Problem 60 - 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. There are \(1,000\) total nuts in the bag, and there are 100 less almonds than pistachios. The cashews weigh 3 \(\mathrm{g}\) , pistachios weigh 4 \(\mathrm{g}\) , and almonds weigh 5 \(\mathrm{g}\) . If the bag weighs \(3.7 \mathrm{kg},\) find out how many of each type of nut is in the bag.

Step-by-Step Solution

Answer

Cashews: 500, Pistachios: 300, Almonds: 200.

Step 1: Identify the Variables

Let \( c \), \( p \), and \( a \) represent the number of cashews, pistachios, and almonds, respectively, in the bag. We need to establish equations based on the given conditions.

Step 2: Set Up Equations

From the problem, we can establish the following equations: 1. The total number of nuts: \( c + p + a = 1000 \) 2. Almonds are 100 less than pistachios: \( a = p - 100 \) 3. The total weight of the nuts in grams is \(3.7 \text{ kg} = 3700 \text{ g}\), thus \( 3c + 4p + 5a = 3700 \).

Step 3: Substitute and Simplify

Substitute \( a = p - 100 \) into the first two equations: 1. \( c + p + (p - 100) = 1000 \) simplifies to \( c + 2p = 1100 \). 2. Substitute in the weight equation: \( 3c + 4p + 5(p - 100) = 3700 \) simplifies to \( 3c + 9p = 4200 \).

Step 4: Set Up the Augmented Matrix and Solve

Write the equations as an augmented matrix: \[\begin{bmatrix} 1 & 2 & 0 & \vert & 1100 \3 & 9 & 0 & \vert & 4200 \end{bmatrix}\] Using row reduction techniques: Divide Row 2 by 3: \[\begin{bmatrix} 1 & 2 & \vert & 1100 \1 & 3 & \vert & 1400 \end{bmatrix}\] Subtract Row 1 from Row 2: \[\begin{bmatrix} 1 & 2 & \vert & 1100 \0 & 1 & \vert & 300 \end{bmatrix}\]This gives \( p = 300 \), then substitute into \( c + 2p = 1100 \) gives \( c = 500 \), and using \( a = p - 100 \), \( a = 200 \).

Step 5: Verify Solution

Re-calculate the totals for cashews \( 500 \), pistachios \( 300 \), and almonds \( 200 \): 1. Total nuts: \( 500 + 300 + 200 = 1000 \), okay. 2. Weight check: \( 3(500) + 4(300) + 5(200) = 1500 + 1200 + 1000 = 3700 \), okay. The solution is verified.

Key Concepts

System of Equations

In this exercise, we're dealing with a classic example of a system of equations. A system of equations is a set of two or more equations that involve the same variables. The main goal is to find a solution that satisfies all the given equations simultaneously. In our problem, the variables are the number of cashews, pistachios, and almonds, represented by \(c\), \(p\), and \(a\), respectively.

The steps include forming equations based on the problem's conditions, like the total number and the total weight of the nuts. Therefore, we have:

• \(c + p + a = 1000\) accounts for the total number of nuts.
• \(a = p - 100\) indicates that there are 100 fewer almonds than pistachios.
• \(3c + 4p + 5a = 3700\) translates the weight conditions into an equation.

These equations must be solved together to find the values of \(c\), \(p\), and \(a\).

Matrix Row Reduction

Matrix row reduction, often referred to as Gaussian elimination, is a method used in linear algebra to solve systems of linear equations. After setting up the equations, we represent them in a matrix form, called an augmented matrix. This matrix encapsulates the system of equations in a compact form, aiding in efficient computation.

In this exercise, the original equations were structured into an augmented matrix: \[\begin{bmatrix}1 & 2 & 0 & \vert & 1100 \3 & 9 & 0 & \vert & 4200\end{bmatrix}\]
Row reduction is performed to simplify the matrix, aiming to get leading 1s in each row. We first modify Row 2 by dividing by 3, simplifying the system and reducing computational complexity:\[\begin{bmatrix}1 & 2 & \vert & 1100 \1 & 3 & \vert & 1400 \end{bmatrix}\]
We then subtract Row 1 from Row 2 to isolate the variable \(p\). This process shows how matrix operations can simplify the calculations and help solve the system.

Variable Substitution

Variable substitution is a technique used to simplify solving systems of equations by replacing one variable with another, often making the system easier to solve. In our exercise, we use substitution early on to reduce the number of variables in the expressions.

Based on the equation \(a = p - 100\), we replace all instances of \(a\) with \(p - 100\) in our other equations:

• Substituting into \(c + p + a = 1000\), results in \(c + 2p = 1100\).
• For the weight equation \(3c + 4p + 5a = 3700\), substitution yields \(3c + 9p = 4200\).

This systematic reduction maintains the integrity of the problem while simplifying equation solving. Often these substitutions bring clarity and aid in the transition to using matrices.

Linear Algebra

Linear algebra is a branch of mathematics focusing on systems of linear equations, vectors, and matrices. It's a foundational tool used to solve complex problems across various scientific disciplines. In this problem, linear algebra aids in structuring the problem through equations and matrices.

Key aspects of linear algebra applied here include:

• Utilizing augmented matrices to represent systems of equations, allowing streamlined processing.
• Employing row reduction techniques to simplify and solve systems more efficiently.

It elegantly handles systems with multiple variables and can be extended to solve much larger scale problems. Understanding the principles of linear algebra empowers you to manipulate and solve multiple equations simultaneously.