#### Chapter 4 System of Linear Equations

Section 4.2 LS in two Variables

We will now discuss another, third method for solving systems of linear equations algebraically. But this method will develop its full potential only for larger systems, i.e. systems of many equations in many variables since it can be systematised very well. Here, we will discuss the general approach. First, let us see an example.
##### Example 4.2.12
Find the solution of the system of linear equations

$\begin{array}{ccccc}\hfill \text{equation} \left(1\right):& \hfill \hfill & \hfill 2x+y& \hfill =\hfill & 9 ,\hfill \\ \hfill \text{equation} \left(2\right):& \hfill \hfill & \hfill 3x-11y& \hfill =\hfill & 1 ,\hfill \end{array}$

where the base set is the range of the real numbers $ℝ$.
This time, the approach is as follows: The first equation is multiplied by the factor $11$ and this results in the equation (1') that is equivalent to equation $\left(1\right)$:

$\begin{array}{cccccc}\hfill \hfill & \hfill \left(2x+y\right)·11& \hfill =\hfill & 9·11\hfill & \hfill \hfill & \hfill \\ \hfill ⇔\hfill & \hfill 22x+11y& \hfill =\hfill & 99\hfill & \hfill \hfill & : \text{equation} \left(1\text{'}\right) .\hfill \end{array}$

Subsequently, the new equation $\left(1\text{'}\right)$ is added to equation $\left(2\right)$, i.e. the sum of the left-hand sides of $\left(1\text{'}\right)$ and $\left(2\right)$ is equated to the sum of the right-hand sides of $\left(1\text{'}\right)$ and $\left(2\right)$. In doing so, the variable $y$ is cancelled out. This was the reason for selecting the factor $11$ in the previous step.

$3x-11y+22x+11y=1+99⇔25x=100⇔x=4 .$

The get the solution for $y$ the just obtained solution for $x$ can be substituted, e.g. into equation $\left(1\right)$:

$2·4+y=9⇔8+y=9⇔y=1 .$

Thus, this system of linear equations has a unique solution $L=\left\{\left(x=4;y=1\right)\right\}$.
As for the other methods,the approach here is not uniquely defined: for example, equation $\left(1\right)$ could have been multiplied by $3$ and equation $\left(2\right)$ by $\left(-2\right)$

$\begin{array}{ccccccccc}\hfill \left(2x+y\right)·3& \hfill =\hfill & 9·3\hfill & \hfill ⇔\hfill & \hfill 6x+3y& \hfill =\hfill & 27\hfill & \hfill \hfill & : \text{equation} \left(1\text{'}\text{'}\right) ,\hfill \\ \hfill \left(3x-11y\right)·\left(-2\right)& \hfill =\hfill & 1·\left(-2\right)\hfill & \hfill ⇔\hfill & \hfill -6x+22y& \hfill =\hfill & -2\hfill & \hfill \hfill & : \text{equation} \left(2\text{'}\text{'}\right) .\hfill \end{array}$

In the subsequent addition of equation $\left(1\text{'}\text{'}\right)$ and equation $\left(2\text{'}\text{'}\right)$ the variable $x$ could have been eliminated:

$6x+3y-6x+22y=27-2⇔25y=25⇔y=1 .$

To get the solution for $x$, the solution for $y$ then could have been substituted, e.g.  into equation $\left(2\right)$

$3x-11·1=1⇔3x=12⇔x=4 .$

##### Info 4.2.13

In the addition method, one of the linear equations is transformed by multiplying it by an arbitrary factor such that in the subsequent addition of the other equation (at least) one variable is eliminated. (Sometimes it is easier to multiply both equations by arbitrary factors before adding them.) As for the substitution method in info box 4.2.7 (or the comparison method in info box 4.2.9), three cases can occur, resulting in a solution set $L$ containing exactly one element, no element, or an infinite number of elements.