91Ó°ÊÓ

When working with systems of linear equations, sometimes you will come across what we call dependent equations. This describes a situation where one equation is a multiple of another. Thus, rather than representing different lines, they actually represent the same line.

In the given exercise, when we look at the system:

• \( x + 2y = 3 \)
• \( -2x - 4y = -6 \)

By simplifying the second equation by dividing everything by -2, we see that it results in the first equation \( x + 2y = 3 \). Therefore, these are dependent equations.

Dependent equations mean that we only have one distinct line on a graph rather than two. All the solutions of this system fall on this single line, illustrating that the equations are not independent. This is a critical step in understanding the nature of the system we are dealing with.

When you have a system of dependent equations, you inherently end up with infinite solutions. This happens because the same line contains infinite points, meaning there are infinite pairs \((x, y)\) that satisfy both equations.

In our exercise, since both equations represent the same line, every point on this line is a potential solution. Therefore, instead of finding a single intersecting point for two lines, which is typical with independent equations, you find many solutions that fit both.

• Example: If \( y = 1 \), you can calculate \( x \) using the equation \( x + 2y = 3 \). This yields \( x = 1 \), resulting in the point (1, 1) which is a solution.
• Similarly, if \( y = 2 \), \( x = -1 \), giving you another solution point (-1, 2).

Each point on the line is a legitimate solution showcasing the infinite nature of such systems.

Linear equations are foundational in algebra, describing the relationship between two variables in a straight line when graphed. A standard form of a linear equation in two variables is given as \( ax + by = c \).

The linear equations in this exercise are straight lines, showing how the dependent nature leads to infinite intersections. Here, each line's slope and intercept determine its path along a graph.

• The slope comes from the coefficient of \( y \), dictating the angle or direction of the line.
• The intercept, specifically the y-intercept derived when \( x = 0 \), shows where the line crosses the y-axis.

In our exercise, both lines follow the equation \( x + 2y = 3 \), meaning they have the same slope and the same y-intercept. Hence, illustrating the elementary quality of linear equations as straight-line representations.

Parameterization is a powerful method used to express infinite solutions of a system in a generalized form. By introducing a parameter, often denoted as \( t \), we can describe all the solutions concisely.

In this example, once it's established that the system equations are dependent, we choose to express \( y \) as \( t \) where \( t \) is any real number. By substituting \( y = t \) in the equation \( x + 2y = 3 \), you can solve for \( x \), yielding \( x = 3 - 2t \).

Consequently, every solution of the system can be written in terms of \( t \) as:

• \((x, y) = (3 - 2t, t)\)

This way, parameterization conveniently handles infinite solutions by mapping any point on the line to a specific value of \( t \). This technique allows all solutions to be written as functions of a single variable, showing the infinite possibilities in a structured manner.