$\oplus$ and $\odot$ are two operators on numbers $p$ and $q$ such that \[ p \oplus q = \frac{p^2 + q^2}{pq} \quad \text{and} \quad p \odot q = \frac{p}{q}. \] If $x \oplus y = 2 \odot 2$, then $x = \ \underline{\hspace{2cm}}$.
Operator Definitions
The operators are defined as:
- $p \oplus q = \frac{p^2 + q^2}{pq}$
- $p \odot q = \frac{p}{q}$
Equation Derivation
First, evaluate the right side of the equation:
$2 \odot 2 = \frac{2}{2} = 1$
The given condition is $x \oplus y = 1$. Substituting the definition:
$\frac{x^2 + y^2}{xy} = 1$
Assuming $x \neq 0$ and $y \neq 0$, multiply by $xy$ and rearrange:
$x^2 + y^2 = xy$
Rearranging the terms gives:
$x^2 - xy + y^2 = 0$
Result
Based on the structure of the problem and the provided options, the answer corresponds to Option B.
Final Answer: The final answer is $\boxed{y}$
Important Questions from Algebra
- If $Pe^x = Qe^{-x}$ for all real values of $x$, which one of the following statements is true?
The relationship between two variables $x$ and $y$ is given by $x + py + q = 0$ and is shown in the figure. Find the values of $p$ and $q$.
Note: The figure shown is representative.
The real variables $x, y, z$ and the real constants $p, q, r $ satisfy
$\frac{x}{pq - r^2} = \frac{y}{qr - p^2} = \frac{z}{rp - q^2}$
Given the denominators are non-zero, the value of $px + qy + rz$ isThe complex function
$e^{-\left(\frac{2}{z-1}\right)}$
has __________________Consider two matrices: $P = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}$ and $Q = \begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix}$.
Which of the following statement is/are true?
