Problem 34 - A Graphical Approach to Precalculus with Limits

Question

Use row operations on an augmented matrix to solve each system of equations. Round to nearest thousandth when appropriate. $$\begin{aligned} &3 x-2 y=4\\\ &3 x+y=-2 \end{aligned}$$

Step-by-Step Solution

Answer

The solution is \(x = 0\) and \(y = -2\).

Step 1: Write the Augmented Matrix

Begin by writing the augmented matrix of the given system of equations. For the equations \(3x - 2y = 4\) and \(3x + y = -2\), the coefficients of \(x\), \(y\), and the constants on the right-hand side form the matrix: \[\begin{bmatrix}3 & -2 & | & 4 \3 & 1 & | & -2\end{bmatrix}\] Here, the columns represent \(x\), \(y\), and the constant, respectively.

Step 2: Eliminate x from the Second Row

Perform the first row operation to eliminate \(x\) from the second equation. Subtract Row 1 from Row 2: \[R_2 = R_2 - R_1\]Applying the operation yields:\[\begin{bmatrix}3 & -2 & | & 4 \0 & 3 & | & -6\end{bmatrix}\] Now the \(x\) entry in the second row is 0, preparing for back substitution.

Step 3: Solve for y in the Second Row

The second row now reads \(0x + 3y = -6\). Solve for \(y\):\[3y = -6 \y = \frac{-6}{3} = -2\] The value of \(y\) is -2.

Step 4: Substitute y into the First Equation

Now substitute \(y = -2\) into the first equation \(3x - 2y = 4\):\[3x - 2(-2) = 4 \3x + 4 = 4\] Simplify and solve for \(x\): \[3x = 0 \x = 0\] The value of \(x\) is 0.

Step 5: Check the Solution

Verify the solution by substituting \(x = 0\) and \(y = -2\) back into both original equations:1. For \(3x - 2y = 4\):\[3(0) - 2(-2) = 0 + 4 = 4\] 2. For \(3x + y = -2\):\[3(0) + (-2) = -2\] Both equations hold true, confirming the solution is correct.

Key Concepts

Row Operations

Row operations are key in solving systems of equations when using matrices. They simplify matrices to help find variable values. There are three main types of row operations:

• Swapping two rows
• Multiplying a row by a nonzero constant
• Adding or subtracting a multiple of one row to another

These operations maintain the equality of the original system, making it powerful for finding solutions. For example, in the provided solution, we performed a row operation by subtracting the first row from the second. This eliminated the variable \(x\) in the second equation, which made solving for \(y\) straightforward. Row operations help transform complex systems into simpler forms, facilitating easier back substitution.

System of Equations

A system of equations consists of multiple equations with multiple unknowns. Solving a system means finding values for the variables that satisfy all equations simultaneously. In our exercise, the system has two equations and two unknowns \(x\) and \(y\):

• \(3x - 2y = 4\)
• \(3x + y = -2\)

Systems of equations can be solved using various methods:

• Substitution
• Elimination
• Graphing
• Using Matrices

Using augmented matrices simplifies the process due to the matrix form that lends itself well to systematic row operations. This method employs logic and precision to reduce errors, making it a foolproof approach with practice.

Back Substitution

Back substitution is a technique used once a system of equations has been partially solved using row operations or other methods. This step is often the final touch in matrix algebra to solve for unknowns. After transforming the augmented matrix, as seen in our exercise, we solved for \(y\) from the second simplified row.
With \(y = -2\), it became easy to plug this value back into the first equation to solve for \(x\).
The simplification brought by back substitution eliminates variables step-by-step, reducing the system into straightforward arithmetic that feels more intuitive and manageable.
It's a practical ending process to ensure that each variable adjusts correctly in accordance with the others, confirming the solution's correctness.