Problem 50 - A Graphical Approach to College Algebra

Question

For what value(s) of \(k\) will the following system of linear equations have no solution? infinitely many solutions? $$\begin{array}{r}x-2 y=3 \\\\-2 x+4 y=k\end{array}$$

Step-by-Step Solution

Answer

No solutions for \(k \neq 6\); infinitely many for \(k = 6\).

Step 1: Analyze the System of Equations

We have the system of linear equations: \( x - 2y = 3 \) and \( -2x + 4y = k \). We need to determine conditions on \( k \) for the system to have no solutions or infinitely many solutions.

Step 2: Identify the Coefficient Matrix

Write the equations in the standard form and arrange the coefficients into a matrix. The coefficients are: \( A = \begin{bmatrix} 1 & -2 \ -2 & 4 \end{bmatrix} \).

Step 3: Check for Consistent or Inconsistent Conditions

Inconsistent systems occur when the coefficient matrix of the equations is such that they correspond to parallel lines when expressed in slope-intercept form. We focus on determining if equations are scalar multiples.

Step 4: Determine Scalar Multiples

Multiply the first equation by 2: \( 2(x - 2y) = 6 \), simplifying to \( 2x - 4y = 6 \). Compare with \(-2x + 4y = k\).

Step 5: Parallel Lines Condition

If the left-hand sides are scalar multiples, compare \(2x - 4y = 6\) with \( -2x + 4y = k \). Factoring out common multiples, we get \( -1 \times (2x - 4y) = -1 imes 6 \). For parallel lines and no solutions, require \(k eq 6\).

Step 6: Infinitely Many Solutions Condition

For infinitely many solutions, the equations are equivalent fully. Thus, the second equation must be \(-1\times\) the first fully, meaning right hand sides must also match, which gives \(k = 6\).

Key Concepts

Understanding the System of Equations

A system of equations consists of two or more equations that have shared variables. In our example, the system includes the equations:

• \(x - 2y = 3\)
• \(-2x + 4y = k\)

These equations involve two variables, \(x\) and \(y\), and one parameter, \(k\). By solving a system of equations, we find values for the variables that satisfy both equations simultaneously.
There are several possibilities when dealing with a system of equations:

• Unique Solution: One specific set of values for \(x\) and \(y\) that satisfy both equations.
• No Solution: The equations represent parallel lines that never intersect.
• Infinitely Many Solutions: The equations represent the same line, hence intersect at infinitely many points.

In this context, understanding how the parameter \(k\) affects the solutions is crucial.

Exploring Solutions of Linear Equations

Solutions of linear equations in a system can reveal whether the equations intersect at a single point, overlap as the same line, or are parallel and never meet.
For our given linear equations:- If the lines intersect at a single point, there is a unique solution.- If they are parallel and never intersect, then the system has no solution.
- If they are exactly the same line, then there are infinitely many solutions.
In our task, we manipulate the first equation to compare with the second. By identifying the coefficients and modifying the equations, we determine the conditions for \(k\) where solutions change:

• No Solution: Occurs when the equations are scalar multiples without equating right-hand sides, hence \(k eq 6\).
• Infinitely Many Solutions: Occurs when they are equivalent equations, leading to \(k = 6\).

Understanding these possible outcomes helps tailor the approach needed based on the altered equation and values.

Role of the Coefficient Matrix

The coefficient matrix provides a compact representation of the system. For our system:\[A = \begin{bmatrix} 1 & -2 \ -2 & 4 \end{bmatrix}\] This matrix captures the coefficients of \(x\) and \(y\) in each equation.
Analyzing this matrix helps in understanding the structure and potential solutions of the system:

• Consistent System: Occurs if the determinant is not zero or the system is algebraically manipulated to equivalent forms like scalar multiples that match entirely.
• Inconsistent System: Observed when the rows indicate parallel lines due to scalar multiples that don't also equate constants.

Using matrix operations or determinant evaluations can guide us to understand whether the system will yield solutions. Here, by multiplying and manipulating the rows, we align the terms to predict for variable \(k\) cases.