Question:
Given that \( \dfrac{1-x}{x-1} = \dfrac{1}{x} \). Compare:
Quantity A: \( x \)
Quantity B: \( -\dfrac{1}{2} \)
Given that \( \dfrac{1-x}{x-1} = \dfrac{1}{x} \). Compare:
Quantity A: \( x \)
Quantity B: \( -\dfrac{1}{2} \)
Quantity A: \( x \)
Quantity B: \( -\dfrac{1}{2} \)
Show Hint
Always check for restrictions on variables when solving rational equations (denominator ≠ 0). This ensures invalid solutions are discarded.
Updated On: Oct 7, 2025
- Quantity A is greater.
- Quantity B is greater.
- The two quantities are equal.
- The relationship cannot be determined from the information given.
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The Correct Option is B
Solution and Explanation
Step 1: Simplify the given equation.
\[ \frac{1-x}{x-1} = \frac{1}{x}. \] Notice that \( \frac{1-x}{x-1} = -1 \) for all \( x \neq 1 \). But let us solve algebraically.
Step 2: Cross-multiply.
Multiply both sides by \( x(x-1) \): \[ x(1-x) = (x-1)(1). \] Step 3: Expand.
\[ x - x^2 = x - 1 \quad \Rightarrow \quad -x^2 = -1 \quad \Rightarrow \quad x^2 = 1. \] Step 4: Find valid values.
Thus, \( x = 1 \) or \( x = -1 \). But \( x \neq 1 \) (denominator restriction). Hence, \( x = -1 \).
Step 5: Compare with Quantity B.
Quantity A = \(-1\). Quantity B = \(-\tfrac{1}{2}\). Since \(-1<-\tfrac{1}{2}\), Quantity B is greater.
Step 6: Conclusion.
\[ \boxed{\text{(B) Quantity B is greater.}} \]
\[ \frac{1-x}{x-1} = \frac{1}{x}. \] Notice that \( \frac{1-x}{x-1} = -1 \) for all \( x \neq 1 \). But let us solve algebraically.
Step 2: Cross-multiply.
Multiply both sides by \( x(x-1) \): \[ x(1-x) = (x-1)(1). \] Step 3: Expand.
\[ x - x^2 = x - 1 \quad \Rightarrow \quad -x^2 = -1 \quad \Rightarrow \quad x^2 = 1. \] Step 4: Find valid values.
Thus, \( x = 1 \) or \( x = -1 \). But \( x \neq 1 \) (denominator restriction). Hence, \( x = -1 \).
Step 5: Compare with Quantity B.
Quantity A = \(-1\). Quantity B = \(-\tfrac{1}{2}\). Since \(-1<-\tfrac{1}{2}\), Quantity B is greater.
Step 6: Conclusion.
\[ \boxed{\text{(B) Quantity B is greater.}} \]
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