#### Chapter 4 System of Linear Equations

**Section 4.3 LS in three Variables**

# 4.3.4 Addition Method

The idea of the addition method, which we already discussed a little before (see section 4.2.3), is to**add**equations of the system such that the number of variables occurring in the system is reduced. For this, one of the equations often has to be multiplied by a cleverly chosen factor before

**adding**these equations.

The addition method for a system of three linear equations in three variables shall be presented in a form that can be easily applied to larger systems. To illustrate the approach, we discuss again the system in the first example 4.3.1, i.e. the example of the three little crooks.

##### **Example 4.3.8 **

The system of linear equations to be solved reads then:

$\begin{array}{ccccc}\text{equation}\hspace{0.5em}(1):\hfill & \hfill \hfill & \hfill x-2y-2z& \hfill =\hfill & -30\hspace{0.5em},\hfill \\ \text{equation}\hspace{0.5em}(2):\hfill & \hfill \hfill & \hfill -3x+y-3z& \hfill =\hfill & -30\hspace{0.5em},\hfill \\ \text{equation}\hspace{0.5em}(3):\hfill & \hfill \hfill & \hfill -5x-5y+z& \hfill =\hfill & -30\hspace{0.5em}.\hfill \end{array}$

Equation $(1)$ is left unchanged in the following. But equation $(2)$ is to be replaced by a new equation resulting from the

$(-3x+y-3z)+3\xb7(x-2y-2z)=-30+3\xb7(-30)\iff -5y-9z=-120\mathrm{\hspace{0.5em}\hspace{0.5em}}:\hspace{0.5em}\text{equation}\hspace{0.5em}(2\text{'})\hspace{0.5em}.$

Likewise, equation $(3)$ will be replaced by $(3)+5\xb7(1)$, i.e. by the

$(-5x-5y+z)+5\xb7(x-2y-2z)=-30+5\xb7(-30)\iff -15y-9z=-180\mathrm{\hspace{0.5em}\hspace{0.5em}}:\hspace{0.5em}\text{equation}\hspace{0.5em}(3\text{'})\hspace{0.5em}.$

The system now reads as follows:

$\begin{array}{ccccc}\text{equation}\hspace{0.5em}(1):\hfill & \hfill \hfill & \hfill x-2y-2z& \hfill =\hfill & -30\hspace{0.5em},\hfill \\ \text{equation}\hspace{0.5em}(2\text{'}):\hfill & \hfill \hfill & \hfill -5y-9z& \hfill =\hfill & -120\hspace{0.5em},\hfill \\ \text{equation}\hspace{0.5em}(3\text{'}):\hfill & \hfill \hfill & \hfill -15y-9z& \hfill =\hfill & -180\hspace{0.5em}.\hfill \end{array}$

Equation $(2\text{'})$ and equation $(3\text{'})$ do not depend on the variable $x$ anymore - that was the intention and the reason for choosing the factors $3$ and $5$ above, respectively.

The subsystem that consists of the two equations $(2\text{'})$ and $(3\text{'})$ in the two variables $y$ and $z$ could now be solved using one of the other methods, e.g. the substitution method. But here, it should be solved completely using the addition method. For this, equation $(2\text{'})$ and equation $(1)$ will be left unchanged in the following. In contrast, equation $(3\text{'})$ has to be replaced, namely by the sum $(3\text{'})+(-3)\xb7(2\text{'})$:

$(-15y-9z)+(-3)\xb7(-5y-9z)=-180+(-3)\xb7(-120)\iff 18z=180\mathrm{\hspace{0.5em}\hspace{0.5em}}:\hspace{0.5em}\text{equation}\hspace{0.5em}(3\text{'}\text{'})\hspace{0.5em}.$

Thus, the system has changed again,

$\begin{array}{ccccc}\text{equation}\hspace{0.5em}(1):\hfill & \hfill \hfill & \hfill x-2y-2z& \hfill =\hfill & -30\hspace{0.5em},\hfill \\ \text{equation}\hspace{0.5em}(2\text{'}):\hfill & \hfill \hfill & \hfill -5y-9z& \hfill =\hfill & -120\hspace{0.5em},\hfill \\ \text{equation}\hspace{0.5em}(3\text{'}\text{'}):\hfill & \hfill \hfill & \hfill 18z& \hfill =\hfill & 180\hspace{0.5em},\hfill \end{array}$

and now has - at least concerning the left-hand side - a kind of

Solving for the variables is now very simple: The last equation (equation $(3\text{'}\text{'})$) only depends on a single variable, namely $z$, and hence can be solved for $z$ immediately: $z=10$.

This value of $z$ is then inserted in the equation in the line above (equation $(2\text{'})$) that immediately provides the value of $y$: $-5y-9\xb710=-120\iff -5y=-30\iff y=6$.

Finally, inserting the values of $y$ and $z$ in the first equation (equation $(1)$) immediately provides the solution for the remaining variable, in the example this is the variable $x$: $x-2\xb76-2\xb710=-30\iff x=2$.

Equation $(1)$ is left unchanged in the following. But equation $(2)$ is to be replaced by a new equation resulting from the

**addition**of equation $(2)$ and equation $(1)$ multiplied by a factor of $3$ - shortly noted as $(2)+3\xb7(1)$:Likewise, equation $(3)$ will be replaced by $(3)+5\xb7(1)$, i.e. by the

**sum**of equation $(3)$ and equation $(1)$ multiplied by a factor of $5$:The system now reads as follows:

Equation $(2\text{'})$ and equation $(3\text{'})$ do not depend on the variable $x$ anymore - that was the intention and the reason for choosing the factors $3$ and $5$ above, respectively.

The subsystem that consists of the two equations $(2\text{'})$ and $(3\text{'})$ in the two variables $y$ and $z$ could now be solved using one of the other methods, e.g. the substitution method. But here, it should be solved completely using the addition method. For this, equation $(2\text{'})$ and equation $(1)$ will be left unchanged in the following. In contrast, equation $(3\text{'})$ has to be replaced, namely by the sum $(3\text{'})+(-3)\xb7(2\text{'})$:

Thus, the system has changed again,

and now has - at least concerning the left-hand side - a kind of

**triangular form**.Solving for the variables is now very simple: The last equation (equation $(3\text{'}\text{'})$) only depends on a single variable, namely $z$, and hence can be solved for $z$ immediately: $z=10$.

This value of $z$ is then inserted in the equation in the line above (equation $(2\text{'})$) that immediately provides the value of $y$: $-5y-9\xb710=-120\iff -5y=-30\iff y=6$.

Finally, inserting the values of $y$ and $z$ in the first equation (equation $(1)$) immediately provides the solution for the remaining variable, in the example this is the variable $x$: $x-2\xb76-2\xb710=-30\iff x=2$.

An attentive reader may ask themselves whether - and if so, why - one is allowed to replace an equation in a system of equations by another equation. In the example above this occurs three times, e.g. if equation $(2)$ is replaced by a combination of equation $(2)$ and three times equation $(1)$, i.e. equation $(2\text{'})$.

Finding the solution of a system of linear equations requires that all equations of the system

**hold simultaneously**, i.e. in example 4.3.8 it is required that equation $(1)$

**and**equation $(2)$ hold which clearly implies that also

holds. If now equation $(1)$

**and**equation $(2\text{'})$ hold simultaneously, then immediately follows that equation $(1)$ and

hold simultaneously as well. Hence, one is allowed to replace equation $(2)$ by equation $(2\text{'})$ in the systems of equations.

Importantly, one can see here: if the

**two**equations - equation $(1)$ and equation $(2)$ - were both replaced by equation (2'), information would be lost and a mistake would be made. (The requirement of

**only**$(2\text{'})$ instead of $(1)$

**and**$(2)$ is much weaker.) This is the reason why in the "new" systems some equations are left unchanged: equation $(1)$

**and**equation $(2)$ are in the corresponding systems equivalent to equation $(1)$ and equation $(2\text{'})$. The same is true for the other replacement in the example above - and generally for such transformations of systems of linear equations by means of the addition method.

##### **Info 4.3.9 **

In the

**addition method**, pairs of linear equations of the system are added while multiplying (at least) one of the equations by a clever chosen factor (or clever chosen factors) such that in the resulting equations (at least) one variable is eliminated. It has to be ensured that in the solution process no information is lost, i.e. the number of (information relevant) equations is fixed. For this, it is most clever to bring the system into

**triangular from**. Then, the solution can be found very easily.