Problem 56 - Algebra A Combined Function

Question

The largest meteorite in the world is the Hoba West, located in Namibia. Its weight is 3 times the weight of the Armanty meteorite, located in Outer Mongolia. If the sum of their weights is 88 tons, find the weight of each.

Step-by-Step Solution

Answer

The Armanty meteorite weighs 22 tons, and the Hoba West meteorite weighs 66 tons.

Step 1: Define Variables

Let's define the variables for the problem. Let \( x \) represent the weight of the Armanty meteorite in tons. Then, the Hoba West meteorite, which is 3 times the weight of the Armanty meteorite, can be represented as \( 3x \).

Step 2: Set Up the Equation

According to the problem, the sum of the weights of the Armanty and Hoba West meteorites is 88 tons. We can express this relationship with the equation: \( x + 3x = 88 \).

Step 3: Simplify the Equation

Combine the terms on the left side of the equation: \( 4x = 88 \). This equation represents the total combined weight of the two meteorites.

Step 4: Solve for x

To find the weight of the Armanty meteorite, divide both sides of the equation by 4: \( x = \frac{88}{4} = 22 \). Thus, the weight of the Armanty meteorite is 22 tons.

Step 5: Calculate the Hoba West Weight

Since the weight of the Hoba West meteorite is 3 times that of the Armanty meteorite, calculate \( 3 \times 22 = 66 \). So, the Hoba West meteorite weighs 66 tons.

Key Concepts

Defining Variables in Algebraic Equations

In algebraic equations, variables are symbols that represent unknown values. In our exercise, we begin by defining the variable for our unknown weight. To solve the problem about the meteorites, we choose to let \( x \) represent the weight of the Armanty meteorite in tons. This is a strategic choice because knowing one variable helps us relate to another quantity easily.

Here, considering the problem mentions that the Hoba West meteorite weighs three times more than the Armanty meteorite, we define the weight of the Hoba West meteorite using the expression \( 3x \). Now, we have expressed both meteorite weights in terms of a single variable \( x \), which simplifies our subsequent calculations. This setup is crucial for forming an equation and systematically approaching the solution.

Steps in Problem Solving

Problem-solving in algebra involves a series of steps designed to simplify and solve equations. Let's break it down:

• Understanding the Problem: Begin by carefully reading the problem statement. Recognize relationships or expressions provided, such as the sum or multiplicative relations given in the text.
• Defining Variables: Identify and define variables that best represent the unknowns in the problem. This helps in constructing equations that reflect those relationships.
• Setting Up the Equation: Translate the word problem into an algebraic equation. In our exercise, we formed the equation \( x + 3x = 88 \) to represent the sum of the meteorite weights.
• Simplifying the Equation: Combine like terms or simplify the equation to make it easier to solve. We simplified \( x + 3x \) to \( 4x \).
• Solving the Equation: Use algebraic methods like addition, subtraction, multiplication, or division to isolate the variable and find its value.
• Interpreting the Solution: Finally, substitute the found variable back into real-world context to find the required answer.

Each step is integral to moving smoothly from problem understanding to solution interpretation.

Understanding and Solving Linear Equations

Linear equations are equations of the first degree, meaning they involve only the single power of the variable (e.g., \( x \) and not \( x^2 \)). In our problem, the equation \( x + 3x = 88 \) is a linear equation.

Linear equations often appear in word problems where one needs to find specific values, much like the weights of our meteorites. Solving a linear equation involves isolating the variable to determine its value. In our case, we simplified \( 4x = 88 \) by dividing both sides by 4, yielding \( x = 22 \).

The key characteristics of linear equations that help in solving them include:

• Constant Slope: These equations graph as straight lines, reflecting a constant rate of change.
• One Solution: Typically, there is one value that satisfies the equation.
• Direct Relation: Variables have a direct and proportional relationship with each other.

By clearly understanding and applying these elements, students can tackle linear equations confidently.