Problem 40 - College Algebra with Modeling and Visualization

Question

Graph each system of equations and find any solutions. Check your answers. Identify the system as consistent or inconsistent. If the system is consistent, state whether the equations are dependent or independent. $$ \begin{array}{r} 2 x-3 y=1 \\ x+y=-2 \end{array} $$

Step-by-Step Solution

Answer

The solution to the system is \((-1, -1)\). The system is consistent and independent.

Step 1: Rewrite Each Equation in Slope-Intercept Form

Convert both equations to the form \( y = mx + b \), where \( m \) is the slope and \( b \) is the y-intercept.**For the first equation**: \[ 2x - 3y = 1 \] Subtract \(2x\) from both sides: \[ -3y = -2x + 1 \] Divide every term by \(-3\): \[ y = \frac{2}{3}x - \frac{1}{3} \] **For the second equation**: \[ x + y = -2 \] Subtract \(x\) from both sides: \[ y = -x - 2 \]

Step 2: Graph Each Equation

Graph the equations derived in Step 1 using their slope-intercept forms.- **First equation**: \( y = \frac{2}{3}x - \frac{1}{3} \). - Start at the y-intercept, \(-\frac{1}{3}\). - Use the slope, \(\frac{2}{3}\), to find another point. From \(-\frac{1}{3}\), go up 2 units and right 3 units, plot the next point.- **Second equation**: \( y = -x - 2 \). - Start at the y-intercept, \(-2\). - Use the slope, \(-1\), to find another point. From \(-2\), go down 1 unit and right 1 unit, plot the next point.

Step 3: Find the Intersection Point

Observe the graph where the two lines intersect. This intersection point is the solution to the system of equations. Based on the graphing:- The lines intersect at \((-1, -1)\).

Step 4: Check the Solution

Substitute \((-1, -1)\) back into the original equations to verify.**First equation**: \[ 2(-1) - 3(-1) = -2 + 3 = 1 \] This is true.**Second equation**: \[ (-1) + (-1) = -2 \] This is also true.Since both equations are satisfied, \((-1, -1)\) is the correct solution.

Step 5: Analyze Consistency and Dependency

Since there is a single intersection point, the system is consistent. The lines are not the same or parallel, indicating the equations are independent.

Key Concepts

Graphing Linear Equations

Graphing linear equations involves drawing the graph of each equation on the same coordinate plane. This process helps to visually identify the solutions to the system.
To start, we convert equations into the slope-intercept form:

• For the first equation, transform it into: \( y = \frac{2}{3}x - \frac{1}{3} \). Start at the y-intercept, which is \( -\frac{1}{3} \), and use the slope \( \frac{2}{3} \) to find another point by moving up 2 units and right 3 units.


• The second equation becomes: \( y = -x - 2 \). Start at \( -2 \) on the y-axis and use the slope \(-1\) to move down 1 unit and right 1 unit.

Once the lines are drawn, observe where they intersect. This intersection point is crucial as it indicates the solution to the system of equations.

Consistent and Inconsistent Systems

When dealing with systems of linear equations, it is essential to determine whether the system is consistent or inconsistent. A consistent system has at least one solution, meaning the lines intersect at least once on the graph. An inconsistent system, on the other hand, has no solutions, as the lines never meet and are parallel to each other.
In our exercise, graphing reveals that the lines intersect at \((-1, -1)\). This means we have at least one solution available, confirming that the system is consistent. Finding such an intersection verifies the possibility of common solutions existing for both equations.

Independent and Dependent Equations

Once it is established that a system is consistent, the next step is to decide whether the equations are independent or dependent. Independent equations imply that the system has exactly one solution, and the lines intersect precisely at one point. This independence results from the lines having different slopes.
Dependent equations occur when the equations basically represent the same line, leading to infinite solutions, as they are one and the same on the graph.
In the exercise, the lines intersect only at \((-1, -1)\), showing they are independent. The differing slopes \( \frac{2}{3} \) and \(-1\) confirm their independence, individual paths, and uniqueness of solutions. This implies that, aside from the single intersection point, the lines continue diverging.