Question

If \(x = \frac{1}{y}\) and \(y = \frac{1}{1-x}\), then y=

• A. 2
• B. \(\frac{1}{2}\)
• C. \(-\frac{1}{2}\)
• D. -1
• E. -2

Hint:

When solving equations that involve variables in the denominator, always check your final answers against the original equations. Solutions that lead to division by zero are extraneous and must be discarded.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This problem presents a system of two equations with two variables, \(x\) and \(y\). To find the value of \(y\), we need to combine the equations to eliminate one variable, in this case, \(x\), and solve for the remaining variable, \(y\).
Step 2: Key Formula or Approach:
The method of substitution is ideal here. We will substitute the expression for \(x\) from the first equation into the second equation. This will result in an equation solely in terms of \(y\), which we can then solve.
Step 3: Detailed Explanation:
We are given two equations:
1) \(x = \frac{1}{y}\)
2) \(y = \frac{1}{1-x}\)
Substitute the expression for \(x\) from equation (1) into equation (2):
\[ y = \frac{1}{1 - \left(\frac{1}{y}\right)} \] To simplify the denominator, find a common denominator:
\[ y = \frac{1}{\frac{y}{y} - \frac{1}{y}} \] \[ y = \frac{1}{\frac{y-1}{y}} \] Now, invert the fraction in the denominator and multiply:
\[ y = 1 \times \frac{y}{y-1} \] \[ y = \frac{y}{y-1} \] To solve for \(y\), multiply both sides by \((y-1)\). Note that this assumes \(y \neq 1\).
\[ y(y-1) = y \] Distribute \(y\) on the left side:
\[ y^2 - y = y \] Move all terms to one side to form a quadratic equation:
\[ y^2 - y - y = 0 \] \[ y^2 - 2y = 0 \] Factor out a common factor of \(y\):
\[ y(y-2) = 0 \] This gives two possible solutions: \(y = 0\) or \(y - 2 = 0 \Rightarrow y = 2\).
We must check these solutions in the original equations. If \(y=0\), the first equation becomes \(x = \frac{1}{0}\), which is undefined. Therefore, \(y=0\) is not a valid solution.
The only valid solution is \(y=2\).
Step 4: Final Answer:
After substituting and solving the resulting equation, we find that the only valid value for \(y\) is 2.