Problem 2 - Precalculus

Question

In Exercises \(1-8\), take a trip down memory lane and solve the given system using substitution and/or elimination. Classify each system as consistent independent, consistent dependent, or inconsistent. Check your answers both algebraically and graphically. $$ \left\\{\begin{aligned} 2 y-3 x &=1 \\ y &=-3 \end{aligned}\right. $$

Step-by-Step Solution

Answer

The system is consistent and independent; the solution is \( x = -\frac{7}{3}, y = -3 \).

Step 1: Solve for y in the first equation

Rearrange the first equation to express one variable in terms of the other:\[ 2y - 3x = 1 \]Express \( y \) in terms of \( x \):\[ y = \frac{3}{2}x + \frac{1}{2} \]

Step 2: Substitute the second equation into the expression for y

Since the second equation is \( y = -3 \), substitute this value into the expression for \( y \) obtained from the first equation:\[ -3 = \frac{3}{2}x + \frac{1}{2} \]

Step 3: Solve for x

Solve the equation from the substitution to find \( x \):Subtract \( \frac{1}{2} \) from both sides:\[ -3 - \frac{1}{2} = \frac{3}{2}x \]Convert \( -3 \) into a fraction with a denominator of 2:\[ -\frac{6}{2} - \frac{1}{2} = \frac{3}{2}x \]Combine the fractions:\[ -\frac{7}{2} = \frac{3}{2}x \]Multiply both sides by \( \frac{2}{3} \) to solve for \( x \):\[ x = -\frac{7}{3} \]

Step 4: Verify the solution algebraically

Substitute \( x = -\frac{7}{3} \) back into the first equation to verify whether it satisfies:\[ 2y - 3(-\frac{7}{3}) = 1 \]This gives:\[ 2y + 7 = 1 \]Subtract 7 from both sides:\[ 2y = -6 \]Divide both sides by 2:\[ y = -3 \]This matches the second equation confirming the solution is consistent.

Step 5: Classify the system

The system is consistent and independent because it has a unique solution. The solution satisfies both equations.

Step 6: Check the solution graphically

Graph the lines represented by the equations:1. The line \( y = \frac{3}{2}x + \frac{1}{2} \)2. The horizontal line \( y = -3 \)The two lines intersect at the point \((-\frac{7}{3}, -3)\), confirming that the solution is correct and verifying the classification.

Key Concepts

Substitution Method

The substitution method is a straightforward approach used to solve systems of equations algebraically. The primary aim is to express one variable in terms of the other and substitute it back into another equation. In essence, you take one equation from the system, rearrange it to make one variable the subject, and then replace this variable in the second equation with the expression you just found.
In the given system of equations, we first express \( y \) in terms of \( x \) from the first equation: \( 2y - 3x = 1 \). Rewriting this gives: \( y = \frac{3}{2}x + \frac{1}{2} \). Then, using the substitution method, we substitute \( y = -3 \) from the second equation into this expression to find \( x \).
Advantages of this method include:

• Simplicity when one equation is easily solvable for one variable.
• Reduces the number of equations needed to solve for each variable.

However, this method might become cumbersome if the equations are not straightforward to manipulate for one of the variables.

Elimination Method

The elimination method, in contrast to substitution, involves combining the equations directly to eliminate one of the variables. By aligning the equations to allow direct subtraction or addition, you can remove one variable and solve for the other seamlessly.
Suppose we didn't use the substitution method and chose to apply elimination here. We could multiply the equations appropriately to align coefficients of either \( y \) or \( x \) so they can be canceled out by subtraction or addition.
This method can be more efficient when:

• Coefficients of the variables are easily manipulated to allow for cancellation.
• Both equations are already in a similar format.

Despite its utility, the elimination method is less suitable for systems with complex coefficients or where the equations aren't easily manipulated to cancel a variable.

Consistent and Independent Systems

A consistent and independent system of equations is one where you have exactly one solution. This means the equations intersect at a single unique point on a graph. In the given example, when solving the equations, we get \( x = -\frac{7}{3} \) and \( y = -3 \), which means our system is consistent, as the solution satisfies both equations.
The characteristics of such systems include:

• Unique solutions, implying no contradictions.
• Graphically, they are represented by intersecting lines at a single point.

Systems are important because knowing whether a system is consistent and independent helps decide which method might be more effective for solving it, and informs us about the relationship between the equations.

Graphical Solution Verification

Verifying solutions graphically involves plotting each equation on a coordinate plane and observing the point of intersection. By visualizing the system on a graph, you can see if the lines intersect (one solution), are parallel (no solution), or are the same line (infinitely many solutions).
In our example, the first equation \( y = \frac{3}{2}x + \frac{1}{2} \) and the second equation \( y = -3 \) are plotted. The intersection confirms the algebraic solution at the point \( \left(-\frac{7}{3}, -3\right)\), indicating that the system is consistent and independent.
The benefits of graphical verification include:

• Providing a visual confirmation of the solution's accuracy.
• Offering an intuitive understanding of systems with multiple solutions or no solutions.

This method is particularly advantageous when teaching or introducing the concept of systems of equations, as it allows for a clear visual demonstration of how they work.