Problem 32 - College Algebra

Question

Solve the system by Gaussian elimination. \(\frac{1}{4} x-\frac{2}{3} y=-1\) \(\frac{1}{2} x+\frac{1}{3} y=3\)

Step-by-Step Solution

Answer

The solution to the system is \(x = 4\) and \(y = 3\).

Step 1: Write the System of Equations in Matrix Form

First, write the given system of linear equations in augmented matrix form. The given equations are: \(\frac{1}{4}x - \frac{2}{3}y = -1\) and \(\frac{1}{2}x + \frac{1}{3}y = 3\). The corresponding augmented matrix is:\[\begin{bmatrix}\frac{1}{4} & -\frac{2}{3} & | & -1 \\frac{1}{2} & \frac{1}{3} & | & 3\end{bmatrix}\]

Step 2: Eliminate the First Column Second Row Entry

To eliminate the \(x\)-term from the second equation, we multiply the first row by 2 and subtract it from the second row, as the coefficient of \(x\) in the second equation is \(\frac{1}{2}\). Multiply the first row by 2:\[ 2 \times \begin{bmatrix} \frac{1}{4} & -\frac{2}{3} & | & -1 \end{bmatrix} = \begin{bmatrix} \frac{1}{2} & -\frac{4}{3} & | & -2 \end{bmatrix}\]Subtract this from the second row:\[\begin{bmatrix} \frac{1}{2} & \frac{1}{3} & | & 3 \end{bmatrix} - \begin{bmatrix} \frac{1}{2} & -\frac{4}{3} & | & -2 \end{bmatrix} = \begin{bmatrix} 0 & \frac{5}{3} & | & 5 \end{bmatrix}\]

Step 3: Solve for y

The modified augmented matrix is now:\[\begin{bmatrix} \frac{1}{4} & -\frac{2}{3} & | & -1 \0 & \frac{5}{3} & | & 5 \end{bmatrix}\]Divide the second row by \(\frac{5}{3}\):\[\begin{bmatrix} 0 & \frac{5}{3} & | & 5 \end{bmatrix}\div \frac{5}{3} = \begin{bmatrix} 0 & 1 & | & 3 \end{bmatrix}\]Thus, \(y = 3\).

Step 4: Substitute y Back to Solve for x

Substitute \(y = 3\) into the first equation:\[\frac{1}{4}x - \frac{2}{3}(3) = -1\]Simplify and solve for \(x\):\[\frac{1}{4}x - 2 = -1 \rightarrow \frac{1}{4}x = 1 \rightarrow x = 4\]The solution is \(x = 4\) and \(y = 3\).

Key Concepts

Linear Equations

Linear equations are like instructions to find points in a plane. Each equation relates two variables, often represented as \(x\) and \(y\). The goal of solving a system of linear equations is to find the values of these variables that make both equations true simultaneously.
The equations can involve fractions and different coefficients, which you need to balance using basic algebraic operations.

• Each linear equation represents a straight line when plotted on a graph.
• For a system of linear equations, we aim to find the intersection point of these lines, which is the solution.

In our exercise, we have two equations involving fractions, making them a bit tricky. Don't worry! We'll convert these equations into a form that's easier to work with—using the magic of Gaussian elimination!

Augmented Matrix

An augmented matrix is a compact way to display a system of linear equations. Imagine compressing the whole equations into a neat grid of numbers! This approach helps in systematically solving the equations, especially when using the Gaussian elimination method.

• The matrix consists of rows and columns, where each row represents an equation.
• Columns represent the coefficients of each variable and the constants from the equations.

In our case, the augmented matrix is:\[\begin{bmatrix} \frac{1}{4} & -\frac{2}{3} & | & -1 \ \frac{1}{2} & \frac{1}{3} & | & 3 \end{bmatrix}\]Notice the vertical line that separates the left-hand side (LHS) coefficients from the constants on the right-hand side (RHS). This visual separation helps to perform row operations more efficiently.

Row Operations

Row operations are the steps we take to transform the augmented matrix into a more manageable form. These operations are essential in Gaussian elimination.
There are three types of row operations:

• Swapping two rows.
• Multiplying a row by a non-zero constant.
• Adding or subtracting a multiple of one row to another row.

For our exercise, we performed these operations to eliminate certain coefficients and reach simpler forms. For instance, we multiplied the first row by 2 to match the coefficient of \(x\) in the second equation, then subtracted one row from another. This operation helped transform an element into zero, simplifying the equation and making it easier to solve for one of the variables.

Solution of System of Equations

Finding the solution of a system of linear equations means discovering the values of \(x\) and \(y\) that satisfy all given equations.
After transforming the augmented matrix through row operations, we get a simplified system, which is easy to solve. In our example:

• From the final row, we determined that \(y = 3\).
• Substituting \(y\) back into one of the original equations allowed us to solve for \(x\), yielding \(x = 4\).

Thus, the solution to the system is the values \(x = 4\) and \(y = 3\). This process demonstrates the effectiveness of Gaussian elimination in breaking down complex systems into solvable parts.