Problem 33
Question
Solve each system of equations by graphing. \(\frac{2}{3} x+y=-3\) \(y-\frac{1}{3} x=6\)
Step-by-Step Solution
Verified Answer
The solution to the system of equations is \((-9, 3)\).
1Step 1: Rewrite each equation in slope-intercept form
The slope-intercept form of a linear equation is given by \( y = mx + b \), where \( m \) is the slope and \( b \) is the y-intercept. Let's rewrite both equations in this form.\For the first equation \( \frac{2}{3}x + y = -3 \), subtract \( \frac{2}{3}x \) from both sides to get: \\[ y = -\frac{2}{3}x - 3 \]\This equation is now in slope-intercept form \( y = mx + b \), with slope \( m = -\frac{2}{3} \) and y-intercept \( b = -3 \).\For the second equation \( y - \frac{1}{3}x = 6 \), add \( \frac{1}{3}x \) to both sides to get: \\[ y = \frac{1}{3}x + 6 \] \This equation is in slope-intercept form \( y = mx + b \), with slope \( m = \frac{1}{3} \) and y-intercept \( b = 6 \).
2Step 2: Plot the graphs of the equations
We will plot both equations on a coordinate plane.\For the first equation \( y = -\frac{2}{3}x - 3 \): \- Start at the y-intercept \( (0, -3) \). \- Use the slope \( -\frac{2}{3} \) to plot another point. From the y-intercept, move down 2 units and right 3 units to plot another point.\For the second equation \( y = \frac{1}{3}x + 6 \):\- Start at the y-intercept \( (0, 6) \). \- Use the slope \( \frac{1}{3} \) to plot another point. From the y-intercept, move up 1 unit and right 3 units to plot another point.\Draw straight lines through the plotted points for each equation.
3Step 3: Identify the point of intersection
Observe where the two lines intersect on the graph. The coordinates of the intersection point represent the solution to the system of equations.\In this case, the lines representing the equations \( y = -\frac{2}{3}x - 3 \) and \( y = \frac{1}{3}x + 6 \) intersect at the point \((-9, 3)\). Thus, the solution to the system is \( x = -9 \) and \( y = 3 \).
Key Concepts
Graphing Linear EquationsSlope-Intercept FormPoint of Intersection
Graphing Linear Equations
Graphing linear equations involves plotting one or more lines on a coordinate plane. Each line represents a solution set of a linear equation. To create the graph, we often convert the equation to its slope-intercept form. This equation format tells us how to plot the graph quickly and efficiently.
- Start by identifying the y-intercept, which is the point where the line crosses the y-axis. This is always at \((0, b)\), with \(b\) being the y-intercept value.
- Next, use the slope of the line to determine additional points. The slope \((m)\) is a crucial piece; it indicates how steep the line is, and describes the change in \(y\) per unit change in \(x\). A positive slope means the line ascends, whereas a negative slope means it descends.
- After plotting these points, draw a straight line through them.
Slope-Intercept Form
The slope-intercept form of a linear equation is a very user-friendly format. It's written as \( y = mx + b \), where \(m\) is the slope of the line and \(b\) is the y-intercept. This form is most useful for graphing and provides clear information:
- **Slope \(m\):** It tells us the rate of increase or decrease. A larger absolute value of \(m\) means a steeper slope.
- **Y-Intercept \(b\):** Marks where the line crosses the y-axis. From here, you use the slope to find other points on the graph.
Point of Intersection
In the graph of a system of linear equations, the point of intersection is a pivotal point. It represents the solution that satisfies all equations in the system simultaneously.
- **Finding the Intersection:** On a graph, this is where the lines cross one another. The coordinates of this point correspond to the values that make both equations true.
- **Significance:** Knowing this point lets us solve systems of equations visually, confirming algebraic solutions. In our exercise, the lines of the two equations intersect at \((-9, 3)\).
- **Practical Uses:** Points of intersection are used extensively in various fields—engineering, economics, and even in simple everyday problems—to find where conditions align.
Other exercises in this chapter
Problem 33
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