Problem 41 - Algebra for College Students

Question

For Problems \(41-58\), use an algebraic approach to solve each problem. Find a number such that one-half of the number is 3 less than two-thirds of the number.

Step-by-Step Solution

Answer

The number is 18.

Step 1: Define the Variable

Let the unknown number be represented by the variable \( x \). We will create an equation using this variable to represent the given conditions.

Step 2: Set Up the Equation

According to the problem, one-half of the number is equal to three less than two-thirds of the number. In mathematical terms, this translates to the equation: \( \frac{1}{2}x = \frac{2}{3}x - 3 \).

Step 3: Eliminate Fractions

To eliminate the fractions, find a common denominator for the terms \( \frac{1}{2}x \) and \( \frac{2}{3}x \). The least common denominator of 2 and 3 is 6. Multiply every term in the equation by 6 to clear the fractions: \( 6 \times \frac{1}{2}x = 6 \times \left( \frac{2}{3}x - 3 \right) \). This simplifies to \( 3x = 4x - 18 \).

Step 4: Simplify and Solve for x

Rearrange the equation to isolate \( x \). Subtract \( 4x \) from both sides to get: \( 3x - 4x = -18 \). This simplifies to \( -x = -18 \). Divide both sides by -1 to solve for \( x \): \( x = 18 \).

Key Concepts

Equation Setup

When faced with an algebra problem, the crucial first step is setting up the equation. Here, we begin by interpreting the given problem into mathematical terms. The problem tells us that one-half of a number is 3 less than two-thirds of the same number. This gives us the framework for our equation. We represent our unknown number with a variable, typically denoted as \( x \). This allows us to translate the statement into the equation: \( \frac{1}{2}x = \frac{2}{3}x - 3 \).

Setting up the equation is like building a bridge from words to numbers. Pay close attention to details in the wording. They will guide you in structuring your equation correctly. Once you have the equation, you’ve mapped out the problem mathematically. You're ready to move forward towards finding the solution.

Fractions Elimination

Fractions can make equations look more complex than they are. Eliminating them simplifies the equation, making it easier to solve. In our problem, we have fractions \( \frac{1}{2} \) and \( \frac{2}{3} \). We eliminate these fractions by finding the least common denominator (LCD).

Determine the LCD of the denominators, which in this case, is 6.

Multiply every term in the equation by 6.

As a result, the equation transforms from \( \frac{1}{2}x = \frac{2}{3}x - 3 \) into \( 3x = 4x - 18 \).

This step reduces fractional complexity, providing a clearer path to solve the equation. Fractions elimination is a vital skill in algebra, especially for beginners who might find fractions intimidating.

Variable Definition

Defining the variable is a foundational step in solving algebraic equations. In this problem, we choose \( x \) to represent the unknown number we are trying to find. This acts as a placeholder for the number we haven't yet calculated.

Choose a simple, relevant variable name, like \( x \), to denote the unknown.

Ensure the variable has a clear definition based on the problem context.

By clearly defining \( x \), we allow ourselves to manipulate the unknown in mathematical operations and transformations. The process of definition helps organize our approach to problem-solving, offering a clearer scope to tackle the equation effectively.

Solving Equations

Once the equation is set up and simplified, the final task is solving it. This involves isolating the variable on one side of the equation. In our scenario, once the fractions are eliminated, we have the simplified equation: \( 3x = 4x - 18 \).

Start by moving all terms involving \( x \) to one side. We subtract \( 4x \) from both sides to get \( 3x - 4x = -18 \).

Simplify this to \( -x = -18 \).

Solve for \( x \) by dividing both sides by -1, resulting in \( x = 18 \).

Solving equations requires consistent operations that maintain equality, aiming towards isolating the variable. Once isolated, the variable unveils the solution to the original problem. Practicing these steps strengthens problem-solving skills in algebra.