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{matrix} equation (1) & : x + y = 50,000 (Euro ) . \end{matrix}

The interest burden from the mortgage agreement at a rate of

5 %

0.05 \cdot x

, and the interest burden from the other one at a rate of

8 %

0.08 \cdot y

. From the problem description we know that the two values add up to

3,700

. This results in a second equation:

\begin{matrix} equation (2) & : 0.05 x + 0.08 y = 3,700 (Euro ) . \end{matrix}

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

(1')

, which is equivalent to equation

(1)

\begin{matrix} equation (1') & : y = 50,000 - x . \end{matrix}

This equation for

y

can be substituted into equation

(2)

. In the resulting equation only the variable

x

occurs, so it can be solved for

x

\begin{matrix} 0.05 x + 0.08 (50,000 - x) = 3,700 \\ \Leftrightarrow & 0.05 x + 4,000 - 0.08 x = 3,700 \\ \Leftrightarrow & 0.03 x = 300 \\ \Leftrightarrow & x = 10,000 . \end{matrix}

Substituting the solution for

x

into equation

(1')

results in

\begin{matrix} y = 50,000 - 10,000 \\ \Leftrightarrow & y = 40,000 . \end{matrix}

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{matrix} equation (1) : & x + y & = & 2 \\ equation (2) : & 2 x + 2 y & = & 1 \end{matrix} .

Solving equation

(1)

for

x

results in

x = 2 - y

. Substituting this equation into equation

(2)

results in:

\begin{matrix} 2 (2 - y) + 2 y = 1 \\ \Leftrightarrow & 4 - 2 y + 2 y = 1 \\ \Leftrightarrow & 4 = 1 . \end{matrix}

This is a contradiction. The LS has no solution.

\begin{matrix} equation (1) : & x + y & = & 2 \\ equation (2) : & 2 x + 2 y & = & 4 \end{matrix} .

Solving equation

(1)

for

y

results in

y = 2 - x

. Substituting this equation into equation

(2)

results in:

\begin{matrix} 2 x + 2 (2 - x) = 4 \\ \Leftrightarrow & 2 x + 4 - 2 x = 4 \\ \Leftrightarrow & 4 = 4 . \end{matrix}

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{matrix} x + y & = 10, \\ x + 2 y & = 13 . \end{matrix}

Both equations are solved for

x

\begin{matrix} x & = 10 - y, \\ x & = 13 - 2 y, \end{matrix}

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

10 - y = 13 - 2 y

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{matrix} 7 x + 2 y & = & 14, \\ 3 x - 5 y & = & 6 \end{matrix}

using the comparison method.

For example, both equations are solved for

x

: For this, the first equation is multiplied by

(1 / 7)

and solved for

x

x = \frac{14}{7} - \frac{2}{7} y \Leftrightarrow x = 2 - \frac{2}{7} y : equation (1') .

In contrast, the second equation is multiplied by

\frac{1}{3}

before solving it for

x

x = \frac{6}{3} + \frac{5}{3} y \Leftrightarrow x = 2 + \frac{5}{3} y : equation (2') .

Equating the two right-hand sides of equation

(1')

and

(2')

results in

2 - \frac{2}{7} y = 2 + \frac{5}{3} y \Leftrightarrow 0 = (\frac{2}{7} + \frac{5}{3}) y \Leftrightarrow y = 0 .

With this solution for

y

, for example, the first equation results in

7 x + 2 \cdot 0 = 14 \Leftrightarrow 7 x = 14 \Leftrightarrow x = 2 .

Hence, the given system of linear equations is solvable uniquely and the solution set is

L = {(x = 2; y = 0)}

.
Alternatively, before equating them, the two equations could have been solved for

y

(or a multiple of

x

or a multiple of

y

). The solution is always the same.

Onlinebrückenkurs Mathematik

1. Elementary Arithmetic

2. Equations in one Variable

3. Inequalities in one Variable

4. System of Linear Equations

5. Geometry

6. Elementary Functions

7. Differential Calculus

8. Integral Calculus

9. Objects in the Two-Dimensional Coordinate System

10. Basic Concepts of Descriptive Vector Geometry

11. Language of Descriptive Statistics

Chapter 4 System of Linear Equations

4.2.2 Substitution Method and Comparison Method

Example 4.2.6

Info 4.2.7

Example 4.2.8

Info 4.2.9

Example 4.2.10

Exercise 4.2.11