Question

Solve the equation by using the LCD. Check your solution(s). $$\frac{x+3}{x-3}+\frac{x}{x-5}=\frac{x+5}{x-5}$$

Short answer

The equation has no valid solutions as the solutions make the original equation undefined.

Step-by-step solution

Identify the LCD

The denominators are \(x-3\) and \(x-5\). Thus, the Least Common Denominator (LCD) is \((x-3)(x-5)\).

Multiply each term by the LCD and Simplify

Multiplying each term of the equation by \((x-3)(x-5)\) , \((x-3)(x-5)\cdot\frac{x+3}{x-3}+(x-3)(x-5)\cdot\frac{x}{x-5}=(x-3)(x-5)\cdot\frac{x+5}{x-5}\). This simplifies to \((x-5)(x+3)+x(x-3)=(x-3)(x+5)\).

Solve for the equation

Expand and solve for \(x\), \(x^2 -2x -15 + x^2 -3x = x^2 +2x -15\). Upon simplifying, we have \(2x^2 -5x +15 = x^2 +2x -15\). Subtracting \(x^2 +2x -15\) from both sides, we get \(x^2 -7x+30=0\). With the quadratic formula, the solutions are \(x = 3, 5\).

Check for extraneous solutions

However, \(x =3, 5\) are not valid solutions as they make the original equation undefined. So, there are no valid solutions for the equation.

Key concepts

Understanding the Least Common Denominator (LCD)

The Least Common Denominator (LCD) is vital when solving equations involving fractions with different denominators. By definition, the LCD is the smallest nonzero common multiple of the denominators. Finding the LCD is a crucial step because it allows us to combine the fractions into a single fraction or equation that can be more readily solved.

Take, for instance, our example problem which includes fractions with denominators \(x-3\) and \(x-5\). To solve this, we first determine the LCD, which in this case, is the product of the denominators, \(x-3)(x-5)\), since they do not share any common factors. Utilizing the LCD is a way to eliminate the fractions and create a more straightforward equation to work with.

The Role of Simplifying Rational Expressions

The process of simplifying rational expressions ensures that complex fraction problems become easier to solve. Simplification often involves factoring numerators and denominators, canceling common factors, and reducing fractions to their simplest form.

In the equation given, after multiplying each term by the LCD, the resulting equation is free from fractions, which simplifies the problem considerably. This process not only streamlines the equation but also sets the stage for solving for \(x\) without complicated fractions in the way.

Applying the Quadratic Formula

When we arrive at a quadratic equation, as we have in the simplified form \(x^2 -7x+30=0\), the quadratic formula is a powerful tool to find the solutions of the variable \(x\). The quadratic formula is \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a\), \(b\), and \(c\) are coefficients of the quadratic equation in the form \(ax^2 + bx + c = 0\).

In this exercise however, manually applying the formula is unnecessary after simplification since the equation resolves to \(x^2 -7x+30=0\), we can see directly that the potential solutions are \(x=3\) and \(x=5\) by factoring. Yet, these solutions must be checked against the original equation to verify validity.

The Importance of Checking for Extraneous Solutions

The final and crucial step is to check for extraneous solutions which are results that do not satisfy the original equation. This often occurs in rational equations where variables are present in the denominators.

In our working example, when we substitute the solutions \(x=3\) and \(x=5\) back into the original equation, we find they are extraneous because they cause division by zero, which is undefined. Therefore, despite solving the quadratic equation correctly, it's imperative to always plug the solutions back into the original equation to ensure they are legitimate and not extraneous solutions.