Problem 57 - College Algebra

Question

For the following exercises, consider this scenario: There is a mound of g pounds of gravel in a quarry. Throughout the day, 400 pounds of gravel is added to the mound. Two orders of 600 pounds are sold and the gravel is removed from the mound. At the end of the day, the mound has 1,200 pounds of gravel. Write the equation that describes the situation.

Step-by-Step Solution

Answer

The equation is \( g - 800 = 1200 \), which simplifies to \( g = 2000 \).

Step 1: Understanding the Initial Amount

The problem states there is an initial amount of gravel, given as \( g \) pounds. This represents the starting amount of gravel in the mound before any actions are taken.

Step 2: Addition of Gravel

During the day, 400 pounds of gravel are added to the initial amount. This addition can be represented by \( g + 400 \).

Step 3: Subtraction of Gravel

Two orders of 600 pounds each are sold, which means 1200 pounds are subtracted from the mound. Represent this as \( g + 400 - 1200 \).

Step 4: Setting up the Equation

At the end of the day, after the additions and subtractions, the mound contains 1200 pounds of gravel. Therefore, the equation becomes \( g + 400 - 1200 = 1200 \).

Step 5: Simplifying the Equation

Simplify the equation by combining terms. Start with \( g + 400 - 1200 = 1200 \). Simplify it to form: \( g - 800 = 1200 \).

Step 6: Solve for g

Add 800 to both sides of the equation to isolate \( g \). This gives \( g = 1200 + 800 \), resulting in \( g = 2000 \).

Key Concepts

Initial Value Problem

An initial value problem in mathematics refers to a situation where you are given a starting point and then factors change that initial value through a process.
This type of problem is about the systematic application of operations to arrive at a final state. Here, the initial amount of gravel is represented as \( g \), which stands for the unknown initial quantity of gravel, before any was added or removed.
Understanding initial value problems is crucial as it forms the foundation for setting up equations in real-life scenarios.
In this exercise, the initial value provided gives us a point to start calculations before external operations—like additions and subtractions—take place.
Tracking how these operations affect the initial value helps us to solve the problem and reach the final value provided.

Addition and Subtraction of Integers

Addition and subtraction of integers involve changing amounts by either increasing or decreasing their value.
In this exercise, you add and subtract large sets of numbers based on the transactions involving the gravel mound.
To add 400 pounds of gravel, you simply calculate \( g + 400 \) to represent the increase.
This operation demonstrates how the initial number increases when amounts are being added.
Subtraction follows the purchase of gravel, where two orders of 600 pounds each mean a total of \( 1200 \) pounds removed, reflected in the expression \( g + 400 - 1200 \).
Additional insights to help understand subtraction:

• It is akin to taking away resources or reducing an amount.
• Ensure careful calculation as errors during subtraction of integers can lead to incorrect results in larger problems.

Understanding these simple arithmetic operations aids in forming correct and reliable equations to reflect real-world changes.

Algebraic Expressions

An algebraic expression is a combination of variables, numbers, and operations that together represent a math quantity.
In this context, the expression \( g + 400 - 1200 \) represents the total amount of gravel throughout the day before reaching the final tally.
Breaking down the expression:

• \( g \) stands for the initial, unknown quantity.
• The operation of addition (+) and subtraction (-) modify the initial value, factoring in the day's changes.

This gives a mathematical representation of the entire scenario where multiple actions affect the total amount.
Writing and understanding these expressions is vital to formulating equations, as they help to visually and logically track the quantities in play.
By adjusting the components of an algebraic expression, we are able to formulate and solve real-life problems by converting them into solvable mathematical equations.

Simplifying Equations

Simplifying equations is the process of making an equation easier to solve or understand by combining like terms and reducing complex expressions.
In this exercise, the equation \( g + 400 - 1200 = 1200 \) becomes simpler by combining the numerical parts: namely, \( 400 - 1200 \).
This reduces the problem to \( g - 800 = 1200 \).
Here's how simplification helps:

• It removes unnecessary complexity, making problems easier to solve.
• Allows you to focus on finding the unknown variables, in this case, \( g \).

The key steps involve:

• Combining like terms — similar numbers or variables.
• Removing parentheses if they exist, through multiplication or distribution.
• Rewriting the equation in its simplest linear form.

Consequently, it permits the effective isolation of the variable, which lets us determine the initial value of the gravel mound, was in fact, 2000 pounds.
This practice of simplifying equations is essential for accurately solving mathematically modeled real-world problems.