Problem 7 - Basic Technical Mathematics with Calculus

Question

Solve the given equations and check the results. $$\frac{x}{6}-\frac{1}{2}=\frac{x}{3}$$

Step-by-Step Solution

Answer

The solution is \( x = -3 \). Verified as correct when substituted back into the original equation.

Step 1: Understand the Equation

The equation given is \( \frac{x}{6} - \frac{1}{2} = \frac{x}{3} \). Our goal is to solve for \( x \). Notice that this equation involves terms with \( x \) on both sides as well as a constant term.

Step 2: Eliminate Fractions

To eliminate the fractions, we need to find a common denominator. The denominators are 6, 2, and 3. The least common multiple of these is 6. Multiply every term in the equation by 6 to clear the fractions:\[ 6 \left( \frac{x}{6} \right) - 6 \left( \frac{1}{2} \right) = 6 \left( \frac{x}{3} \right) \] This simplifies to:\[ x - 3 = 2x \]

Step 3: Rearrange the Equation

Now, we have the equation \( x - 3 = 2x \). To solve for \( x \), move the \( x \) on the left side to the right side of the equation by subtracting \( x \) from both sides:\[ x - x - 3 = 2x - x \]This simplifies to:\[ -3 = x \]

Step 4: Verify the Solution

We have found that \( x = -3 \). To verify, substitute \( x = -3 \) back into the original equation:\[ \frac{-3}{6} - \frac{1}{2} = \frac{-3}{3} \]Simplify each term:\[ -\frac{1}{2} - \frac{1}{2} = -1 \]Both sides equal \(-1\), verifying that the solution is indeed correct.

Key Concepts

Solving Equations

Solving equations is a fundamental concept in algebra that involves finding the value of the unknown variable that makes the equation true. When solving, our goal is to isolate the variable on one side of the equation. This often involves a series of algebraic manipulations such as addition, subtraction, multiplication, and division.

To solve an equation:

• First, simplify both sides if necessary.
• Next, attempt to get all terms containing the variable on one side and constants on the other.
• Use inverse operations to isolate the variable.

These steps transform a complex problem into a simpler form, eventually revealing the value of the variable. Verifying the solution by plugging it back into the original equation is also crucial to ensure accuracy. By practicing these steps regularly, solving equations becomes an automatic and intuitive process.

Fractions in Algebra

When dealing with algebraic equations, fractions often appear and must be managed correctly. A fraction consists of two parts: a numerator (top) and a denominator (bottom).

• The numerator represents how many parts we have.
• The denominator tells us into how many parts the whole is divided.

In algebra, fractions can be intimidating, but they are manageable with some strategic approaches.

To work with algebraic fractions, it's often useful to perform operations that eliminate the fractions altogether. This typically involves using a common denominator. By identifying a common denominator for the terms in the equation, you can multiply each term by this value, effectively clearing the fractions. This is seen in the given solution, where all terms were multiplied by the least common multiple of the denominators to streamline the equation.

Least Common Multiple

The least common multiple (LCM) is the smallest number that is a multiple of two or more numbers. It plays a crucial role in algebra, particularly when working with fractions.

• Finding the LCM helps to combine fractions.
• It is essential when clearing fractions from an algebraic equation.

To identify the LCM, consider the prime factorization of each number. For example, to find the LCM of 2, 3, and 6:

• 2 has the factorization of 2.
• 3 has the factorization of 3.
• 6 has the factorization of 2 x 3.

Here, each prime factor should be taken at its highest power, which results in 6 being the LCM. Once the LCM is determined, it can be used to eliminate fractions by ensuring that each term in the equation is expressed with this consistent denominator. This makes the equation easier to solve, as seen in our original problem, where multiplying through by the LCM resolved the fractional components.