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Chapter 10: Problem 24

Five pounds of potatoes, 1 pound of onions, and 2 pounds of apples cost $$\$ 3.80$$. Two pounds of potatoes, 3 pounds of onions, and 4 pounds of apples cost $$\$ 5.78$$. Three pounds of potatoes, 4 pounds of onions, and 1 pound of apples cost $$\$ 4.08$$. Find the price per pound for each item.

Short Answer

Expert verified
Potatoes: $0.30, Onions: $0.58, Apples: $0.86 per pound.

Step by step solution

01

Define the Variables

Let's denote the price per pound of potatoes as \( p \), the price per pound of onions as \( o \), and the price per pound of apples as \( a \).
02

Write the System of Equations

Using the provided information, set up the system of equations based on the cost of the items:1. \( 5p + 1o + 2a = 3.80 \)2. \( 2p + 3o + 4a = 5.78 \)3. \( 3p + 4o + 1a = 4.08 \)
03

Eliminate One Variable

We will use the elimination method to remove one variable. Start by eliminating \( a \). Multiply the first equation by 2 and the second equation by 1 (no change needed) and subtract:\[(10p + 2o + 4a = 7.60) - (2p + 3o + 4a = 5.78) \]This results in: \( 8p - o = 1.82 \).
04

Continue the Elimination Process

Now eliminate \( a \) using equations 2 and 3. Multiply equation 3 by 4 and subtract equation 2:\[(12p + 16o + 4a = 16.32) - (2p + 3o + 4a = 5.78) \]This gives: \( 10p + 13o = 10.54 \).
05

Solve the Two-Variable System

Now we have a simpler system:1. \( 8p - o = 1.82 \)2. \( 10p + 13o = 10.54 \)Solve for \( o \) from the first equation: \( o = 8p - 1.82 \). Substitute this in the second equation:\[10p + 13(8p - 1.82) = 10.54 \]Simplify and solve: \( 10p + 104p - 23.66 = 10.54 \), leading to \( 114p = 34.20 \), thus \( p = 0.30 \).
06

Find the Remaining Variables

Use \( p = 0.30 \) to find \( o \) from \( o = 8p - 1.82 \):\[o = 8(0.30) - 1.82 = 0.58 \]Use one of the original equations to find \( a \). Use \( 5p + o + 2a = 3.80 \) with the known values and solve for \( a \):\[5(0.30) + 0.58 + 2a = 3.80 \]\[1.50 + 0.58 + 2a = 3.80 \]\[2.08 + 2a = 3.80 \]\[2a = 1.72 \], \( a = 0.86 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Price per Pound Calculation
Calculating the price per pound of a particular item involves determining the cost for a single unit of weight鈥攖ypically, a pound. This is useful when you're given a total cost for multiple items and want to find out how much one unit of a specific item costs. To solve such problems, we often gather various combinations of these items with given costs, as seen in the provided exercise.
  • The key is to set up a system of equations based on the data you have, which represents different combinations of items with their total cost.
  • By solving these equations, you can determine the price per pound for each individual item.
Breaking the process down step-by-step allows you to clearly see how each variable (like the price per pound of onions, potatoes, or apples) is isolated and calculated. Once you solve these equations, you obtain the cost for 1 pound of each item.
Elimination Method in Algebra
The elimination method in algebra is a powerful tool for solving systems of linear equations. It involves strategically adding or subtracting equations in a system to eliminate one of the variables. This makes it easier to solve for the remaining variables.
  • The goal is to manipulate the equations so that when you add or subtract them, one variable cancels out.
  • In our example, we started by trying to remove the variable representing apples (\( a \)).
This is achieved by making the coefficients (numbers in front of variables) of \( a \) the same in two equations and then subtracting one from the other. After \( a \) was eliminated, a simpler system with just two variables remained, allowing further analysis. By continuing this process, equations become simpler, eventually allowing all variables to be solved.
Solving Linear Equations
Solving linear equations means finding the values of variables that make the equation true. These types of equations allow us to model real-world relationships where variables are linearly proportional to each other.
  • The process typically involves isolating the variable you are solving for on one side of the equation.
  • For the problem given, once we had our simplified system of equations, solving them involved typical algebraic operations鈥攕uch as addition, multiplication, and substitution.
For example, once one of the variables鈥攑otatoes in this case鈥攚as isolated and found to be \( 0.30 \), substitution allowed us to easily find the other variables. Solving linear equations in systems often requires a combination of strategies, such as elimination and substitution, to methodically break down and solve for each unknown.

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Most popular questions from this chapter

Use Cramer's rule to find the solution set for each of the following systems. (Objective 2) $$ \left(\begin{array}{rl} x+\frac{2}{3} y & =-6 \\ -\frac{1}{4} x+3 y & =-8 \end{array}\right) $$

Evaluate each of the following determinants. $$ \left|\begin{array}{rr} -7 & -2 \\ -2 & 4 \end{array}\right| $$

Evaluate each of the following determinants. $$ \left|\begin{array}{rr} \frac{3}{2} & -\frac{1}{2} \\ \frac{1}{2} & -\frac{2}{5} \end{array}\right| $$

Evaluate each of the following determinants. $$ \left|\begin{array}{rr} 8 & -3 \\ 6 & 4 \end{array}\right| $$

For Problems \(1-10\), use expansion by minors to evaluate each determinant. (Objective 1) $$ \left|\begin{array}{rrr} -3 & -2 & 1 \\ 5 & 0 & 6 \\ 2 & 1 & -4 \end{array}\right| $$
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