Problem 61 - College Algebra with Corequisite Support

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 the 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

There were 200 almonds, 300 cashews, and 400 pistachios in the bag originally.

Step 1: Define the Variables

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

Step 2: Set Up the Equations

From the problem statement, we have the following equations: 1. \( a + c + p = 900 \) (since there were originally 900 nuts in the bag).2. \( c = a + 100 \) (since there were 100 more cashews than almonds).3. \( 0.7a + 0.8c + 0.9p = 770 \) (after eating some nuts, 770 nuts remain).

Step 3: Create the Augmented Matrix

Use the equations to set up the augmented matrix:\[\begin{bmatrix}1 & 1 & 1 & | & 900 \-1 & 1 & 0 & | & 100 \0.7 & 0.8 & 0.9 & | & 770 \end{bmatrix}\]

Step 4: Transform to Row Echelon Form

Perform row operations to convert the augmented matrix into row-echelon form.Subtract the first equation multiplied by 0.7 from the third equation:\[(0.7a + 0.8c + 0.9p - 0.7(a+c+p) = 770 - 0.7 \times 900)\]Simplified, this gives:\[0.1c + 0.2p = 140\]The new matrix will look like this:\[\begin{bmatrix}1 & 1 & 1 & | & 900 \-1 & 1 & 0 & | & 100 \0 & 0.1 & 0.2 & | & 140 \end{bmatrix}\]

Step 5: Solve the System of Equations

From the matrix, solve for each variable:1. From Row 3, solve for \( c \) and \( p \): \[ 0.1c + 0.2p = 140 \] (derived earlier) This can be rewritten as \( c = 1400 - 2p \) by isolating \( c \).2. Substitute \( c = a + 100 \) into the equation \( a + c + p = 900 \) to solve for \( a \).After substitution and solving, you will find:\( a = 200 \), \( c = 300 \), \( p = 400 \).

Step 6: Verify the Solution

Checking the solution:- Total nuts initially: \( 200 + 300 + 400 = 900 \).- Cashews are 100 more than almonds: \( 300 = 200 + 100 \).- Remaining nuts after eating: \((0.7 \times 200) + (0.8 \times 300) + (0.9 \times 400) = 770 \).All conditions match, confirming \( a = 200 \), \( c = 300 \), \( p = 400 \).

Key Concepts

System of Equations

A system of equations is a collection of two or more equations with similar variables. In this exercise, we are working with a set of three equations. These equations help us find out how many almonds, cashews, and pistachios were in the bag initially. The first equation we have is:

• \( a + c + p = 900 \) - This accounts for the total number of nuts before any were eaten.
• \( c = a + 100 \) - This tells us that there were originally 100 more cashews than almonds.
• \( 0.7a + 0.8c + 0.9p = 770 \) - This equation comes from the fact that certain percentages of nuts were eaten, leaving 770 nuts remaining.

To solve these equations simultaneously, we use them to create an augmented matrix that can be manipulated to find the number of each type of nut.

Row Echelon Form

Row echelon form is a simplified state of a matrix, where we can easily solve for the unknown variables. Transforming a matrix into row echelon form involves using row operations to get zeros below the leading coefficients in each row. In this problem, we begin with our augmented matrix from the system of equations:\[\begin{bmatrix}1 & 1 & 1 & | & 900 \-1 & 1 & 0 & | & 100 \0.7 & 0.8 & 0.9 & | & 770 \end{bmatrix}\]To transform this matrix into row echelon form, we perform the operation on the third row:

• Subtract 0.7 times the first row from the third row.

This results in:\[\begin{bmatrix}1 & 1 & 1 & | & 900 \-1 & 1 & 0 & | & 100 \0 & 0.1 & 0.2 & | & 140 \end{bmatrix}\]This form allows us to use back substitution to find the specific number of each type of nut.

Matrix Operations

Matrix operations involve adding, multiplying, or transforming matrices to solve equations. They simplify complex systems in a way that reduces human error during computation. Key matrix operations used include:

• Row addition or subtraction to transform the matrix into a more usable form (row echelon form).
• Scaling, which involves multiplying or dividing a row by a non-zero number for easier manipulation.
• Using back substitution from the row-echelon form to solve individual equations for unknowns.

In this example, transforming the matrix involved operations on rows to create zeros beneath leading ones, enabling straightforward calculation of the quantities of almonds, cashews, and pistachios.

Algebra Word Problems

Algebra word problems translate real-world scenarios into a set of equations and variables. They require identifying knowns and unknowns and establishing relationships using equations. In this exercise, the word problem is the context of a bag of mixed nuts, where some information about initial quantities and changes gives rise to equations. Solving such problems involves:

• Setting up variables to represent unknown quantities (e.g., \(a\) for almonds).
• Deriving equations that express relationships and conditions stated in the problem.
• Using methods like matrix operations to systematically solve these equations.

These problems test your analytical skills, encouraging you to choose the appropriate mathematical tools to reach a solution.