Problem 44
Question
(a) find the slope and y-intercept (if possible) of the equation of the line algebraically, and (b) sketch the line by hand. Use a graphing utility to verify your answers to parts (a) and (b). $$2 x+3 y-9=0$$
Step-by-Step Solution
Verified Answer
The slope of the given line is \(-\frac{2}{3}\) and the y-intercept is \(3\).
1Step 1: Convert to slope-intercept form
Rearrange the given equation \(2x + 3y - 9 = 0\) to get it into the form \(y = mx + b\). That is, isolate \(y\) by subtracting \(2x\) from both sides and then dividing by \(3\). Doing so, you get: \(y = -\frac{2}{3}x + 3\)
2Step 2: Identify the Slope and Y-Intercept
Compare the rearranged equation with the standard slope-intercept form \(y = mx + b\). The slope (m) is represented by the coefficient of \(x\), which is \(-\frac{2}{3}\). The y-intercept (b) is simply the constant term, which is \(3\) in this case.
3Step 3: Sketch the Line on a Graph
Start by marking the y-intercept (\(b = 3\)) on the y-axis. Then, from the y-intercept, use the slope (\(m = -\frac{2}{3}\)) to find the next point. This means move 2 units down (as it's negative) and 3 units to the right (as it's the denominator of the fraction). Draw the line that passes through these points.
4Step 4: Verify with a Graphing Utility
Input the equation (\(y = -\frac{2}{3}x + 3\)) into a graphing utility to verify the manual sketching. The lines should overlap, confirming the accuracy of the graph and the slope and y-intercept values.
Key Concepts
Finding SlopeY-InterceptGraphing Linear EquationsAlgebraic Manipulation
Finding Slope
In algebra, one of the fundamental concepts is the slope of a line. The slope is a measure of how steep or flat a line is on a graph. To find it, you need two points on the line. You can calculate the slope by using the formula:
\( m = \frac{{y_2 - y_1}}{{x_2 - x_1}} \), where \( m \) is the slope, and \( (x_1, y_1) \) and \( (x_2, y_2) \) are the coordinates of the two points.
In the exercise where you have the equation in standard form, you can find the slope by transforming the equation into slope-intercept form, as shown in the provided example. The coefficient of \( x \) in the equation \( y = mx + b \) is your slope. It's important to remember that the slope tells us the rise over run for the line, which delineates how many units you go up or down for each unit you move to the right on a graph.
\( m = \frac{{y_2 - y_1}}{{x_2 - x_1}} \), where \( m \) is the slope, and \( (x_1, y_1) \) and \( (x_2, y_2) \) are the coordinates of the two points.
In the exercise where you have the equation in standard form, you can find the slope by transforming the equation into slope-intercept form, as shown in the provided example. The coefficient of \( x \) in the equation \( y = mx + b \) is your slope. It's important to remember that the slope tells us the rise over run for the line, which delineates how many units you go up or down for each unit you move to the right on a graph.
Y-Intercept
The y-intercept of a line is the point where the line crosses the y-axis on a graph. It is reflected in the slope-intercept form of a line's equation as the constant term, \( b \), in \( y = mx + b \). Finding the y-intercept is straightforward when you have the equation in this form: it's simply the number that is by itself on one side of the equation, without any variables attached.
In our example, by converting the given standard form equation into slope-intercept form, we identified the y-intercept as \( 3 \). This means the line crosses the y-axis at the point \( (0, 3) \). When graphing a line, you typically start by plotting the y-intercept, since it's a fixed point that you can count on to begin drawing your line.
In our example, by converting the given standard form equation into slope-intercept form, we identified the y-intercept as \( 3 \). This means the line crosses the y-axis at the point \( (0, 3) \). When graphing a line, you typically start by plotting the y-intercept, since it's a fixed point that you can count on to begin drawing your line.
Graphing Linear Equations
The process of graphing linear equations involves plotting points on a graph that represent solutions to the equation and then connecting these points with a straight line. A linear equation can always be plotted by finding the x and y intercepts, or by using the slope and y-intercept if it's in slope-intercept form, as demonstrated in the solution.
By starting at the y-intercept and using the slope, you determine the next points by moving vertically and horizontally according to the slope's rise and run values. Repeat the process to find as many points as needed, and then draw the line through these points to complete the graph.
By starting at the y-intercept and using the slope, you determine the next points by moving vertically and horizontally according to the slope's rise and run values. Repeat the process to find as many points as needed, and then draw the line through these points to complete the graph.
Algebraic Manipulation
To solve algebraic equations, or to express them in different forms, algebraic manipulation is used. This includes operations such as adding, subtracting, multiplying, and dividing both sides of an equation by the same number or term to maintain equality. It also includes factoring, expanding expressions, and simplifying complex fractions.
In the given exercise, we manipulated the original equation from standard form into slope-intercept form through a series of steps. We subtracted \( 2x \) from both sides and then divided each term by \( 3 \) to isolate the \( y \)-variable. Mastering these manipulation skills allows for more convenient forms of an equation, enabling easier interpretation and graphing of lines.
In the given exercise, we manipulated the original equation from standard form into slope-intercept form through a series of steps. We subtracted \( 2x \) from both sides and then divided each term by \( 3 \) to isolate the \( y \)-variable. Mastering these manipulation skills allows for more convenient forms of an equation, enabling easier interpretation and graphing of lines.
Other exercises in this chapter
Problem 44
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