Problem 45 - Algebra 2

Question

Solve each equation. Check each solution. $$ \frac{1}{x-5}=\frac{x}{x^{2}-25} $$

Step-by-Step Solution

Answer

The given equation has no solution.

Step 1: Identify the Common Denominator

Identify the common denominator of all terms in the equation. Here, the common denominator is \((x-5)(x+5)\) as \(x^{2}-25\) can be factored into \((x-5)(x+5)\).

Step 2: Eliminate the Denominators

Multiply both sides of the equation by the common denominator to eliminate the fractions: \((x-5)(x+5) \times \frac{1}{x-5} = (x-5)(x+5) \times \frac{x}{x^{2}-25}\). This simplifies to \(x+5 = x\).

Step 3: Solve the Simplified Equation

Next, the equation \(x+5 = x\) is solved for 'x'. But, subtracting 'x' from both sides gives '5=0', which is a false statement, indicating that the equation has no solution.

Step 4: Check the Solution

As there is no solution obtained in the previous step, there is nothing to check here. Hence, we can conclude that the given equation has no solution.

Key Concepts

Understanding Common Denominator in Rational Equations

When solving rational equations, identifying a common denominator is a crucial first step. A common denominator is the least expression that can be multiplied by both sides of an equation to eliminate the fractions. This simplifies the equation and makes it easier to solve.

In the given problem, we have the equation \(\frac{1}{x-5} = \frac{x}{x^{2}-25}\). To solve this, find a denominator common to both sides. Notice that \(x^2 - 25\) can be factored as \((x-5)(x+5)\). So, the common denominator here is \((x-5)(x+5)\).

By multiplying every term by \( (x-5)(x+5) \), we eliminate the denominators. This transforms the complex rational equation into a simpler form, paving the way for us to find a solution or determine if a solution exists. Remember, using a common denominator helps address rational equations more efficiently.

Factoring: Breaking Down Complex Expressions

Factoring is a technique used to simplify polynomial expressions. It involves expressing a polynomial as a product of its simpler factors. For instance, the expression \(x^2 - 25\) can be written as \((x-5)(x+5)\). This is known as factoring into binomials—expressing a difference of squares.

When dealing with rational equations, factoring plays an essential role in making these problems more manageable. In the highlighted equation, identifying and factoring \(x^2 - 25\) allows us to find the common denominator. This process not only simplifies the equation but also reveals potential solutions and restrictions.

Always check for the possibility of factoring expressions, especially when presented with quadratic terms. Factoring can open pathways to simpler equations and more straightforward calculations.

Handling No Solution in Equations

In algebra, sometimes after simplifying an equation, you may encounter a situation where there is no valid solution. This can happen when an equation simplifies to a false statement such as \(5=0\). For the problem at hand, after manipulation, the equation \(x+5 = x\) simplifies to \(5 = 0\), which is clearly false.

No solution means there's no value of the variable that can satisfy the equation. This could occur when there are restrictions inherent to the original equation, such as division by zero or invalid operations within its domain.

If you find that an equation simplifies to an untrue statement, it's a sign to step back and re-check any assumptions made during factorization or manipulation. Verify if potential restrictions preclude any values, reaffirming that no solution is justly concluded.

Working with Algebraic Fractions

Algebraic fractions, like regular fractions, consist of a numerator and a denominator, but they involve polynomials or algebraic expressions. Solving equations with algebraic fractions requires careful steps to maintain mathematical integrity.

Let's consider \(\frac{1}{x-5}\) and \(\frac{x}{x^2-25}\). Each term is an algebraic fraction because of their non-numeric expressions. Always identify the restrictions that algebraic fractions introduce—such as zero denominators.

Before eliminating fractions by multiplying through by a common denominator, recognize which values make denominators zero and discard them from potential solutions. Here, exclusions like \(x=5\) and \(x=-5\) must be considered as they make the original denominators zero. Thus, working with algebraic fractions demands a detailed look at the expression to ensure valid manipulations while being mindful of any resulting solution constraints.