Problem 58
Question
Solve each equation. $$\frac{5}{x+1}-\frac{1}{1-x}=\frac{1}{x^{2}-1}$$
Step-by-Step Solution
Verified Answer
x = \( \frac{5}{6} \).
1Step 1: Identify and Simplify
Notice that the denominators \(x+1\, ,1-x\), and \(x^2-1\) can be factored. Rewrite \(1-x\) as \( -(x-1)\). Also, note \(x^2-1\) factors to \( (x+1)(x-1) \).
2Step 2: Find a Common Denominator
The common denominator for the fractions \( \frac{5}{x+1}, \frac{1}{1-x}, \frac{1}{x^2-1} \) is \( (x+1)(x-1) \). Rewrite each fraction with this common denominator: \( \frac{5(x-1)}{(x+1)(x-1)} - \frac{-1}{(x+1)(x-1) } = \frac{1}{ (x+1)(x-1) } \) .
3Step 3: Combine the Fractions
Since all fractions now have the same denominator, combine them: \( \frac{5(x-1) + (x+1)}{(x+1)(x-1)} = \frac{1}{(x+1)(x-1)} \).
4Step 4: Simplify the Numerator
Expand and simplify the numerator: \( 5(x-1) + (x+1) = 5x - 5 + x + 1 = 6x - 4 \). Thus, the equation simplifies to: \( \frac{6x - 4}{(x+1)(x-1)} = \frac{1}{(x+1)(x-1)} \).
5Step 5: Solve for x
Since the denominators are equal, equate the numerators: \( 6x - 4 = 1 \). Solve this linear equation by isolating x: \( 6x = 5 \) and \( x = \frac{5}{6} \).
6Step 6: Verify the Solution
Plug \( x = \frac{5}{6} \) back into the original equation to verify it does not make any denominator zero and satisfies the equation.
Key Concepts
Common DenominatorFactoringLinear Equation
Common Denominator
When solving rational equations, one of the first steps is finding a common denominator for all the fractions involved. The common denominator is a shared multiple of the original denominators. This step simplifies the equation by allowing all the terms to be combined into a single fraction.
In the provided exercise, we have the fractions \ \(\frac{5}{x+1}\), \ \(\frac{1}{1-x}\), and \ \(\frac{1}{x^2-1}\). To find a common denominator, first rewrite \ \(1-x\) as \ \(-(x-1)\). Notice that \ \(x^2-1\) can be factored into \ \((x+1)(x-1)\), which is a product of the other two denominators.
Thus, our common denominator will be \ \((x+1)(x-1)\). By rewriting each fraction with this common denominator, we can combine and simplify the equation more easily.
In the provided exercise, we have the fractions \ \(\frac{5}{x+1}\), \ \(\frac{1}{1-x}\), and \ \(\frac{1}{x^2-1}\). To find a common denominator, first rewrite \ \(1-x\) as \ \(-(x-1)\). Notice that \ \(x^2-1\) can be factored into \ \((x+1)(x-1)\), which is a product of the other two denominators.
Thus, our common denominator will be \ \((x+1)(x-1)\). By rewriting each fraction with this common denominator, we can combine and simplify the equation more easily.
Factoring
Factoring plays a crucial role in simplifying and solving rational equations. In the given problem, we identified that \ \(x^2-1\) can be factored. This common algebraic technique, called difference of squares, expresses a difference between squared terms as a product of their sum and difference:
\ \(a^2 - b^2 = (a + b)(a - b)\)
Applying this to our equation, \ \(x^2-1\) becomes \ \((x+1)(x-1)\).
We also needed to handle \ \(\frac{1}{1-x}\). By recognizing that \ \(1-x\) is the negative of ̆\ \(x-1\), it can be rewritten as \ \(\frac{-1}{x-1}\).
Factoring helps break down complex expressions into simpler, more manageable parts, allowing us to solve the equation efficiently.
\ \(a^2 - b^2 = (a + b)(a - b)\)
Applying this to our equation, \ \(x^2-1\) becomes \ \((x+1)(x-1)\).
We also needed to handle \ \(\frac{1}{1-x}\). By recognizing that \ \(1-x\) is the negative of ̆\ \(x-1\), it can be rewritten as \ \(\frac{-1}{x-1}\).
Factoring helps break down complex expressions into simpler, more manageable parts, allowing us to solve the equation efficiently.
Linear Equation
The final step in solving rational equations involves solving the resulting linear equation. After combining and simplifying the fractions using their common denominator, we reach a simpler equation:
\ \(\frac{6x-4}{(x+1)(x-1)} = \frac{1}{(x+1)(x-1)}\).
Since the denominators are identical, we equate the numerators, resulting in a linear equation:
\ \(6x - 4 = 1\).
Solving this linear equation requires isolating the variable \ \(x\). First, add \ \(4\) to both sides:
\ \(6x - 4 + 4 = 1 + 4\),
giving us \ \(6x = 5\).
Next, divide both sides by \ \(6\):
\ \(x = \frac{5}{6}\).
Finally, verify the solution by substituting \ \(x = \frac{5}{6}\) back into the original equation to ensure it doesn’t make any denominator zero and satisfies the rational equation.
\ \(\frac{6x-4}{(x+1)(x-1)} = \frac{1}{(x+1)(x-1)}\).
Since the denominators are identical, we equate the numerators, resulting in a linear equation:
\ \(6x - 4 = 1\).
Solving this linear equation requires isolating the variable \ \(x\). First, add \ \(4\) to both sides:
\ \(6x - 4 + 4 = 1 + 4\),
giving us \ \(6x = 5\).
Next, divide both sides by \ \(6\):
\ \(x = \frac{5}{6}\).
Finally, verify the solution by substituting \ \(x = \frac{5}{6}\) back into the original equation to ensure it doesn’t make any denominator zero and satisfies the rational equation.
Other exercises in this chapter
Problem 57
Simplify. $$\frac{1-8 x^{-3}}{x^{-1}+2 x^{-2}+4 x^{-3}}$$
View solution Problem 57
Convert each rational expression into an equivalent rational expression that has the indicated denominator. $$\frac{1}{2 x+2}, \frac{?}{-6 x-6}$$
View solution Problem 58
Perform the indicated operations. When possible write down only the answer. $$\frac{5 x-5 y}{x} \cdot \frac{1}{x-y}$$
View solution Problem 58
Simplify. $$\frac{a+27 a^{-2}}{1-3 a^{-1}+9 a^{-2}}$$
View solution