Problem 31 - College Algebra

Question

Solve the system by Gaussian elimination. \(\frac{3}{4} x-\frac{3}{5} y=4\) \(\frac{1}{4} x+\frac{2}{3} y=1\)

Step-by-Step Solution

Answer

The solution is \(x = \frac{220}{39}\), \(y = \frac{5}{13}\).

Step 1: Write the System in Augmented Matrix Form

We start by writing the given system of equations in augmented matrix form. The system is: \[ \begin{align*} \frac{3}{4}x - \frac{3}{5}y &= 4 \\frac{1}{4}x + \frac{2}{3}y &= 1 \end{align*} \]This gives us the following augmented matrix:\[\begin{bmatrix} \frac{3}{4} & -\frac{3}{5} & \vert & 4 \\frac{1}{4} & \frac{2}{3} & \vert & 1 \end{bmatrix} \]

Step 2: Eliminate x from the Second Row

To eliminate the variable \(x\) from the second row, we need to make the first element of the second row zero. We do this by multiplying the first row by \(\frac{1}{3}\) and subtracting it from the second row:Multiply first row by \(\frac{1}{3}\):\[\begin{bmatrix} \frac{1}{4} & -\frac{1}{5} & \vert & \frac{4}{3} \end{bmatrix} \]Subtract from the second row:\[\begin{bmatrix}\frac{3}{4} & -\frac{3}{5} & \vert & 4 \0 & \frac{13}{15} & \vert & \frac{1}{3} \end{bmatrix}\]

Step 3: Solve for y in Second Equation

Now that we have a 0 in the first position of the second row, solve for \(y\) from the second equation:\[\frac{13}{15}y = \frac{1}{3} \]Multiply both sides by \(\frac{15}{13}\) to solve for \(y\):\[y = \frac{1}{3} \times \frac{15}{13} = \frac{5}{13} \]

Step 4: Substitute y Value into First Equation

Substitute \(y = \frac{5}{13}\) back into the first equation to solve for \(x\):\[\frac{3}{4}x - \frac{3}{5}\left(\frac{5}{13}\right) = 4\]Simplify and solve for \(x\):\[\frac{3}{4}x - \frac{15}{65} = 4\]\[\frac{3}{4}x = 4 + \frac{15}{65}\]Adjust the equations to have a common denominator and solve for \(x\):\[\frac{3}{4}x = 4 + \frac{3}{13}\]\[\frac{3}{4}x = \frac{52}{13} + \frac{3}{13} = \frac{55}{13}\]\[x = \frac{55}{13} \times \frac{4}{3} = \frac{220}{39}\]

Step 5: Simplify the Solution

The values of \(x\) and \(y\) can be simplified further:\(y = \frac{5}{13}\) is already in its simplest form.For \(x\), simplify \(\frac{220}{39}\):Convert to a decimal to simplify manually or verify through GCD methods:\(x \approx 5.64\).Though lack of further simplification, in rational format, maintain: \(x = \frac{220}{39}\).

Key Concepts

System of Equations

A system of equations is like a puzzle where you have several mathematical expressions that share the same set of variables. The goal is to find the values of these variables that satisfy all the equations simultaneously. In our example, the system consists of two equations:

• \( \frac{3}{4}x - \frac{3}{5}y = 4 \)
• \( \frac{1}{4}x + \frac{2}{3}y = 1 \)

These equations are linear, meaning the variables \(x\) and \(y\) are raised only to the first power. Typically, these equations intersect at a point that represents the solution—a pair \((x, y)\) that satisfies both equations. Solving them involves understanding their relationship and using techniques like elimination or substitution.

Augmented Matrix

An augmented matrix is a compact way of representing a system of linear equations. It combines the coefficient matrix with a column that includes the constants from the equations' right-hand sides. In our example:

• The coefficients for \(x\) and \(y\) are extracted from the equations.
• The constant terms (4 and 1) appear on the right side, separated by a vertical line.

The corresponding augmented matrix for this system is:\[\begin{bmatrix} \\frac{3}{4} & -\frac{3}{5} & \vert & 4 \\frac{1}{4} & \frac{2}{3} & \vert & 1 \\end{bmatrix}\]The augmented matrix format simplifies operations like row manipulations, aiding in systematic approaches to solve the system, such as Gaussian elimination.

Solving Equations Step-by-Step

Solving a system step-by-step ensures clarity and reduces errors. The step-by-step process for Gaussian elimination includes:

• Writing the system as an augmented matrix for a clear visual representation.
• Applying row operations to reach a row-echelon form.
• Further simplifying to reduce any possible variables.

Starting with the augmented matrix lets us target and eliminate variables systematically. The goal is to convert the system into a form that is easy to solve by back substitution, step-by-step ensuring we achieve a simpler, easier matrix form.

Variable Elimination

Variable elimination is a critical part of solving equations in Gaussian elimination. The process involves using mathematical operations to remove variables from specific equations, enabling us to solve for the remaining ones. For our exercise:

• We start by trying to eliminate the variable \(x\) from the second equation. We achieve this by manipulating the rows—specifically making sure a zero appears in one of the coefficients for \(x\).
• After eliminating \(x\), we are left with an equation that involves only \(y\), making it straightforward to solve for \(y\).

Once \(y\) is known, you can substitute it back into any of the original equations to find \(x\). This method ensures that the equations become progressively simpler, making it easier to zero in on a solution.