Chapter 6: Problem 31
Solve the given equations and check the results. $$\frac{1}{x^{2}-x}-\frac{1}{x}+\frac{1}{1-x}=0$$
Short Answer
Expert verified
There are no solutions; x = 1 is invalid due to division by zero.
Step by step solution
01
Simplify the Equation
Start by combining the fractions on the left-hand side. The expression is \( \frac{1}{x^{2}-x} - \frac{1}{x} + \frac{1}{1-x} \). To combine these, find a common denominator. Notice that \( x^{2}-x = x(x-1) \) and \( 1-x = -(x-1) \). Therefore, the common denominator is \( x(x-1)(1-x) \).
02
Rewrite Each Term
Rewrite each term with the common denominator: - \( \frac{1}{x^{2}-x} = \frac{1}{x(x-1)} = \frac{1-x}{x(x-1)(1-x)} \)- \( \frac{1}{x} = \frac{(1-x)^{2}}{x(x-1)(1-x)} \)- \( \frac{1}{1-x} = \frac{x}{x(x-1)(1-x)} \)
03
Combine the Fractions
Combine the fractions over the common denominator: \[ \frac{1-x}{x(x-1)(1-x)} - \frac{(1-x)^{2}}{x(x-1)(1-x)} + \frac{x}{x(x-1)(1-x)} = \frac{-x^{2} + 2x - 1}{x(x-1)(1-x)} \].
04
Solve the Numerator
Since the overall equation equals zero, we focus on the numerator: Solve \(-x^{2} + 2x - 1 = 0\), by factoring or using the quadratic formula: Using the quadratic formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},\] where \( a = -1, b = 2, c = -1\).Compute the discriminant: \( 2^2 - 4(-1)(-1) = 4 - 4 = 0 \). Therefore, \( x = \frac{-2 \pm 0}{-2} = 1 \).
05
Verify the Solution
Check for potential restrictions and verify solution if it doesn’t make any denominator zero. The solution \( x = 1 \) makes the term \( \frac{1}{x^2-x} \) undefined because it leads to division by zero. Thus, there are no viable solutions.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Common Denominator
When dealing with rational equations such as \( \frac{1}{x^2-x} - \frac{1}{x} + \frac{1}{1-x} = 0 \), finding a common denominator is essential.This means determining a shared base for all fractions involved.
In the given equation, each denominator is unique: \( x^2-x \), \( x \), and \( 1-x \).
We'll transform each expression to have the same denominator, simplifying the process to add or subtract the fractions.
In the given equation, each denominator is unique: \( x^2-x \), \( x \), and \( 1-x \).
We'll transform each expression to have the same denominator, simplifying the process to add or subtract the fractions.
- First, notice that \( x^2-x \) can be factored into \( x(x-1) \).
- Also, recognize \( 1-x \) can be written as \( -(x-1) \).
- Thus, the common base becomes \( x(x-1)(1-x) \), as it incorporates all individual factors from the denominators.
Quadratic Formula
The quadratic formula is a fundamental method to solve equations of the form \( ax^2 + bx + c = 0 \).
Our equation simplifies to solving the quadratic \(-x^2 + 2x - 1 = 0\).
Using the quadratic formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},\]we identify the coefficients from our equation:
The solution is \(x = \frac{-2}{-2} = 1\). This critical step is often employed in algebra when factoring is complex or impossible.
Our equation simplifies to solving the quadratic \(-x^2 + 2x - 1 = 0\).
Using the quadratic formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},\]we identify the coefficients from our equation:
- \(a = -1\)
- \(b = 2\)
- \(c = -1\)
- \(2^2 - 4(-1)(-1) = 4 - 4 = 0\).
The solution is \(x = \frac{-2}{-2} = 1\). This critical step is often employed in algebra when factoring is complex or impossible.
Fraction Simplification
Simplifying fractions is a crucial skill for solving rational equations.It lets us reduce expressions to their most basic form, which is necessary for addition or subtraction.
For example, let's look at our fractions with a common denominator:
\[\frac{1-x - (1-x)^2 + x}{x(x-1)(1-x)} = \frac{-x^2 + 2x - 1}{x(x-1)(1-x)}.\]By combining them, we focus on the numerator, as the denominator is consistent.Through simplification, handling fractions becomes straightforward.
This reduces errors and provides clarity, making it a vital process in algebraic manipulation.
For example, let's look at our fractions with a common denominator:
- The individual terms become \(\frac{1-x}{x(x-1)(1-x)}\), \(\frac{(1-x)^2}{x(x-1)(1-x)}\), and \(\frac{x}{x(x-1)(1-x)}\).
\[\frac{1-x - (1-x)^2 + x}{x(x-1)(1-x)} = \frac{-x^2 + 2x - 1}{x(x-1)(1-x)}.\]By combining them, we focus on the numerator, as the denominator is consistent.Through simplification, handling fractions becomes straightforward.
This reduces errors and provides clarity, making it a vital process in algebraic manipulation.