Problem 32 - College Algebra with Corequisite Support

Question

For the following exercises, solve the system by Gaussian elimination. $$ \begin{array}{l} \frac{1}{4} x-\frac{2}{3} y=-1 \\ \frac{1}{2} x+\frac{1}{3} y=3 \end{array} $$

Step-by-Step Solution

Answer

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

Step 1: Write the System in Matrix Form

To solve the system using Gaussian elimination, first write the system of equations in augmented matrix form:\[ \begin{bmatrix} \frac{1}{4} & -\frac{2}{3} & | & -1 \ \frac{1}{2} & \frac{1}{3} & | & 3 \end{bmatrix} \]

Step 2: Eliminate the Leading Coefficient of the Second Row

Multiply the first row by 2 to make the calculation easier for elimination:\[ \begin{bmatrix} \frac{1}{2} & -\frac{4}{3} & | & -2 \ \frac{1}{2} & \frac{1}{3} & | & 3 \end{bmatrix} \]Subtract the new first row from the second row to eliminate the leading coefficient of the second row:\[ \begin{bmatrix} \frac{1}{2} & -\frac{4}{3} & | & -2 \ 0 & \frac{5}{3} & | & 5 \end{bmatrix} \]

Step 3: Solve for y

Given that the second row of the matrix is equivalent to \[ 0x + \frac{5}{3}y = 5 \], solve for \( y \):\[ \frac{5}{3}y = 5 \quad \Rightarrow \quad y = \frac{5 \times 3}{5} \quad \Rightarrow \quad y = 3 \]

Step 4: Back-Substitute to Solve for x

Substitute \( y = 3 \) back into the first equation to solve for \( x \):\[ \frac{1}{2}x - \frac{4}{3}(3) = -2 \]Simplify and solve for \( x \):\[ \frac{1}{2}x - 4 = -2 \quad \Rightarrow \quad \frac{1}{2}x = 2 \quad \Rightarrow \quad x = 4 \]

Step 5: Verify the Solution

Check that the solution satisfies both original equations.For the first equation: \( \frac{1}{4}(4) - \frac{2}{3}(3) = 1 - 2 = -1 \) (True)For the second equation: \( \frac{1}{2}(4) + \frac{1}{3}(3) = 2 + 1 = 3 \) (True)Both equations are satisfied, confirming the solution is correct.

Key Concepts

System of Equations

A system of equations is a set of two or more equations that share the same variables. In our example, we have two equations involving two variables: \( x \) and \( y \). The goal is to find values for \( x \) and \( y \) that make both equations true at the same time.
Understanding the format of these equations is crucial. Notice the variables are lined up in the same order in each equation:

• The first equation: \( \frac{1}{4} x - \frac{2}{3} y = -1 \)
• The second equation: \( \frac{1}{2} x + \frac{1}{3} y = 3 \)

Systems of equations can be solved using various methods like substitution, elimination, or the method we're focusing on here: Gaussian elimination. This method transforms the equations into a matrix format, making it easier to manipulate and solve.

Matrix Form

Transforming a system of equations into matrix form involves representing the equations as rows in a matrix. This is called an augmented matrix. It includes both the coefficients of the variables and the constants from each equation.
For our system:
\[ \begin{bmatrix} \frac{1}{4} & -\frac{2}{3} & | & -1 \ \frac{1}{2} & \frac{1}{3} & | & 3 \end{bmatrix} \]
The left side of the vertical line contains the coefficients of \( x \) and \( y \), while the right side lists the constants on the right-hand side of the equations.
Matrix form simplifies computation and helps us perform row operations more efficiently. These operations aim to achieve an upper triangular matrix, eventually leading us to an easy solution through back-substitution.

Back-Substitution

Once the matrix is converted into an upper triangular form, back-substitution begins. This process involves solving the equations from the bottom row upwards.
In the solution, the matrix was reduced to:
\[ 0x + \frac{5}{3}y = 5 \]
We can easily solve this for \( y \) because this equation now has only one variable.
After finding \( y \), back-substitute the value of \( y \) into the earlier equations to solve for \( x \). This approach simplifies solving sequentially as each step relies on previously calculated values. In our example, once \( y = 3 \) was found, substituting back into the first equation enabled quick calculation of \( x = 4 \).

Solution Verification

Verification is a critical but often overlooked step in solving systems of equations. After finding solutions for the variables, it's essential to check if they satisfy the original equations.
Substitute the values found for \( x \) and \( y \) into each original equation:

• For the first equation: \( \frac{1}{4}(4) - \frac{2}{3}(3) = 1 - 2 = -1 \), which holds true.
• For the second equation: \( \frac{1}{2}(4) + \frac{1}{3}(3) = 2 + 1 = 3 \), also correct.

This confirmation ensures that no errors occurred during the process and that the solutions are indeed valid. It's a good practice to make verification a routine part of solving any system of equations.