

$x - 2y$ \t ${}= -8$ \\ $2x + 3y$ \t ${}=5$

#[To eliminate $y,$ we want to make the coefficients of $y$ the same numerically but with opposite sign. We can accomplish this without using fractions by multiplying the first equation by $3$ and the second by $2$:][Para eliminar $y,$ queremos que los coeficientes de $y$ sea el mismo numéricamente pero con signos opuestos. Podemos lograr esto sin utilizar fracciones multiplicando la primera ecuación por $3$ y el segundo por $2.$]#

$3x -6y$ \t ${}=-24$ \t $\color{blue}{(3)}(x - 2y) = \color{blue}{(3)}(-8)$ \\ $4x + 6y$ \t ${}= 10$ \t $\color{blue}{(2)}(2x + 3y) = \color{blue}{(2)}(5)$

#[Now if we add them, the $y$s cancel, and we get:][Ahora si las sumamos, las $y$ se cancelan y obtenemos:]#

$7x = -14$

#[So][Por lo tanto]#

$x = \dfrac{-14}{7} = -2.$

%%Solution

$(x, y) = (-2, 3)$

#[Redundant and inconsistent systems][Systemas redundantes y inconsistentes]#

#[Next, we look at what happens in the elimination method when we encounter redundant and inconsistent systems:][A continuación, nos fijamos en lo que sucede en el método de eliminación cuando nos encontramos con sistemas redundantes e inconsistentes:]#

#[Solving by elimination in redundant or inconsistent systems][Resolución por eliminación en sistemas redundantes o inconsistentes]#

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.]#

#[Examples][Ejemplos]#

	#[Redundant system][Sistema redundante]#:	\t !r!$x - 2y$ \t $=$ \t $-8$ \\ \t !r!$-2x + 4y$ \t $=$ \t $16$
	#[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:]#
		\t !r!$2x - 4y$ \t $=$ \t $-16$ \t $\color{blue}{(2)}(x - 2y) = \color{blue}{(2)}(-8)$ \\ \t !r!$-2x + 4y$ \t $=$ \t $5$
	#[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:]#
		$0 = 0$

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

$\left(x, \dfrac{x+8}{2}\right)$; $x$ #[arbitrary][arbitrario]#. \t \gap[40]

#[General solution (free variable x)][Solción general (variable libre x)]#

(In the game version you can 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][arbitrario]#. \t \gap[40]

#[General solution (free variable y)][Solción general (variable libre y)]#

#[However, it sometimes also happens that we don\'t get $0 = 0$. For instance, suppose our original system was the following:][Sin embargo, a veces también sucede que no obtenemos $0 = 0$. Por ejemplo, supongamos que nuestro sistema original era el siguiente:]#

	#[Inconsistent system][Sistema inconsistente]#:	\t !r!$2x - 4y$ \t $=$ \t $-2$ \\ \t !r!$-3x + 6y$ \t $=$ \t $-1$
	#[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 $2$:]#
		\t !r!$6x - 12y$ \t $=$ \t $-6$ \\ \t !r!$-6x + 12y$ \t $=$ \t $-2$
	#[Now if we add them, the $y$s cancel, and we get:][Ahora si las sumamos, las $y$ se cancelan y obtenemos:]#
		$0 = -8.$

#[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 other words, 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.]#

%%Q #[What happens when one of the coefficients is zero, as in, for instance,][ ¿Qué sucede cuando uno de los coeficientes es cero, como en, por ejemplo,]#

$3x - y$ \t ${}= 2$ \\ $5y$ \t ${}= 6$?

%%A #[Nothing different, really. The second equation has $x$ already eliminated, and says that $y = 6/5.$ We can then obtain $x$ by substituting this value of $y$ in the first equation, or by eliminating $y$ in the usual way: Multiply the first equation by 5 and then add.][Nada diferente, de verdad. La segunda ecuación tiene $x$ ya eliminado, y nos dice que $y = 6/5.$ A continuación, podemos obtener $x$ sustituyendo este valor de $y$ en la primera ecuación, o por eliminar $y$ de la forma habitual: multiplicar la primera ecuación por 5 y luego sumar.]#

Now try some of the exercises in Section 3.1 in Finite Mathematics or Section 4.1 in Finite Mathematics and Applied Calculus. or move ahead to the Part B of this tutorial by pressing "Next tutorial" on the sidebar.

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