Problem 47 - College Algebra

Question

Exercises \(31-50\) contain equations with variables in denominators. For each equation, a. Write the value or values of the variable that make a denominator zero. These are the restrictions on the variable. b. Keeping the restrictions in mind, solve the equation. $$ \frac{2}{x+1}-\frac{1}{x-1}=\frac{2 x}{x^{2}-1} $$

Step-by-Step Solution

Answer

The restrictions are \(x\) cannot be -1 or 1. And the solution to the equation is \(x = -3\).

Step 1: Identify Restrictions

Set each denominator equal to zero and solve for \(x\): For numerator \(x+1\), \(x+1=0\) gives \(x=-1\). For numerator \(x-1\) and \(x^{2}-1\), \(x-1=0\) and \(x^{2}-1=0\) gives \(x=1\). So, \(x\) cannot be -1 or 1.

Step 2: Simplify the Equation

The equation can be simplified. However, note that \(x^{2}-1\) can be factored into \((x-1)(x+1)\). Substitute \((x-1)(x+1)\) for \(x^{2}-1\) in the equation to get: \[ \frac{2}{x+1}-\frac{1}{x-1}=\frac{2x}{(x-1)(x+1)} \] Multiply all terms by \( (x-1)(x+1) \) to clear the denominators, resulting in the equation \(2(x-1) - (x+1) = 2x\).

Step 3: Solve the Equation

Expand the left side to get \(2x-2-x-1=2x\). Simplify to get \(x-3=2x\). Solve for \(x\), and you get \(x = -3\). The solution is \(x=-3\).

Key Concepts

Restrictions on Variables

When solving rational equations, it's crucial to note the values that variables cannot take, known as restrictions. This is because division by zero is undefined in mathematics, leading to an invalid equation. For example, in the equation \(\frac{2}{x+1}-\frac{1}{x-1}=\frac{2 x}{x^{2}-1}\), denominators are expressions that include the variable \(x\).

To find the restrictions, set each denominator equal to zero and solve for \(x\). In this case, \(x+1=0\) gives a restriction of \(x=-1\), and \(x-1=0\) gives a restriction of \(x=1\). These values make the denominators zero and must be excluded from all potential solutions. It is essential to always start with this step before attempting to solve a rational equation.

Clearing Denominators

Clearing denominators is a vital step in solving rational equations. It involves getting rid of the fractions by multiplying every term by a common denominator. The benefit of this is transforming the rational equation into a simpler algebraic equation without fractions, which is often easier to manage and solve.

For the problem at hand, we can use the denominators \(x+1\) and \(x-1\) to create a common denominator, which is, in fact, their product \( (x-1)(x+1) \). By multiplying every term of the equation by this common denominator, we eliminate the fractions, allowing us to solve for \(x\) using typical algebraic methods. It’s important to note that clearing denominators doesn’t change the solutions of the equation as long as we include the restrictions previously identified.

Factoring Polynomials

Factoring polynomials is a skill that simplifies the process of solving rational equations by breaking down complex expressions into simpler ones. In the case of \(x^{2}-1\), we can factor this polynomial into \( (x-1)(x+1) \), a product of two binomials. These simpler expressions mirror the denominators we already have in the original equation.

Recognizing these factors allows us to manipulate the equation more comfortably, as it reveals the underlying structure and relationships between its terms. Factoring is not just a tool for simplification but it's also immensely helpful in identifying potential solutions and further understanding the behavior of the equation.

Solving Algebraic Equations

After clearing denominators and simplifying the equation, we arrive at a traditional algebraic equation which can be solved using standard procedures. These include combining like terms, isolating the variable on one side, and finding the value of the variable that satisfies the equation.

In our example, \(2(x-1) - (x+1) = 2x\) simplifies to \(-3 = x\), arriving at the solution \(x = -3\). It’s crucial to verify that this solution doesn't violate any of the variable restrictions identified earlier. Since \(x = -3\) is not one of the restricted values (\(xeq-1\) and \(xeq1\)), it is the valid solution to the equation.