Problem 46 - Algebra and Trigonometry

Question

\(9- 46\) The given equation is either linear or equivalent to a linear equation. Solve the equation. $$ \frac{1}{x}-\frac{2}{2 x+1}=\frac{1}{2 x^{2}+x} $$

Step-by-Step Solution

Answer

The solution is all real numbers except \(x = 0\) and \(x = -\frac{1}{2}\).

Step 1: Identify the Least Common Denominator (LCD)

First, observe the fractions in the equation. We need a common denominator to combine these fractions. The denominators are \(x\), \(2x+1\), and \(2x^2+x\). Factoring \(2x^2+x\) gives \(x(2x+1)\), which is the least common denominator (LCD).

Step 2: Multiply Through by the LCD

Multiply every term in the equation by the LCD, which is \(x(2x+1)\), to eliminate the fractions:\[x(2x+1) \cdot \left(\frac{1}{x} - \frac{2}{2x+1} = \frac{1}{2x^2+x}\right)\]Simplifying, we get \((2x+1) - 2x = 1\).

Step 3: Simplify the Resulting Equation

The equation \((2x+1) - 2x = 1\) simplifies as follows: - Distribute: \(2x + 1 - 2x = 1\)- Simplify: \(1 = 1\).This indicates that the equation is consistent and true for any \(x\) not causing division by zero.

Step 4: Determine Restrictions and Validity

Check where the original fractions are undefined: \(x eq 0\), \(2x+1 eq 0\) (so \(x eq -\frac{1}{2}\)). Thus, the solution is valid for all \(x\) except \(x = 0\) and \(x = -\frac{1}{2}\).

Key Concepts

Least Common Denominator

In equations involving fractions, finding the Least Common Denominator (LCD) is crucial. The LCD is the smallest expression that each denominator divides evenly. This allows us to combine fractions more easily.

In the given problem, the denominators are:

• \( x \)
• \( 2x + 1 \)
• \( 2x^2 + x \)

Notice that \( 2x^2 + x \) can be factored into \( x(2x + 1) \). This insight shows that \( x(2x + 1) \) is the LCD. Why? Because it's the simplest expression that contains all factors of each original denominator. By multiplying each term of the equation by this LCD, we eliminate the fractions, simplifying the process of solving the equation.

Fractions in Equations

Fractions can make equations seem complex, but understanding how to handle them makes a big difference. When equations include fractions, finding the LCD helps in rewriting these fractions into a more manageable form. This is essential because:

• It allows you to eliminate denominators by multiplying through with the LCD.
• Once the fractions are removed, you're usually left with simpler algebraic expressions.

Consider the original equation: \[\frac{1}{x} - \frac{2}{2x+1} = \frac{1}{2x^2+x}\]After multiplying by the LCD \( x(2x+1) \), the fractions disappear, leading to a simpler linear equation. This step is crucial for transforming a potentially complicated problem into a straightforward solution.

Factoring

Factoring is the process of breaking down an expression into products of simpler expressions. It is a key tool in algebra, helping to simplify and solve equations.

In this exercise, factoring is needed to identify the LCD. You may notice that the expression \( 2x^2 + x \) can be factored into \( x(2x + 1) \). Here's how:

• Look for a common factor in the terms. In this case, both terms have \( x \) as a factor.
• Factor out the \( x \), leaving \( 2x + 1 \) inside the parentheses.

This reveals the structure of the polynomial, allowing us to find the LCD, which simplifies solving the equation. Factoring also helps identify possible points where the equation is undefined, guiding us in finding valid solutions.

Restrictions in Equations

Restrictions in equations occur when certain values of variables make any denominator zero, which is undefined in mathematics. Recognizing these restrictions ensures that the solutions are valid for the equation.

• If \( x = 0 \), the denominator \( x \) becomes zero.
• If \( 2x + 1 = 0 \), then \( x = -\frac{1}{2} \) causes division by zero in the fraction \( \frac{-2}{2x+1} \).

Therefore, it's crucial to determine which values of \( x \) make the equation undefined. For the original equation, both \( x = 0 \) and \( x = -\frac{1}{2} \) are excluded from the solution, ensuring that you don't mistakenly include invalid solutions while solving the equation. Properly addressing these restrictions is a fundamental part of solving equations with fractions correctly.