Problem 20 - Elementary Algebra

Question

\(y=-\frac{2}{3} x+\frac{1}{3} ;\left(1, \frac{1}{3}\right),(0,1),(5,-3),\left(\frac{1}{2}, 0\right),(-1,0)\)

Step-by-Step Solution

Answer

The points (5, -3) and (\frac{1}{2}, 0) are on the line.

Step 1: Understanding the Function

The given function is a linear equation in the form of \( y = mx + b \), where \( m \) is the slope and \( b \) is the y-intercept. In this case, \( m = -\frac{2}{3} \) and \( b = \frac{1}{3} \).

Step 2: Verify the Points

We need to check whether each point satisfies the equation \( y = -\frac{2}{3}x + \frac{1}{3} \). This means verifying if substituting \( x \) and \( y \) from each point into the equation holds true.

Step 3: Check Point (1, \frac{1}{3})

Substitute \( x = 1 \) and \( y = \frac{1}{3} \) into the equation: \( \frac{1}{3} = -\frac{2}{3} * 1 + \frac{1}{3} \). Calculating, \( -\frac{2}{3} + \frac{1}{3} = -\frac{1}{3} \), which does not equal \( \frac{1}{3} \). So, this point is not on the line.

Step 4: Check Point (0, 1)

Substitute \( x = 0 \) and \( y = 1 \) into the equation: \( 1 = -\frac{2}{3} * 0 + \frac{1}{3} \). Calculating, \( 0 + \frac{1}{3} = \frac{1}{3} \), which does not equal 1. So, this point is not on the line.

Step 5: Check Point (5, -3)

Substitute \( x = 5 \) and \( y = -3 \) into the equation: \( -3 = -\frac{2}{3} * 5 + \frac{1}{3} \). Calculating, \( -\frac{10}{3} + \frac{1}{3} = -\frac{9}{3} = -3 \). So, this point is on the line.

Step 6: Check Point (\frac{1}{2}, 0)

Substitute \( x = \frac{1}{2} \) and \( y = 0 \) into the equation: \( 0 = -\frac{2}{3} * \frac{1}{2} + \frac{1}{3} \). Calculating, \( -\frac{1}{3} + \frac{1}{3} = 0 \). So, this point is on the line.

Step 7: Check Point (-1, 0)

Substitute \( x = -1 \) and \( y = 0 \) into the equation: \( 0 = -\frac{2}{3} * -1 + \frac{1}{3} \). Calculating, \( \frac{2}{3} + \frac{1}{3} = 1 \), which does not equal 0. So, this point is not on the line.

Key Concepts

Slope-Intercept Form

Slope-intercept form is a straightforward way to describe a linear equation. It is written as \( y = mx + b \). Here, \( m \) represents the slope of the line, and \( b \) is the y-intercept. The y-intercept is the point where the line crosses the y-axis. This form makes it easy to understand and graph the line quickly.
In the given equation \( y = -\frac{2}{3}x + \frac{1}{3} \), the slope \( m \) is \(-\frac{2}{3}\). This indicates that for every unit increase in \( x \), the value of \( y \) decreases by \( \frac{2}{3} \).
The y-intercept \( b \) is \( \frac{1}{3} \), meaning the line crosses the y-axis at the point \( (0, \frac{1}{3}) \). Understanding these components will help you graph the line accurately.

Verifying Points

Verifying points involves checking whether a given point lies on the line. This is done by substituting the \( x \) and \( y \) values of the point into the equation and seeing if the equation holds true. When substituting, if the left side of the equation equals the right side, then the point is on the line.
For example, for the point \((5, -3)\): substitute \( x = 5 \) and \( y = -3 \) into the equation \( y = -\frac{2}{3}x + \frac{1}{3} \).

• Calculating: \(-3 = -\frac{2}{3} \times 5 + \frac{1}{3} = -\frac{10}{3} + \frac{1}{3} = -\frac{9}{3} = -3 \).

This result shows the point \((5, -3)\) is on the line.
Repeating this process for different points helps verify which ones lie on the line.

Graphing Lines

Graphing lines is a visual representation of a linear equation on a coordinate plane. The process typically begins with plotting the y-intercept, followed by using the slope to identify additional points on the line. This equation \( y = -\frac{2}{3}x + \frac{1}{3} \), for example, starts at \((0, \frac{1}{3})\) on the y-axis.
From that point, apply the slope \(-\frac{2}{3}\) to find another point. Because the slope is negative, the line will tilt downward. Start at \((0, \frac{1}{3})\), move right 3 units, and down 2 units. Plot the new point and draw a straight line through both points.
This line represents all solutions to the equation. Graphing provides a clear understanding of the relationship between \( x \) and \( y \) in the equation. It also helps visually confirm which points lie on the line.