Problem 41 - College Algebra with Corequisite Support

Question

For the following exercises, graph both straight lines (left-hand side being y1 and right-hand side being \(y 2\) ) on the same axes. Find the point of intersection and solve the inequality by observing where it is true comparing the \(y\) -values of the lines. $$ \frac{1}{2} x+1>\frac{1}{2} x-5 $$

Step-by-Step Solution

Answer

The inequality is always true, as the lines are parallel and \( y_1 > y_2 \) for any \( x \).

Step 1: Identify the Equations of Lines

The given inequality is \( \frac{1}{2} x + 1 > \frac{1}{2} x - 5 \). Identify the left-hand side as the first line, \( y_1 = \frac{1}{2} x + 1 \), and the right-hand side as the second line, \( y_2 = \frac{1}{2} x - 5 \).

Step 2: Graph Both Lines

Graph both lines on the same axes. Notice that both lines have the same slope \( \frac{1}{2} \), indicating they are parallel. The line \( y_1 = \frac{1}{2} x + 1 \) passes through the point (0,1) and \( y_2 = \frac{1}{2} x - 5 \) passes through the point (0,-5). Since they are parallel, they do not intersect.

Step 3: Compare Y-values of the Lines

Since the lines are parallel and never intersect, compare their \(y\)-values. For any \(x\), \( y_1 = \frac{1}{2}x + 1 \) is always greater than \( y_2 = \frac{1}{2}x - 5 \).

Step 4: Solve the Inequality

Determine where the inequality \( \frac{1}{2} x + 1 > \frac{1}{2} x - 5 \) holds true. The inequality simplifies to \( 1 > -5 \). Hence, it is always true for any value of \( x \).

Key Concepts

Parallel Lines

Parallel lines are lines in the same plane that never meet. This is because they have the same slope, meaning they rise and run at the same angle. Two lines with identical slopes will always be equidistant from each other, regardless of the direction they extend.
In the exercise, we see two lines: \( y_1 = \frac{1}{2}x + 1 \) and \( y_2 = \frac{1}{2}x - 5 \). Both have the slope \( \frac{1}{2} \), indicating that they are parallel.

• A constant slope ensures parallelism.
• Parallel lines never cross paths at any point.

The only difference between these lines is their y-intercepts, making them parallel but distinct in position.

Point of Intersection

In general, the point of intersection refers to where two lines on a graph meet, which can help us solve equations by finding common values of \( x \) and \( y \). However, for parallel lines, a point of intersection doesn't exist because they do not share any common point.
In the given exercise, although we graph both lines, we notice they do not meet; thus, there is no intersection point.
This concept highlights the distinct path that each parallel line takes, maintaining a constant distance from each other over the entire graph.

Graphing Techniques

Graphing is a valuable technique in understanding linear equations and inequalities. To graph any linear equation, we start by finding two key points: the y-intercept and an additional point using the slope.
For the equations in our exercise:

• Start at the y-intercept, the point where the line crosses the y-axis. For \( y_1 \), it's \( (0, 1) \), and for \( y_2 \), it's \( (0, -5) \).
• Use the slope (\( \frac{1}{2} \)) to determine the direction of the line. It means for every 1 unit you move horizontally, you move up by 0.5 units vertically.

By following these steps, you can accurately portray and analyze linear inequalities on a graph, making it easier to interpret their relationships.

Linear Inequalities

Linear inequalities display the relative sizes between two algebraic expressions involving variables. They can be visually represented using graphs, similar to equations, but with shading to indicate the solution set.
The inequality \( \frac{1}{2}x + 1 > \frac{1}{2}x - 5 \) can be understood by comparing the two lines. As we saw, the line representing \( y_1 \) is always above the line for \( y_2 \) on the graph, meaning its y-values are greater for all \( x \) values.

• The solution is determined by which region is shaded, depending on the inequality sign.
• In our example, this inequality is always true, as shown by the simplified inequality \( 1 > -5 \).

By understanding and graphing linear inequalities, we comprehend the extensive area of solutions that satisfy the inequality rather than a singular line.