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Chapter 9: Problem 94

Solve each problem. At the Brendan Berger ranch, 6 goats and 5 sheep sell for \(\$ 305,\) while 2 goats and 9 sheep sell for \(\$ 285 .\) Find the cost of a single goat and of a single sheep.

Short Answer

Expert verified
The cost of a single goat is \( \$30 \) and the cost of a single sheep is \( \$25 \).

Step by step solution

01

- Define variables

Let the cost of one goat be denoted by \( G \) and the cost of one sheep be denoted by \( S \).
02

- Formulate the equations

Based on the provided information, form the system of linear equations:1. \( 6G + 5S = 305 \) (for 6 goats and 5 sheep costing \$305)2. \( 2G + 9S = 285 \) (for 2 goats and 9 sheep costing \$285)
03

- Solve by elimination or substitution

To eliminate one of the variables, multiply the second equation by 3 to align the coefficients of \( G \):\[ 6G + 27S = 855 \]Now subtract the first equation from this result:\[ (6G + 27S) - (6G + 5S) = 855 - 305 \]\[ 22S = 550 \]\[ S = 25 \]
04

- Find the value of the other variable

Substitute \( S = 25 \) back into either equation to solve for \( G \). Using the first equation:\[ 6G + 5(25) = 305 \]\[ 6G + 125 = 305 \]\[ 6G = 180 \]\[ G = 30 \]
05

- Verify the solution

Substitute \( G = 30 \) and \( S = 25 \) into the second equation to verify:\[ 2(30) + 9(25) = 285 \]\[ 60 + 225 = 285 \]The solution is verified.

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Key Concepts

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

System of Equations
A system of equations is a set of two or more equations with the same variables. In this exercise, we worked with two equations involving goats and sheep.
Each equation represents a linear relationship between the cost of goats and sheep.
The goal is to find the values of the variables (cost of a single goat and a single sheep) that satisfy both equations simultaneously.
This is done by finding common solutions for the variables in all equations in the system.
Elimination Method
The elimination method involves manipulating the equations to eliminate one of the variables, making it easier to solve for the other.
In this exercise, we used elimination by aligning the coefficients of one variable. We multiplied the second equation by 3 to align the coefficients of the goats:
  • Original second equation: 2G + 9S = 285
  • After multiplication: 6G + 27S = 855
Next, we subtracted the first equation from this modified equation:
  • First equation: 6G + 5S = 305
  • Subtraction: (6G + 27S) - (6G + 5S) = 855 - 305
  • Result: 22S = 550, which simplifies to S = 25
This method allowed us to find the value of one of the variables (S) directly.
Substitution Method
After finding one variable (in this case, S = 25) using elimination, the substitution method helps you find the other variable.
We substitute the known value of S back into one of the original equations to solve for G:
Using the first equation 6G + 5(25) = 305, we get:
  • 6G + 125 = 305
  • Simplifying: 6G = 180
  • Finally: G = 30
Substitution helps us use the known variable to find the unknown variable.
This method is especially useful when dealing with more complex systems of equations.
Verification of Solutions
To ensure our solution is correct, we must verify it by substituting the found values of the variables back into the original equations.
This step ensures the solutions satisfy both equations in the system.
For our exercise, we verify by substituting G = 30 and S = 25 into the second equation:
  • 2(30) + 9(25) = 285
  • 60 + 225 = 285
Both sides of the equation equal 285, confirming our solution is correct.
Always verifying the solution is crucial to avoid errors and ensure the integrity of your solution.

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