/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 470 A rancher sold 25 hogs and 60 sh... [FREE SOLUTION] | 91Ó°ÊÓ

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A rancher sold 25 hogs and 60 sheep to Mr. Kay for \(\$ 3450 .\) At the same prices, he sold 35 hogs and 50 sheep to Mr. Bea for \(\$ 3300\). Find the price of each hog and sheep.

Short Answer

Expert verified
Thus, the price of each hog is \$30 and the price of each sheep is \$45.

Step by step solution

01

Create the system of linear equations

We are given the total number of hogs and sheep bought by Mr. Kay and Mr. Bea along with their respective payments. Let's represent the price of each hog by x and the price of each sheep by y. Then we can create the following equations based on the provided information: 1. \( 25x + 60y = 3450 \) (Total payment by Mr. Kay) 2. \( 35x + 50y = 3300 \) (Total payment by Mr. Bea) These are our two linear equations.
02

Solve the system of linear equations

Now, let's solve these equations to find the values of x and y, which represent the prices of hogs and sheep. There are several methods to solve a system of linear equations, like substitution, elimination, and matrix method. Here, we will use the elimination method. First, multiply equation (1) by 7 and equation (2) by 5 to make the coefficients of x equal: 1. \( 175x + 420y = 24150 \) (Multiplying equation 1 by 7) 2. \( 175x + 250y = 16500 \) (Multiplying equation 2 by 5) Now, subtract equation (2) from equation (1): \( 170y = 7650 \) Next, divide by 170 on both sides to get the value of y: \( y = 45 \) (Price of each sheep) Now that we have found the price of each sheep, we can find the price of each hog by substituting the value of y in either of the equations. We will substitute it in equation (1): \( 25x + 60(45) = 3450 \) Simplify the equation: \( 25x + 2700 = 3450 \) Subtract 2700 from both sides: \( 25x = 750 \) Finally, divide by 25 on both sides to get the value of x: \( x = 30 \) (Price of each hog)
03

Write the final answer

So, the price of each hog is \$30 and the price of each sheep is \$45.

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Key Concepts

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

Linear Algebra
Linear algebra is a branch of mathematics that deals with vectors, vector spaces, linear mappings, and systems of linear equations – like the one presented in our rancher problem. In practical terms, linear algebra allows us to solve for unknown values when we have a set of equations that relate those unknowns linearly. When we talk about solving a system of linear equations, we're often looking for the values of variables that make all the equations true simultaneously.

In our example, 'linear algebra' plays a foundational role in determining the price of each hog and sheep. We use two linear equations representing two transactions to establish a relationship between the prices of the two goods. The essence of linear algebra in this situation is to provide a methodical approach to finding the most accurate values of hog and sheep prices (denoted by x and y in the equations) that satisfy both transactions' conditions. Understanding the basics of linear algebra is crucial to not only solve such problems but also to apply this knowledge in various scientific and engineering disciplines.
Elimination Method
The elimination method is one of the techniques in linear algebra used to solve systems of linear equations. To fully grasp this method, it's important to note that it involves combining equations to remove one of the variables, making it possible to solve the resulting equation for the remaining variable.

In the rancher's problem, we demonstrate the elimination method by manipulating the equations so that when we subtract one from another, one variable is eliminated. This is achieved by ensuring the coefficients of one of the variables are opposites or the same. In simpler terms, we made the number in front of x the same in both equations through multiplication, allowing us to eliminate x and find y. Then, using back-substitution, we find the value of x. This method is popular because of its straightforward approach and can be applied in various situations ranging from simple to more complex systems of equations. Familiarity with the elimination method is a powerful skill for students studying algebra, as it's one of the fundamental tools for solving real-world problems.
Algebraic Equations
Algebraic equations are the bread and butter of algebra. They are mathematical statements that show two expressions are equal, often containing one or more unknowns or variables. An essential part of working with algebraic equations is finding the value of these unknowns.

In the context of the exercise we're discussing, we are dealing precisely with this: finding the values for x (the price of a hog) and y (the price of a sheep). We formed algebraic equations based on the transactions provided (25 hogs and 60 sheep for \(3450 and 35 hogs and 50 sheep for \)3300). Each equation symbolized a real-world scenario and represents a piece of the puzzle. Understanding algebraic equations is not only key to unlocking such numerical puzzles but also a stepping stone to advancing in more complex mathematics and applying mathematical thinking to everyday problems.

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Most popular questions from this chapter

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