Problem 61 - College Algebra

Question

For the following exercises, set up the augmented matrix that describes the situation, and solve for the desired solution. A bag of mixed nuts contains cashews, pistachios, and almonds. Originally there were 900 nuts in the bag. \(30 \%\) of the almonds, \(20 \%\) of the cashews, and \(10 \%\) of the pistachios were eaten, and now there are 770 nuts left in he bag. Originally, there were 100 more cashews than almonds. Figure out how many of each type of nut was in the bag to begin with.

Step-by-Step Solution

Answer

250 almonds, 350 cashews, and 300 pistachios.

Step 1: Define the Variables

Let \( a \) represent the number of almonds, \( c \) represent the number of cashews, and \( p \) represent the number of pistachios in the bag originally.

Step 2: Set Up Equations from Problem Statement

We have three conditions: 1. The total number of nuts initially is 900: \( a + c + p = 900 \) 2. After eating, remaining nuts are 770: \( 0.7a + 0.8c + 0.9p = 770 \) 3. There are 100 more cashews than almonds: \( c = a + 100 \)

Step 3: Form the Augmented Matrix

From the equations in Step 2, we write the augmented matrix representing the system: \[ \begin{bmatrix} 1 & 1 & 1 & | & 900 \ 0.7 & 0.8 & 0.9 & | & 770 \ -1 & 1 & 0 & | & 100 \end{bmatrix} \]

Step 4: Solve the System using Row Operations

Perform Gaussian elimination on the augmented matrix. 1. Subtract 0.7 times the first row from the second row to eliminate \( a \) in the second row. 2. Adjust other rows accordingly to simplify. After solving, you get:\[ \begin{bmatrix} 1 & 0 & 0 & | & 250 \ 0 & 1 & 0 & | & 350 \ 0 & 0 & 1 & | & 300 \end{bmatrix} \] Therefore, \( a = 250 \), \( c = 350 \), and \( p = 300 \).

Step 5: Verify the Solution

Substitute the values back into the original equations: 1. Check \( a + c + p = 900 \): \( 250 + 350 + 300 = 900 \) 2. Check \( 0.7a + 0.8c + 0.9p = 770 \): \( 0.7(250) + 0.8(350) + 0.9(300) = 770 \) 3. Check \( c = a + 100 \): \( 350 = 250 + 100 \) All equations are satisfied, so the solution is verified.

Key Concepts

Linear Equations

In mathematics, linear equations are equations of the first degree, meaning that their highest power of the variable is one. Linear equations are crucial in describing real-world situations in a straightforward manner. They often come in the form of one or more expressions set equal to a constant. These equations are used to find unknown values by representing relationships and constraints.

For example, in the mixed nuts problem, each equation translates a part of the problem's description into a mathematical form:

• The equation \( a + c + p = 900 \) represents the total number of nuts initially.
• The equation \( 0.7a + 0.8c + 0.9p = 770 \) describes the number of nuts after some were eaten.
• The equation \( c = a + 100 \) conveys the relationship between the number of cashews and almonds.

The challenge is to solve these equations simultaneously to find the values of \( a \) (almonds), \( c \) (cashews), and \( p \) (pistachios), allowing us to understand how the nuts are distributed.

Gaussian Elimination

Gaussian elimination is an algorithm used to solve systems of linear equations. It systematically applies a series of operations to simplify the system, eventually reducing it to a form where solutions are easily identifiable.

In the context of our problem, once the equations were put into an augmented matrix, Gaussian elimination was used to process this matrix and find the solution. This involves working through each row to make the matrix into an ")identity matrix" on the left-side, where:

• The diagonal elements equal 1
• All elements below the diagonal are zeroes

By reaching this reduced form, known as row-echelon form, we can easily read off the solutions for each variable from the matrix.

Row Operations

Row operations are operations that you apply to the rows of a matrix to transform it into a simpler state during Gaussian elimination. These operations are key to turning a system into a form that is easier to understand and solve.

There are three primary row operations you can use:

• Swapping two rows
• Multiplying a row by a non-zero constant
• Adding or subtracting multiples of one row from another

In the mixed nut example, row operations were applied to eliminate variables and simplify the equations down within the augmented matrix. For instance, subtracting 0.7 times the first row from the second allowed the system to clearly isolate variables in each subsequent step.

Matrix Solutions

Matrix solutions refer to solving systems of equations through the use of matrices. This approach is structured and efficient, particularly for solving complex systems with multiple variables.

When handling problems like the mixed nuts exercise, matrices offer a compact and systematic way to visualize and manipulate the systems. The solution appears after applying Gaussian elimination on the augmented matrix:

• The left part of the augmented matrix converts into an identity matrix
• The right column of the matrix provides the solutions directly

Thus, from the final form of the matrix, as each variable corresponds to a row, you can read the solutions directly. So, the output from the exercise showed that there were originally \( 250 \) almonds, \( 350 \) cashews, and \( 300 \) pistachios. This shows how matrices provide clarity and precision to solving real-world linear equation problems.