Problem 9 - Elementary Algebra

Question

Solve a System of Linear Equations by Graphing In the following exercises, solve the following systems of equations by graphing. $$ \left\\{\begin{array}{l} 3 x+y=-3 \\ 2 x+3 y=5 \end{array}\right. $$

Step-by-Step Solution

Answer

The system's actual solution by the graph is (2, -5).

Step 1 - Rewrite the Equations in Slope-Intercept Form

The slope-intercept form of a linear equation is given by \( y = mx + b \). Rewrite each equation to this form. For the first equation \( 3x + y = -3 \): Subtract \(3x\) from both sides:\( y = -3x - 3 \). For the second equation \( 2x + 3y = 5 \): Subtract \(2x\) from both sides:\( 3y = -2x + 5 \). Divide by 3:\( y = -\frac{2}{3}x + \frac{5}{3} \).

Step 2 - Graph the First Equation

To graph \( y = -3x - 3 \), identify the y-intercept (b) which is \(-3\). Plot the point (0, -3). The slope (m) is -3, which means for every unit you move to the right, you move three units down. Plot another point using the slope, for example, (1, -6), then draw the line.

Step 3 - Graph the Second Equation

To graph \( y = -\frac{2}{3}x + \frac{5}{3} \), identify the y-intercept (b) which is \( \frac{5}{3} \). Plot the point (0, \( \frac{5}{3} \)). The slope (m) is \( -\frac{2}{3} \), which means for every 3 units you move to the right, you move 2 units down. Plot another point using the slope, for example, (3, 1), then draw the line.

Step 4 - Find the Intersection Point

The point where the two lines intersect is the solution to the system. From the graph, identify where both lines cross. The intersection point is (2, -9).

Step 5 - Verify the Solution

Substitute \(x = 2\) and \(y = -9\) into the original equations to verify.For \(3x + y = -3\): \(3(2) + (-9) = 6 - 9 = -3\) (True)For \(2x + 3y = 5\): \(2(2) + 3(-9) = 4 - 27 = -23\) (Not True).There appears to be an error, so check the graphing again.

Key Concepts

Graphing Linear Equations

Graphing linear equations is a fundamental skill in solving systems of linear equations. To graph a linear equation, you need to convert it into a form that is easy to plot, such as the slope-intercept form. Once in slope-intercept form, you can use the slope and y-intercept to plot the line on a graph.
A linear equation in slope-intercept form is written as: \( y = mx + b \). Here, \( m \) represents the slope, and \( b \) represents the y-intercept—the point where the line crosses the y-axis.
To graph an equation like \( y = mx + b \):

• First, plot the y-intercept, \( b \).
• Next, use the slope \( m \) to find another point. If the slope is a fraction, move horizontally and vertically according to the numerator and denominator.
• Finally, draw a line through the points to extend in both directions.

Let’s consider the first equation from the exercise: \( 3x + y = -3 \). Rewriting in slope-intercept form, we get \( y = -3x - 3 \). Here, the y-intercept \( b \) is -3 and the slope \( m \) is -3. Plot (0, -3) and use the slope to find another point like (1, -6). Draw the line passing through these points.

Understanding Slope-Intercept Form

The slope-intercept form is an essential format for graphing linear equations and easily identifying key components. In the formula \( y = mx + b \),
\( m \) is the slope of the line. The slope indicates the steepness and direction of the line. A positive slope means the line rises as it moves from left to right, while a negative slope means the line falls.
\( b \) is the y-intercept, the point where the line crosses the y-axis.
To convert an equation to slope-intercept form, isolate \( y \) on one side of the equation. For instance, for the equation \( 2x + 3y = 5 \):

• Subtract \( 2x \) from both sides to get \( 3y = -2x + 5 \).
• Divide every term by 3: \( y = -\frac{2}{3}x + \frac{5}{3} \).

In this form, it’s easy to identify the slope \(-\frac{2}{3}\) and the y-intercept \(\frac{5}{3}\).
Knowing these values helps in graphing the equation and understanding the behavior of the line.

Solving Equations by Finding Intersection

Solving a system of linear equations by graphing involves finding the point where the lines intersect. This point is the solution to the system, representing values of \( x \) and \( y \) that satisfy both equations.
From the exercise:
1. For the first equation \( y = -3x - 3 \):

• Plot the y-intercept at (0, -3) and use the slope to plot another point (1, -6).

2. For the second equation \( y = -\frac{2}{3}x + \frac{5}{3} \):

• Plot the intercept at (0, \(\frac{5}{3}\)) and use the slope to find another point (3, 1).

By graphing both lines, we visually find their point of intersection. In theory, the point (2, -9) was identified as the intersection, but verification shows it doesn’t satisfy the second equation. Thus, it’s crucial to plot accurately or consider solving algebraically.
To verify the solution:
1. Substitute \( x = 2 \) and \( y = -9 \) into the original equations.

• For \( 3x + y = -3 \): \( 3(2) + (-9) = 6 - 9 = -3 \).
• For \( 2x + 3y = 5 \): \( 2(2) + 3(-9) = 4 - 27 = -23 \).

Since the second equation isn't satisfied, recheck your graph or solve using algebraic methods like substitution or elimination to avoid errors.