Problem 26 - College Algebra

Question

For the following exercises, solve the system by Gaussian elimination. $$ \begin{aligned} 3 x+4 y &=12 \\\\-6 x-8 y &=-24 \end{aligned} $$

Step-by-Step Solution

Answer

Infinitely many solutions: \(x = \frac{12 - 4t}{3}\), \(y = t\).

Step 1: Write the System as an Augmented Matrix

The system of equations \(3x + 4y = 12\) and \(-6x - 8y = -24\) can be written as an augmented matrix:\[\begin{bmatrix}3 & 4 & | & 12 \-6 & -8 & | & -24\end{bmatrix}\]

Step 2: Begin the Elimination Process

To start the elimination, add 2 times the first row to the second row in order to create a zero in the first column of the second row. The new second row becomes:\(-6 + 2(3) = 0\) and \(-8 + 2(4) = 0\) and \(-24 + 2(12) = 0\).The new augmented matrix is:\[\begin{bmatrix}3 & 4 & | & 12 \0 & 0 & | & 0\end{bmatrix}\]

Step 3: Interpret the Matrix

The second row \(0x + 0y = 0\) is a true statement and doesn't provide new information. This implies that the original system of equations is dependent and has infinitely many solutions. The solutions can be expressed in terms of a parameter.

Step 4: Parameterize the Solution

Express \(y\) in terms of \(t\), where \(t\) is any real number. Then solve for \(x\):From the first equation \(3x + 4y = 12\), let \(y = t\).Substitute \(y\) into the equation: \[3x + 4t = 12\]Solve for \(x\):\[x = \frac{12 - 4t}{3}\]The solution set is \(x = \frac{12 - 4t}{3}\) and \(y = t\) for any real number \(t\).

Key Concepts

Augmented Matrix

In solving systems of equations, an augmented matrix is a very useful tool because it simplifies the process by stripping away variables and using coefficients. You'll find it easier to manipulate equations this way. An augmented matrix collects all the coefficients of the variables and the constants from each equation into a matrix with rows representing equations and columns representing variables. For our system:

• The equations are \(3x + 4y = 12\) and \(-6x - 8y = -24\).
• It transforms to the augmented matrix: \[ \begin{bmatrix} 3 & 4 & | & 12 \ -6 & -8 & | & -24 \end{bmatrix} \]

The vertical line separates the coefficients and the constants. The augmented matrix prepares the system for simplified arithmetic manipulations and eventual solutions through Gaussian elimination.

Dependent System

A dependent system occurs when two or more equations in the system essentially describe the same line or plane, meaning they aren't independent of one another. In simple terms, this means that the information provided by one equation is already contained within the other(s). In our exercise, after applying Gaussian elimination, the matrix simplifies to:

• \(0x + 0y = 0\)
• This row suggests that all variables cancel out, leaving a true statement \(0 = 0\).

This redundancy implies that the system is dependent, without unique solutions tied to specific values of the variables. Instead, it offers infinite possibilities for solutions.

Infinite Solutions

When we talk about infinite solutions in a system of equations, it means there are countless numbers of points that satisfy all equations simultaneously. For the given system, eliminating variables led to a scenario where no new information was added—instead, every possible solution is included. This happens because the system's equations plot as the same line in two-dimensional space. Any point on that line is a solution to the system:

• Both equations simplify to represent the same relationship between \(x\) and \(y\).
• All solutions lie along the same line, which leads us directly to parameterization.

Using infinite solutions effectively means all solutions must satisfy the initial set-up of the problem.

Parameterization

To express infinite solutions succinctly and conveniently, we use a parameter to solve such a system. Parameterization involves designating one variable as a parameter (usually \(t\)), and expressing other variables in terms of this parameter. From our system, we have:

• Choose \(y = t\), where \(t\) can be any real number.
• Substituting \(y = t\) into the remaining equation \(3x + 4y = 12\) gives us \(3x + 4t = 12\).

Now solve for \(x\):

• \(x = \frac{12 - 4t}{3}\)

The parametrized solution thus becomes:

• \(x = \frac{12 - 4t}{3}\)
• \(y = t\)

By using parameterization, we describe the entire line of solutions compactly and clearly, expressing a range of solutions in terms of a single variable \(t\). This approach is essential when dealing with infinite solutions.