Problem 82 - Algebra for College Students

Question

Solve each equation. Identify each equation as a conditional equation, an inconsistent equation, or an identity. State the solution sets to the identities using interval notation. $$\frac{2}{x+1}+\frac{3}{x-1}=\frac{5 x+1}{x^{2}-1}$$

Step-by-Step Solution

Answer

The equation is an identity. Solution set: \( (-∞, -1) ∪ (-1, 1) ∪ (1, ∞) \).

Step 1 - Simplify the Equation

Recognize that the denominators on each side of the equation can be factored. The left side is already simplified. For the right side of the equation, notice: \(x^2 - 1 = (x + 1)(x - 1)\). This allows for common denominators.

Step 2 - Combine Fractions

Combine the fractions on the left-hand side: \(\frac{2}{x+1}\) and \(\frac{3}{x-1}\). Find a common denominator: \(\frac{2(x-1) + 3(x+1)}{(x+1)(x-1)} = \frac{2x - 2 + 3x + 3}{(x+1)(x-1)} = \frac{5x + 1}{(x+1)(x-1)}\).

Step 3 - Compare Both Sides

Compare the simplified left-hand side \(\frac{5x + 1}{(x+1)(x-1)}\) to the right-hand side \(\frac{5 x + 1}{x^{2}-1}\). Since both fractions are identical \(\frac{5x + 1}{(x+1)(x-1)} = \frac{5 x + 1}{(x^2 - 1)}\).

Step 4 - Determine the Type of Equation

Since both sides of the equation are identical and the denominators \((x+1)(x-1)\) and \(x^2 - 1\) match, the equation holds for all x in the domain. The equation is an identity.

Step 5 - State the Solution Set

x can be any real number except for the values that make the denominator zero, which are \(x = 1\) and \(x = -1\). Thus, the solution set in interval notation is: \( (-∞, -1) ∪ (-1, 1) ∪ (1, ∞) \).

Key Concepts

conditional equation

A conditional equation is one that is true only under certain conditions or for specific values of the variable involved. Simply put, not every x will satisfy the equation, but there's a particular set of values that will. To identify if an equation is conditional, solve it step-by-step and check if there are specific solutions for the variable. If these specific values are the only ones that solve the equation, you're dealing with a conditional equation.

inconsistent equation

An inconsistent equation is an equation that has no solution because it leads to a contradiction. This implies that no value for the unknown variable(s) will satisfy the equation. An example in a simpler form would be something like:

• 0x + 1 = 0
• 1 = 0

Clearly, a statement like 1 = 0 is false, making the equation inconsistent. When solving rational equations, if you end up with such contradictions, you can conclude that the initial equation is inconsistent.

identities in algebra

Identities in algebra are equations that hold true for any value within the domain of the variable. For example, in our exercise, $$\frac{2}{x+1}+\frac{3}{x-1}=\frac{5 x+1}{x^{2}-1},$$the equation simplifies to $$ \frac{5x + 1}{(x+1)(x-1)} = \frac{5 x + 1}{x^{2}-1}$$which is true for all x, except for values that make the denominator zero. Therefore, such an equation is an identity. Always ensure to factorize and cancel out terms correctly to verify if both sides of the equation are equal for all allowed x.

interval notation

In algebra, interval notation is a way to represent a range of values for which an equation or inequality holds true. When representing the solution set for identities or inequalities, you use round brackets to denote values that are not included, and square brackets for values that are included. For instance, if the solution for a rational equation excludes -1 and 1, the interval notation would be: $$ (-∞, -1) ∪ (-1, 1) ∪ (1, ∞) $$This tells us the variable x can take any real number except -1 and 1.

common denominator in fractions

Finding a common denominator is crucial when adding or subtracting fractions, particularly in rational equations. The common denominator is the least common multiple of the denominators. For example, to add $$ \frac{2}{x+1}$$ and $$\frac{3}{x-1},$$we need a common denominator, which here is $$ (x+1)(x-1) $$or equivalently $$ x^2 - 1 $$.Thus, you convert each fraction: $$ \frac{2(x-1) + 3(x+1)}{(x+1)(x-1)}.$$ This step ensures that you are dealing with like terms and can simplify the equation effectively. Identifying and using common denominators properly helps in arriving at the correct simplified form of the equation.