Problem 10 - Elementary and Intermediate Algebra

Question

By what should both sides of the equation be multiplied to clear it of fractions? $$ \text { a. } \frac{x}{5}=\frac{3 x}{10}+\frac{7}{2 x} $$ $$ \text { b. } \frac{2 x}{x-6}=4+\frac{1}{x-6} $$

Step-by-Step Solution

Answer

(a) Multiply by \(10x\) to clear fractions; (b) multiply by \(x-6\) to clear fractions.

Step 1: Identify the Denominators for Part (a)

The given equation is \( \frac{x}{5} = \frac{3x}{10} + \frac{7}{2x} \). Identify all the denominators: 5, 10, and \(2x\).

Step 2: Find the Least Common Denominator for Part (a)

For the denominators 5, 10, and \(2x\), find the least common multiple (LCM). The LCM of 5 and 10 is 10. Considering \(2x\) as well, the LCM becomes \(10x\).

Step 3: Multiply Part (a) by the LCM

To clear the fractions, multiply every term in the equation by \(10x\). This gives: \(10x \cdot \frac{x}{5} = 10x \cdot \frac{3x}{10} + 10x \cdot \frac{7}{2x} \).

Step 4: Simplify the Equation for Part (a)

Simplify each term: \(2x^2 = 3x^2 + 35\). Fractions are now eliminated.

Step 5: Identify the Denominators for Part (b)

The given equation is \( \frac{2x}{x-6} = 4 + \frac{1}{x-6} \). The denominators here are \(x-6\).

Step 6: Find the Least Common Denominator for Part (b)

Since 4 is an integer, the LCM of \(x-6\) and any integer is \(x-6\) itself.

Step 7: Multiply Part (b) by the LCM

To clear the fractions, multiply every term in the equation by \(x-6\). This gives: \((x-6) \cdot \frac{2x}{x-6} = (x-6) \cdot 4 + (x-6) \cdot \frac{1}{x-6}\).

Step 8: Simplify the Equation for Part (b)

Simplify each term: \(2x = 4(x-6) + 1\). Fractions are now eliminated.

Key Concepts

Least Common Multiple

When dealing with algebraic equations that include fractions, the objective is often to eliminate these fractions to simplify the equation. This is where the concept of the Least Common Multiple (LCM) comes into play. The LCM is the smallest number that multiple denominators can equally divide without leaving a remainder. By identifying the LCM of the denominators, you can multiply each term of the equation by this number to clear the fractions effectively.

• For example, consider the denominators 5, 10, and \(2x\) from the equation \( \frac{x}{5} = \frac{3x}{10} + \frac{7}{2x} \).
• The LCM of 5 and 10 is 10, and when considering \(2x\), the LCM becomes \(10x\).

By multiplying each term of the equation by \(10x\), every fraction is eliminated because each denominator now divides into the product without remainder. This results in a more manageable equation without fractions, which eases further simplification.

Simplifying Equations

Once fractions are eliminated by multiplying with the LCM, the next step is to simplify the equation. Simplification involves performing arithmetic operations to combine like terms and isolate variables. This makes solving the equation much simpler.

• After multiplying by the LCM, \(10x \cdot \frac{x}{5} = 2x^2\), \(10x \cdot \frac{3x}{10} = 3x^2\), and \(10x \cdot \frac{7}{2x} = 35\).
• The equation \(2x^2 = 3x^2 + 35\) is now free of fractions.

Next, rearrange and simplify to isolate \(x\) or bring similar terms together. In part (b), multiplying by \(x-6\) immediately simplifies \(\frac{2x}{x-6} = 2x\) and \(\frac{1}{x-6} = 1\), potentially centralizing the equation format around whole numbers and intact variable terms, aiding the solving process.

Algebraic Equations

Algebraic equations are mathematical statements of equality that involve variables, constants, and arithmetic operations. The goal is typically to find the value of the variables that hold the equation true. In the context of solving equations with fractions, we often rely on established arithmetic techniques to simplify the equation initially, so these variables are easier to solve for.

• In our exercises, the approach involved was identifying denominators, calculating the LCM, using this to eliminate fractions, and then simplifying.
• For the equation \(\frac{2x}{x-6} = 4 + \frac{1}{x-6}\), multiplying through by \(x-6\) allowed us to immediately deal with the fractions, consolidating the equation.

Ultimately, it's about breaking down problems into manageable steps, understanding the role of each mathematical component, and applying these consistently. This mastery of fundamental processes paves the way for tackling increasingly complex algebraic challenges with confidence.