#### Chapter 4 System of Linear Equations

Section 4.2 LS in two Variables

# 4.2.2 Substitution Method and Comparison Method

Until now we studied the solvability and the graphical solution of systems of linear equations of the form 4.2.1. Next, we need to study these systems algebraically. To this end, we investigate a further example.
##### Example 4.2.6
For the renovation of their house, the Müller family had to take out two mortgages with a total amount of $50,000$ Euro. The interest they have to pay annually is in total $3,700$ Euro. The interest rate for one mortgage agreement is $5%$ annually, and $8%$ annually for the other. What are the amounts of the individual mortgages?
Let the amounts of the individual mortgages be denoted by $x$ and $y$. As we know from the problem description, the sum of the two amounts is $50,000$ Euro. Hence, the first equation reads:

$\begin{array}{cc}\multicolumn{1}{c}{\text{equation} \left(1\right)}& :x+y=50,000\mathrm{ }\mathrm{ }\mathrm{ }\text{(Euro )} .\hfill \end{array}$

The interest burden from the mortgage agreement at a rate of $5%$ is $0.05·x$, and the interest burden from the other one at a rate of $8%$ is $0.08·y$. From the problem description we know that the two values add up to $3,700$. This results in a second equation:

$\begin{array}{cc}\multicolumn{1}{c}{\text{equation} \left(2\right)}& :0.05x+0.08y=3,700\mathrm{ }\mathrm{ }\mathrm{ }\text{(Euro )} .\hfill \end{array}$

Again, one ends up with a system of linear equations as in example 4.2.1.
To solve the system algebraically, the first equation is solved for $y$. This results in an equation $\left(1\text{'}\right)$, which is equivalent to equation $\left(1\right)$:

$\begin{array}{cc}\multicolumn{1}{c}{\text{equation} \left(1\text{'}\right)}& :y=50,000-x .\hfill \end{array}$

This equation for $y$ can be substituted into equation $\left(2\right)$. In the resulting equation only the variable $x$ occurs, so it can be solved for $x$:

$\begin{array}{ccc}\multicolumn{1}{c}{}& \hfill & 0.05x+0.08\left(50,000-x\right)=3,700\hfill \\ \multicolumn{1}{c}{⇔}& \hfill & 0.05x+4,000-0.08x=3,700\hfill \\ \multicolumn{1}{c}{⇔}& \hfill & 0.03x=300\hfill \\ \multicolumn{1}{c}{⇔}& \hfill & x=10,000 .\hfill \end{array}$

Substituting the solution for $x$ into equation $\left(1\text{'}\right)$ results in

$\begin{array}{ccc}\multicolumn{1}{c}{}& \hfill & y=50,000-10,000\hfill \\ \multicolumn{1}{c}{⇔}& \hfill & y=40,000 .\hfill \end{array}$

Hence, the mortgage amounts are $10,000$ Euro (mortgage with interest of $5%$ annually) and $40,000$ Euro (mortgage with interest of $8%$ annually).
The previous example illustrates the characteristics of the so-called substitution method:
##### Info 4.2.7

In the substitution method, as a first step one of the two linear equations is solved for one of the variables - or a multiple of one of the variables. As a second step the solution is substituted into the other linear equation. Only three cases can occur:
• In the resulting equation (after collecting like terms) the other variable still occurs. Solving the resulting equation for this other variable results in the first part of the solution. The second part is obtained, for example, by substituting the solution of the first part into the equation from the first step. The solution is unique. (If this solution does not belong to the base set, it has to be excluded.)
• In the resulting equation (after collecting like terms) the other variable does not occur any more and the equation is a contradiction. Then the system of linear equations has no solution.
• In the resulting equation (after collecting like terms) the other variable does not occur any more and the equation is always true. Then the system of linear equations has an infinite number of solutions (unless the definition of the base set results in some restrictions).
This approach involves certain details. It is not defined which of the linear equations is to be solved for which of the variables - or multiples of the variable. As long as the applied transformations are equivalent any of the possible ways will result in the same solution. Preferring a specific way is partly a matter of taste and partly a matter of skills: a clever choice can simplify some intermediate calculations.
Concerning the substitution method, cases (ii) and (iii) mentioned above shall be illustrated again by means of the systems of linear equations in example 4.2.4:
##### Example 4.2.8
For both systems of linear equations the base set is $ℝ$.
 $\begin{array}{ccccc}\hfill \text{equation} \left(1\right):& \hfill \hfill & \hfill x+y& \hfill =\hfill & 2\hfill \\ \hfill \text{equation} \left(2\right):& \hfill \hfill & \hfill 2x+2y& \hfill =\hfill & 1\hfill \end{array} .$ Solving equation $\left(1\right)$ for $x$ results in $x=2-y$. Substituting this equation into equation $\left(2\right)$ results in: $\begin{array}{ccc}\multicolumn{1}{c}{}& \hfill & 2\left(2-y\right)+2y=1\hfill \\ \multicolumn{1}{c}{}& ⇔\hfill & 4-2y+2y=1\hfill \\ \multicolumn{1}{c}{}& ⇔\hfill & 4=1 .\hfill \end{array}$ This is a contradiction. The LS has no solution.
 $\begin{array}{ccccc}\hfill \text{equation} \left(1\right):& \hfill \hfill & \hfill x+y& \hfill =\hfill & 2\hfill \\ \hfill \text{equation} \left(2\right):& \hfill \hfill & \hfill 2x+2y& \hfill =\hfill & 4\hfill \end{array} .$ Solving equation $\left(1\right)$ for $y$ results in $y=2-x$. Substituting this equation into equation $\left(2\right)$ results in: $\begin{array}{ccc}\multicolumn{1}{c}{}& \hfill & 2x+2\left(2-x\right)=4\hfill \\ \multicolumn{1}{c}{}& ⇔\hfill & 2x+4-2x=4\hfill \\ \multicolumn{1}{c}{}& ⇔\hfill & 4=4 .\hfill \end{array}$ This is always true. The LS has an infinite number of solutions.
The substitution method is not the only approach for solving systems of linear equations. In the following section another method is discussed, which is closely related to the graphical solution of a LS.
##### Info 4.2.9

In the comparison method, as a first step both linear equations are solved for one of the variables - or a multiple of one of the variables. As a second step the two resulting equations will be equated. Then the three cases discussed for the substitution method can occur.
This approach involves certain details as well. For example, it is not defined for which variable the linear equations are to be solved.
For illustration, the first example is solved again, this time by means of the comparison method:
##### Example 4.2.10
The system of linear equations in the first example reads:

$\begin{array}{cc}\multicolumn{1}{c}{x+y}& =10 ,\hfill \\ \multicolumn{1}{c}{x+2y}& =13 .\hfill \end{array}$

Both equations are solved for $x$

$\begin{array}{cc}\multicolumn{1}{c}{x}& =10-y ,\hfill \\ \multicolumn{1}{c}{x}& =13-2y ,\hfill \end{array}$

and the right-hand-sides of the two equations are equated

$10-y=13-2y$

which results in $y=3$. The solution for $y$ can be substituted into one of the equations solved for $x$ which results in $x=7$.
##### Exercise 4.2.11
Find the solution set of the following system of linear equations

$\begin{array}{ccc}\multicolumn{1}{c}{7x+2y}& =\hfill & 14 ,\hfill \\ \multicolumn{1}{c}{3x-5y}& =\hfill & 6\hfill \end{array}$

using the comparison method.