Problem 42 - College Algebra

Question

For the following exercises, graph both straight lines (left-hand side being \(y_{1}\) and right-hand side being \(y_{2}\) ) on the same axes. Find the poin of intersection and solve the inequality by observing where it is true comparing the \(y\) -values of the lines. $$ 4 x+1<\frac{1}{2} x+3 $$

Step-by-Step Solution

Answer

The inequality is true for \( x < \frac{4}{7} \).

Step 1: Express Lines as Equations

First, express both sides of the inequality as separate equations. We have:\[ y_1 = 4x + 1 \]and\[ y_2 = \frac{1}{2}x + 3 \].

Step 2: Find the Intersection Point

Set the expressions for \( y_1 \) and \( y_2 \) equal to each other to find the point of intersection.\[ 4x + 1 = \frac{1}{2}x + 3 \]Solve for \( x \):\[ 4x - \frac{1}{2}x = 3 - 1 \]\[ \left(4 - \frac{1}{2}\right)x = 2 \]\[ \frac{7}{2}x = 2 \]\[ x = \frac{4}{7} \].Substitute \( x = \frac{4}{7} \) back into either \( y_1 \) or \( y_2 \) to find the corresponding \( y \)-coordinate:\[ y_1 = 4\left(\frac{4}{7}\right) + 1 = \frac{16}{7} + 1 = \frac{23}{7} \].Thus, the point of intersection is \( \left( \frac{4}{7}, \frac{23}{7} \right) \).

Step 3: Graph the Lines

Graph the lines \( y_1 = 4x + 1 \) and \( y_2 = \frac{1}{2}x + 3 \) on the same axes. The line \( y_1 = 4x + 1 \) has a steeper slope compared to \( y_2 = \frac{1}{2}x + 3 \), which is shallower and has a higher \( y \)-intercept.

Step 4: Solve the Inequality

Based on the graphs: \( y_1 \) is the line that is higher (greater) than \( y_2 \) when \( x \) is less than \( \frac{4}{7} \).Thus, \( 4x + 1 < \frac{1}{2}x + 3 \) is true for \( x < \frac{4}{7} \).

Key Concepts

graphing inequalities

Graphing inequalities is a technique used to visualize where one mathematical expression is greater or less than another. This is done by graphing each side of the inequality as a separate line on the Cartesian coordinate system. To start, you need to change the inequality into two equations. In this context:

• Equation 1: \(y_1 = 4x + 1\)
• Equation 2: \(y_2 = \frac{1}{2}x + 3\)

Once both equations are graphed, you can observe how these lines intersect and compare their positions. The inequality is solved by finding which segments of the graph satisfy the inequality condition, telling us where one line is above the other. Using the visual approach of graphing not only clarifies the solution but also helps in understanding the relationship between the expressions involved.

intersection point

Finding the intersection point is essential when solving systems of equations graphically, especially for inequalities. The intersection is the point where the graphs of two equations meet, sharing both \(x\) and \(y\) coordinates. To calculate this:

• Set the equations equal: \(4x + 1 = \frac{1}{2}x + 3\)
• Solve for \(x\): Simplify and rearrange terms to isolate \(x\)
• Determine the \(y\) value by substituting \(x\) back into either equation

For our specific equations, the calculated intersection point is \(\left(\frac{4}{7}, \frac{23}{7}\right)\). This point is critical as it will divide the graph into areas where one line is greater than the other, aiding in solving the inequality.

linear equations

Linear equations are equations of the first degree, meaning they graph as straight lines. The general form is \(y = mx + b\), where:

• \(m\) is the slope, indicating the steepness and direction of the line
• \(b\) is the y-intercept, showing where the line crosses the y-axis

For the given exercise:

• \(y_1 = 4x + 1\): The slope \(m = 4\) indicates a steeper incline
• \(y_2 = \frac{1}{2}x + 3\): The slope \(m = \frac{1}{2}\) denotes a gentler slope, with a higher starting point on the y-axis

Understanding linear equations is fundamental, as they form the basis for graphing inequalities and finding intersection points.

algebraic solutions

Algebraic solutions involve manipulating equations to find specific values for variables that satisfy a system of equations or inequalities. For solving the inequality \(4x + 1 < \frac{1}{2}x + 3\), we rely on algebra:

• Start by equating the two expressions to find the intersection; this involves basic algebraic steps of subtraction and rearrangement.
• Once \(x = \frac{4}{7}\) is found, test regions on the number line to understand where the inequality holds.

The algebraic manipulation provides a clear mathematical pathway to the solutions, while graphing offers a visual confirmation. Together, these methods form a comprehensive approach to understanding and solving inequalities.