Let's solve the rational equation $\frac{2}{x^2 + 1} - \frac{1}{x - 1} = \frac{1}{2}$.

Step 1: Find the Least Common Denominator (LCD)

The denominators in the given equation are:

• $x^2 + 1$
• $x - 1$
• 2

The Least Common Denominator (LCD) of these terms is their product since there is no simple relationship between them. So the LCD is:

$\text{LCD} = (x^2 + 1)(x - 1)(2)$

Step 2: Multiply both sides of the equation by the LCD

Now multiply every term in the equation by the LCD to eliminate the denominators.

$\text{LCD} \cdot \left( \frac{2}{x^2 + 1} - \frac{1}{x - 1} \right) = \text{LCD} \cdot \frac{1}{2}$

Expanding this, we get:

$2(x - 1)(2) - (x^2 + 1)(2) = (x^2 + 1)(x - 1)$

Step 3: Solve the resulting equation

Let's now simplify each term:

$2 \cdot (x - 1) \cdot 2 = 4(x - 1)$ $(x^2 + 1) \cdot 2 = 2(x^2 + 1)$ $(x^2 + 1)(x - 1) = (x^3 - x^2 + x - 1)$

Now substitute the simplified expressions into the equation:

$4(x - 1) - 2(x^2 + 1) = x^3 - x^2 + x - 1$

Expand and simplify both sides:

Left side: $4x - 4 - 2x^2 - 2 = 4x - 2x^2 - 6$

Right side: $x^3 - x^2 + x - 1$

The equation becomes:

$4x - 2x^2 - 6 = x^3 - x^2 + x - 1$

Step 4: Rearrange and solve for $x$

Move all terms to one side of the equation:

$0 = x^3 - x^2 + x - 1 - 4x + 2x^2 + 6$

Combine like terms:

$x^3 + x^2 - 3x + 5 = 0$

This is now a cubic equation. Solving cubic equations typically involves either factoring (if possible) or using numerical methods such as synthetic division or the cubic formula.

At this point, you can proceed by trying to find rational roots using the Rational Root Theorem, or you may use a graphing tool or calculator to approximate the roots.

Want details or have any questions?

Relative Questions:

1. What are the general steps for solving rational equations?
2. How can the Rational Root Theorem help in solving cubic equations?
3. Why is finding the Least Common Denominator (LCD) necessary in solving rational equations?
4. What methods can be used to solve cubic equations?
5. How does the distributive property help in solving complex algebraic expressions?

Tip:

Always check for extraneous solutions when solving rational equations, as multiplying by the LCD can sometimes introduce solutions that are not valid in the original equation.