Problem 37 - College Algebra

Question

contain rational 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. $$ \frac{x-2}{2 x}+1-\frac{x+1}{x} $$

Step-by-Step Solution

Answer

The equation \((x-2)/(2x) + 1 - (x+1)/x\) does not have a solution, because the calculated solution \(x = 0\) is a restriction for this equation.

Step 1: Identify Restrictions

To find the values of x that make the denominator zero, set the denominator of each fraction equal to zero and solve for x. Denominators here are \(2x\) and \(x\). Solving \(2x = 0\) gives \(x = 0\), and solving \(x = 0\) gives \(x = 0\) again. Therefore, the restriction on the variable x is that \(x ≠ 0\).

Step 2: Simplify the Equation

First, find a common denominator for the fractions in the equation. Since the denominators are \(2x\) and \(x\), a common denominator will be \(2x\). Rewrite each fraction in the equation with \(2x\) as the denominator: \((x-2)/2x + 2x/2x - (x+1)*2/2x\). This simplifies to \((x-2+4x-2x+2)/2x\), and further simplifies to \((3x)/2x\).

Step 3: Solve the Simplified Equation

Now that all terms are over a common denominator, the rational equation \((3x)/2x = 0\) can be solved like a linear equation. Multiply both sides of the equation by \(2x\) to eliminate the denominator, leading to the equation \(3x = 0\). Then, divide both sides by 3, which gives the solution \(x = 0\) to this equation.

Step 4: Validate the Solution Against Restrictions

The calculated solution \(x = 0\) cannot be accepted because it violates the calculated restriction from Step 1, which stated that \(x ≠ 0\). Therefore, the given equation does not have a solution under the restriction initially identified.

Key Concepts

Denominator Restrictions

When working with rational equations, especially those containing variables in the denominators, it's critical to identify any potential restrictions. The values that make the denominator zero cause the rational expression to be undefined. In our exercise, we begin by determining these restrictions. Consider the denominators in the expression \(2x\) and \(x\). By setting each denominator equal to zero, and solving, we find:

• \(2x = 0\) implies \(x = 0\)
• \(x = 0\) implies \(x = 0\)

Thus, the restriction is \(x eq 0\). This means any solution to the equation must respect this limitation to be valid.

Common Denominator

Finding a common denominator is a crucial step in adding or subtracting rational expressions. A common denominator allows you to combine the fractions by rewriting them with the same denominator. In the exercise, we worked with the denominators \(2x\) and \(x\). The least common multiple in this instance is \(2x\), since \(2x\) is a multiple of \(x\).
Using the common denominator \(2x\), rewrite each term of the equation:

• \( \frac{x-2}{2x} \) remains \( \frac{x-2}{2x} \)
• 1 becomes \( \frac{2x}{2x} \), because \(1 = \frac{2x}{2x} \)
• \(-\frac{x+1}{x}\) becomes \(-\frac{2(x+1)}{2x}\) or equivalently \(-\frac{2x+2}{2x}\)

Having done this, we can now combine the expressions easily by summing the numerators over the common denominator.

Solving Linear Equations

Once all the rational expressions share a common denominator, you can treat the equation as a linear equation in the numerators. In this exercise, after rewriting over a common denominator, the simplified expression becomes \((3x)/(2x) = 0\).
The next step is to solve this as a linear equation. To remove the fraction, multiply both sides by \(2x\), neutralizing the denominator to simplify the solution process. This results in the equation \(3x = 0\).
To isolate \(x\), divide both sides by 3, obtaining \(x = 0\). However, keep in mind the restrictions discovered earlier to verify if this proposed solution fits.

Rational Expressions

Rational expressions, much like fractions, involve a ratio of polynomials. When combined, as seen in this problem, it's essential to handle each part properly. We first identify the parts of the equation that results in undefined behavior, particularly values that cause the denominator to be zero.
Through understanding how these expressions work, you'll learn how to solve them systematically:

• Identify restrictions early to avoid invalid solutions.
• Simplify by finding a common denominator for operations.
• Consolidate terms with these common denominators for easier solution steps.
• Ensure final solutions uphold the initial constraints discovered.

This methodical approach is vital whenever you work with rational equations, ensuring clarity and correctness in results.