#### Chapter 4 System of Linear Equations

**Section 4.3 LS in three Variables**

# 4.3.3 Substitution Method

The**substitution method**was already discussed in section 4.2.2 for systems of two linear equations. For systems of three linear equations like the system in example 4.3.2, the approach is basically the same.

##### **Example 4.3.6 **

Let us return to the first example 4.3.1 of this section. The system of linear equations for the puzzle of the three tempted children reads:

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

For example, one can start by solving equation $(1)$ for $x$:

$x=2y+2z-30\mathrm{\hspace{0.5em}\hspace{0.5em}}:\hspace{0.5em}\text{equation}\hspace{0.5em}(1\text{'})\hspace{0.5em}.$

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

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

This step reduces the initial system to a system of

$y=24-\frac{9}{5}z\hspace{0.5em}.$

This relation can be

$-15(24-\frac{9}{5}z)-9z=-180\iff 360-27z+9z=180\iff 18z=180\iff z=10\hspace{0.5em}.$

So, the value of $y$ is

$y=24-\frac{9}{5}\xb710=24-9\xb72=6$

and finally the value of $x$ is (using, for example, equation $(1\text{'})$):

$x=2\xb76+2\xb710-30=12+20-30=2\hspace{0.5em}.$

Hence, the system of linear equations has a unique solution. The solution set consists of exactly one element, namely $L=\{(x=2;y=6;z=10)\}$.

For example, one can start by solving equation $(1)$ for $x$:

**Substituting**this equation in equation $(2)$ and equation $(3)$ eliminates the variable $x$ from these equations:This step reduces the initial system to a system of

**two**linear equations in the**two**variables $y$ and $z$ which can be solved using the methods from the previous section 4.2. For example, equation $(2\text{'})$ can be solved for $y$:This relation can be

**substituted**into equation $(3\text{'})$:So, the value of $y$ is

and finally the value of $x$ is (using, for example, equation $(1\text{'})$):

Hence, the system of linear equations has a unique solution. The solution set consists of exactly one element, namely $L=\{(x=2;y=6;z=10)\}$.

Since the right-hand sides of the three equations are all equal to $-30$, the one and other reader may ask himself whether it wouldn't be more practical to equate all the right-hand sides in pairs and to continue with the resulting equations.

But this approach is not helpful and - if you are not careful - possibly even wrong. In any case, the number of variables would not decrease by this approach. But this is exactly what the substitution method and the comparison method are for: In both approaches (and in the addition method as well), as a first and second step one variable is eliminated such that the initial system reduces to a system of two linear equations (and hence to a simpler problem).

By the way, the approach for the solution of this reduced problem does not depend on the initial approach. In other words, it is allowed and can be even clever to start with one method, e.g. the substitution method, to reduce the system of three linear equations to a system of two linear equations and to solve this simpler system using another method, e.g. the substitution method. In this sense, the methods can be mixed.

##### **Info 4.3.7 **

In the substitution method, as a first step one of the three linear equations is solved for one of the variables - or for a multiple of one of the variables. As a second step the resulting relation is

**substituted**into one of the two other linear equations. It results a

**system of only two linear equations**in the (remaining)

**two**variables. This system can be solved using one of the methods described in section 4.2.