Problem 42 - Algebra 2

Question

Give an example of a system of equations that is consistent and independent.

Step-by-Step Solution

Answer

The system \( x + y = 5 \) and \( x - y = 1 \) is consistent and independent.

Step 1: Understanding Consistent and Independent Systems

A consistent system of equations has at least one solution, while an independent system has exactly one solution. This means the lines in a 2-variable system intersect at one point.

Step 2: Creating a Simple Linear System

We need to create a system where two lines intersect at one point. Consider the equations: \( x + y = 5 \) and \( x - y = 1 \).

Step 3: Solving the System Using Substitution

To solve, substitute for \( x \) in the second equation: from \( x + y = 5 \), we find \( x = 5 - y \). Substitute \( x = 5 - y \) into \( x - y = 1 \): \((5 - y) - y = 1\).

Step 4: Simplify and Solve for y

Combine like terms: \(5 - 2y = 1\). Subtract 5 from both sides: \(-2y = -4\). Divide by -2: \(y = 2\).

Step 5: Solve for x

Substitute \( y = 2 \) back into \( x + y = 5 \): \(x + 2 = 5\). Subtract 2: \(x = 3\).

Step 6: Verify the Solution

Substitute \( x = 3 \) and \( y = 2 \) into both equations to verify. For \( x + y = 5 \): \(3 + 2 = 5\) is true. For \( x - y = 1 \): \(3 - 2 = 1\) is also true. Thus, (3, 2) is the solution, confirming the system is consistent and independent.

Key Concepts

Solving Systems by Substitution

In mathematics, solving systems of equations is a common task that helps us find where different lines intersect. One effective method to solve such systems is called substitution. On this journey, we start by finding an expression for one variable in terms of the other. Let's use a simple example to illustrate this method.
Consider the system of linear equations given by:

• Equation 1: \( x + y = 5 \)
• Equation 2: \( x - y = 1 \)

To solve using substitution, first, express one variable in terms of the other in either equation. From Equation 1, we can express \( x \) as \( x = 5 - y \). Now that we have \( x \) isolated, we substitute \( 5 - y \) for \( x \) in Equation 2: \((5 - y) - y = 1\)
This results in a single equation with one variable \( y \), simplifying the system. After simplification, you can solve for \( y \), and once you have \( y \), substitute back into one of the original equations to find \( x \). This way, substitution helps transition from a system of equations to a more straightforward algebra problem.

Linear Equations

Linear equations are the foundation of algebra and define a straight line when plotted on a graph. They are powerful tools for modeling relationships between variables. Linear equations are generally in the form \( ax + by = c \), where \( a \), \( b \), and \( c \) are constants. In our example, we have two linear equations:

• \( x + y = 5 \)
• \( x - y = 1 \)

Each equation represents a line on the Cartesian plane where all solutions fall along the line it defines. The beauty of linear equations is in their simplicity: the degree of the variable is one, and their graph is a straight line. These lines can interact in different ways that determine the system's solution type. The two lines may intersect at a point, be parallel, or coincide, indicating different scenarios such as having one solution, no solution, or infinitely many solutions.

Intersection of Lines

The intersection of lines is a crucial concept in understanding systems of linear equations. When two lines intersect, they cross at a specific point showing a solution to both equations simultaneously. In our example, we see that:

• The point \((3,2)\) is where the two lines, \(x + y = 5\) and \(x - y = 1\), meet.

This intersection point is distinct and highlights that the lines are neither parallel nor coincident, categorizing the system as consistent and independent.

• A consistent system implies there is at least one solution; in this case, the intersecting point.
• The term "independent" further refines this system, indicating precisely one intersection point.

In essence, the intersection gives us the key to unlocking the solution to the equations, confirming with real-world meaning, like pinpointing the exact moment when two trains might meet on their paths.