Problem 104 - College Algebra with Modeling and Visualization

Question

Clear fractions and solve. $$ \frac{1}{x-2}+\frac{3}{x+1}=0 $$

Step-by-Step Solution

Answer

The solution is \(x = \frac{5}{4}\).

Step 1: Identify the Common Denominator

To clear fractions, we first need to find a common denominator. Here, the denominators are \(x - 2\) and \(x + 1\). The common denominator is the product of these two terms: \((x-2)(x+1)\).

Step 2: Multiply Through by the Common Denominator

Multiply each term of the equation by the common denominator \((x-2)(x+1)\) to eliminate the fractions.\[(x-2)(x+1) \cdot \frac{1}{x-2} + (x-2)(x+1) \cdot \frac{3}{x+1} = 0 \cdot (x-2)(x+1)\]This simplifies to:\[(x+1) + 3(x-2) = 0\]

Step 3: Simplify the Resulting Equation

Distribute the terms inside the parentheses:\[x + 1 + 3x - 6 = 0\]Combine like terms:\[4x - 5 = 0\]

Step 4: Solve for x

Add 5 to both sides of the equation:\[4x = 5\]Divide both sides by 4 to solve for \(x\):\[x = \frac{5}{4}\]

Key Concepts

Clearing Fractions

Clearing fractions from an equation is like sweeping away obstacles to make it easier to solve. By eliminating fractions, you transform the equation into a simpler form, making it easier to manage. Here's how it works:

• First, identify the fractions in the equation. These are terms that have denominators under the numerators.
• To clear these fractions away, we need to find a common denominator.
• Once you have determined the common denominator, multiply every term in the equation by it.

This process will help each fraction cancel out, transforming the equation into a linear one without fractions. This makes it much easier to handle further steps, as we won't have to deal with fractions in our calculations. Example: For the equation \(\frac{1}{x-2} + \frac{3}{x+1} = 0\), multiplying through by the common denominator, \((x-2)(x+1)\), clears the fractions, transforming the equation into: \[(x+1) + 3(x-2) = 0\]

Common Denominator

Finding a common denominator is a crucial step in solving equations involving fractions. It's about finding a shared 'base' that all the fractions can adhere to, allowing you to eliminate them from the equation.

• The common denominator is essentially a multiple that all denominators in the equation can divide into without leaving a remainder.
• In equations with rational expressions, find the least common denominator (LCD) by multiplying the distinct algebraic expressions found in the denominators.

For the example equation \(\frac{1}{x-2} + \frac{3}{x+1} = 0\), the denominators are \(x-2\) and \(x+1\). Thus, the common denominator becomes the product, \((x-2)(x+1)\). Using this, every fraction in the equation is multiplied by this common denominator, allowing all divisors to cancel their respective terms, leaving an equation free of fractions, ready for simplification.

Simplifying Equations

Simplifying equations is like tidying up; it involves combining and reducing terms to easily reveal the solutions. Once you've cleared the fractions, you'll have an equation that's simpler and more straightforward to handle.

• Start by distributing terms: apply the multiplication to terms within parentheses.
• Next, combine like terms to condense the equation further.

In our example, after clearing the fractions, we simplify by distributing and combining:\[x + 1 + 3(x-2) = 0\]Distribute the 3 across the \((x-2)\):\[x + 1 + 3x - 6 = 0\],then combine like terms:\[4x - 5 = 0\]. Through these steps, the equation becomes easier to solve. Now, with simplified linear expressions, you can effortlessly isolate the variable by performing basic algebraic operations. Following this, solving for \(x\) is straightforward: add 5 to both sides, then divide by 4 to get \(x = \frac{5}{4}\). This new form reveals the solution clearly and concisely.