Problem 52 - Precalculus

Question

Solve the system of linear equations using Gaussian elimination with back- substitution. $$\begin{array}{r} 3 x+2 y=11 \\ x-y=12 \end{array}$$

Step-by-Step Solution

Answer

The solution is \(x = 7\) and \(y = -5\).

Step 1: Write the Augmented Matrix

First, let’s convert the system of equations into an augmented matrix. Our system is: \[\begin{array}{r} 3x + 2y = 11 \ x - y = 12 \end{array}\] The augmented matrix will be: \[\begin{bmatrix} 3 & 2 & | & 11 \ 1 & -1 & | & 12 \end{bmatrix} \]

Step 2: Eliminate x from the second row

We aim to make the element under the first pivot (which is 3) equal to zero. Use row operations to achieve this. We can replace Row 2 (R2) with \(R2 - \frac{1}{3}R1\). \[\begin{bmatrix} 3 & 2 & | & 11 \ 1 - \frac{1}{3}(3) & -1 - \frac{1}{3}(2) & | & 12 - \frac{1}{3}(11) \end{bmatrix}\] Simplifying gives: \[\begin{bmatrix} 3 & 2 & | & 11 \ 0 & -\frac{5}{3} & | & \frac{25}{3} \end{bmatrix}\]

Step 3: Solve for y using Back-Substitution

Now, take the second row of the matrix \(0\, -\frac{5}{3}y = \frac{25}{3}\) to solve for \(y\). This simplifies to \[-\frac{5}{3}y = \frac{25}{3}\] Dividing both sides by \(-\frac{5}{3}\) gives: \[y = -5\]

Step 4: Substitute y to find x

Substitute \(y = -5\) back into the first equation \(3x + 2y = 11\) to solve for \(x\). \[3x + 2(-5) = 11\] This simplifies to: \[3x - 10 = 11\] Add 10 to both sides to obtain: \[3x = 21\] Finally, divide by 3: \[x = 7\]

Key Concepts

Linear Equations

Linear equations are mathematical statements that involve variables, usually in the form of \( x, y, \) or \( z, \) and constants. Each term in a linear equation is either constant or the product of a constant and a single variable. These equations are linear, meaning they graph as straight lines on the coordinate plane.

A system of linear equations consists of multiple linear equations. The solution to the system is a specific set of values for the variables, making each equation true. In our exercise, we deal with two linear equations:

• \(3x + 2y = 11\)
• \(x - y = 12\)

These equations share the same variables \(x\) and \(y\) and represent two lines. The intersection of these lines will provide the solution or set of solutions for the variables. Using techniques like substitution, elimination, or matrix operations such as Gaussian elimination, lets us find these solutions.

Augmented Matrix

An augmented matrix is a convenient way to represent and solve systems of linear equations using matrix operations. To form an augmented matrix, you place the coefficients of the variables and the constants (the right-hand sides of the equations) into a matrix format.

For our problem, converting the system of equations into an augmented matrix involves organizing the coefficients and constants into a matrix separated by a vertical line:
\[\begin{bmatrix} 3 & 2 & | & 11 \ 1 & -1 & | & 12 \\end{bmatrix}\]

In this matrix, each row corresponds to an equation, each column corresponds to coefficients of a variable, and the vertical line separates the left side of the equations from the right side. This makes it easier to apply row operations systematically to solve the equations.

Back-Substitution

Back-substitution is a method used to find the solution of a system of linear equations after transforming the augmented matrix into a simpler form, such as row-echelon form, through Gaussian elimination. Once you have a matrix in this form, you start solving the system from the bottom up.

In our scenario, after performing row operations, the augmented matrix becomes:
\[\begin{bmatrix} 3 & 2 & | & 11 \ 0 & -\frac{5}{3} & | & \frac{25}{3} \\end{bmatrix}\]

Here, the second row gives the equation \(-\frac{5}{3}y = \frac{25}{3}\). Solving for \(y\) gives \(y = -5\). Then, using this value of \(y\), we substitute back into the first equation to find \(x\):

• Substituting \(y = -5\) into \(3x + 2y = 11\)
• We get \(3x + 2(-5) = 11\)
• Which simplifies to \(3x = 21\)
• Finally, \(x = 7\)

Thus, using back-substitution, we find that the solution to the system is \(x = 7\) and \(y = -5\).