Question

Consider the pair of equations: \(x^2 - xy - x = 22\) and \(y^2 - xy + y = 34\). If x > y, then x- y equals

• A. 8
• B. 6
• C. 7
• D. 4

Answer 1

The Correct Option is A

We are given the system of equations:

\[ x^2 - xy - x = 22 \quad \text{and} \quad y^2 - xy + y = 34 \]

Step 1: Add the two equations

Adding the two given equations, we get: \[ x^2 - xy - x + y^2 - xy + y = 56 \] Simplifying: \[ x^2 - 2xy + y^2 - x + y = 56 \]

Step 2: Recognize the quadratic form

Notice that the left-hand side can be written as: \[ (x - y)^2 - (x - y) = 56 \]

Step 3: Substitute \( t = (x - y) \)

Let \( t = (x - y) \). Then, the equation becomes: \[ t^2 - t = 56 \] Rearranging: \[ t^2 - t - 56 = 0 \]

Step 4: Solve the quadratic equation

We solve the quadratic equation: \[ (t - 8)(t + 7) = 0 \] So, we have two possible solutions: \[ t = 8 \quad \text{or} \quad t = -7 \]

Step 5: Conclusion

Since \( t = (x - y) \), we conclude: \[ x - y = 8 \]

Answer 2

Given Equations: 

\[ y^2 - xy + y = 34 \] \[ x^2 - xy - x = 22 \]

Step 1: Add the two equations:

\[ (y^2 - xy + y) + (x^2 - xy - x) = 34 + 22 \Rightarrow x^2 + y^2 - 2xy + y - x = 56 \]

Step 2: Recognize the squared difference:

\[ x^2 + y^2 - 2xy = (x - y)^2 \] So, \[ (x - y)^2 + (y - x) = 56 \Rightarrow (x - y)^2 - (x - y) = 56 \]

Step 3: Let \( n = x - y \), then:

\[ n^2 - n = 56 \Rightarrow n^2 - n - 56 = 0 \]

Step 4: Solve quadratic for n:

\[ n = \frac{1 \pm \sqrt{1 + 4 \times 56}}{2} = \frac{1 \pm \sqrt{225}}{2} = \frac{1 \pm 15}{2} \] Thus, \[ n = 8 \quad \text{or} \quad n = -7 \]

Step 5: Determine if \( x > y \):

Since \( n = x - y \), if \( n = 8 \), then \( x > y \).
If \( n = -7 \), then \( x < y \).

Conclusion:

Given the problem context, the solution \( n = 8 \) is acceptable, so: \[ \boxed{x - y = 8} \] which means \( x > y \).