Problem 55 - Precalculus

Question

Solve the system of linear equations using Gaussian elimination with back- substitution. $$\begin{aligned} &\frac{2}{3} x+\frac{1}{3} y=\frac{8}{9}\\\ &\frac{1}{2} x+\frac{1}{4} y=\frac{3}{4} \end{aligned}$$

Step-by-Step Solution

Answer

\(x = \frac{4}{3}\), \(y = 0\)

Step 1: Convert to Row Echelon Form

We start by converting the system of equations into an augmented matrix. The system is:\[\begin{aligned} \frac{2}{3}x + \frac{1}{3}y &= \frac{8}{9} \ \frac{1}{2}x + \frac{1}{4}y &= \frac{3}{4} \end{aligned}\]The corresponding augmented matrix is:\[\begin{bmatrix} \frac{2}{3} & \frac{1}{3} & | & \frac{8}{9} \ \frac{1}{2} & \frac{1}{4} & | & \frac{3}{4}\end{bmatrix}\]Next, we eliminate the first element of the second row by scaling and subtracting a suitable multiple of the first row from it. Multiply the first row by \(\frac{3}{4}\) and subtract from the second row:\[\begin{bmatrix} \frac{2}{3} & \frac{1}{3} & | & \frac{8}{9} \ 0 & -\frac{1}{12} & | & 0\end{bmatrix}\]Now, the matrix is in row-echelon form.

Step 2: Back Substitution

We perform back substitution starting from the last row. The second row now reads:\[-\frac{1}{12}y = 0\]From which we get:\[y = 0\]Substitute \(y = 0\) into the first row equation:\[\frac{2}{3}x + \frac{1}{3} \cdot 0 = \frac{8}{9}\]Which simplifies to:\[\frac{2}{3}x = \frac{8}{9}\]Solve for \(x\):\[x = \frac{8}{9} \cdot \frac{3}{2} = \frac{4}{3}\]

Step 3: Verification Step

Finally, verify the solution by substituting \(x = \frac{4}{3}\) and \(y = 0\) back into the original equations:1. \[\frac{2}{3}x + \frac{1}{3}y = \frac{8}{9}\] becomes \[\frac{2}{3} \cdot \frac{4}{3} + \frac{1}{3} \cdot 0 = \frac{8}{9}\] which is true.2. \[\frac{1}{2}x + \frac{1}{4}y = \frac{3}{4}\] becomes \[\frac{1}{2} \cdot \frac{4}{3} + \frac{1}{4} \cdot 0 = \frac{3}{4}\] which is also true.Hence, the solution \(x = \frac{4}{3}\), \(y = 0\) is correct.

Key Concepts

System of Linear Equations

When we talk about a **system of linear equations**, we're referring to a collection of two or more equations that involve the same set of variables. The goal is to find values for these variables that satisfy all the equations in the system. Linear equations are characterized by variables raised to the power of one, which forms straight lines when plotted on a graph.

In this context, we have a system of two equations with two variables, namely \(x\) and \(y\). These equations show up in the popular standard form:

• \(ax + by = c\)

where \(a\), \(b\), and \(c\) are constants. For our system:

• \(\frac{2}{3}x + \frac{1}{3}y = \frac{8}{9}\)
• \(\frac{1}{2}x + \frac{1}{4}y = \frac{3}{4}\)

To solve this, one effective method is **Gaussian elimination**, which transforms the equations and helps us find the variables' values.

Augmented Matrix

Understanding an **augmented matrix** is crucial as we proceed with Gaussian elimination. An augmented matrix is a compact way to represent a system of linear equations. It combines the coefficients of the variables and the constants from the equations into a single rectangular grid.

Here's how it works for our system:

• Take the coefficients from the equations \(\frac{2}{3}x + \frac{1}{3}y = \frac{8}{9}\) and \(\frac{1}{2}x + \frac{1}{4}y = \frac{3}{4}\).
• Write them in matrix form, along with the constants separated by a line to signify the augmented part.

This gives us:o \[ \begin{bmatrix} \frac{2}{3} & \frac{1}{3} & | & \frac{8}{9} \ \frac{1}{2} & \frac{1}{4} & | & \frac{3}{4} \end{bmatrix}\]

Each row corresponds to an equation, and each column corresponds to a variable or the constants on the right-hand side. This form makes it easier to apply row operations needed for Gaussian elimination.

Row-Echelon Form

The **row-echelon form** is an intermediate step in Gaussian elimination. This form simplifies the process of solving a system of linear equations by transforming the augmented matrix into a set of triangular-like equations.

In achieving row-echelon form, we focus on creating zeros below the leading coefficients of each row. For our matrix:\[ \begin{bmatrix} \frac{2}{3} & \frac{1}{3} & | & \frac{8}{9} \ 0 & -\frac{1}{12} & | & 0 \end{bmatrix}\]To reach this form, we perform operations to eliminate the first element in the second row—achieved here by scaling and subtracting the first row from the second. The resulting matrix is structured so that we have zero below the leading coefficient of the first row. This triangular nature helps us in subsequent back substitution to find the variable's values effectively.

Back Substitution

**Back substitution** is the final stage in solving a system of linear equations after reaching row-echelon form. The concept is simple: solve the equations starting from the bottom where there are fewer variables, and work upwards.

In our context:

• The second row becomes \(-\frac{1}{12}y = 0\), leading directly to \(y = 0\).
• Substitute \(y = 0\) into the first row \(\frac{2}{3}x + \frac{1}{3} \cdot 0 = \frac{8}{9}\), simplifying to \(\frac{2}{3}x = \frac{8}{9}\).
• Solve for \(x\), finding \(x = \frac{4}{3}\).

This step-by-step reduction helps pinpoint the solution to the entire system. The strength of back substitution lies in its ability to wrap up the process simply after transforming the equations into a manageable form.