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

Given:\( x^2-xy-x=22 \) and \(y^2-xy+y=34\)
Adding the two given equations we get:
\(x^2-2xy+y^2-x+y=56\)
\(⇒ (x-y)^2-(x-y)=56\)

Let \((x-y) =t\)
\(⇒ t^2-t=56\)
\(t^2-t-56=0\)
\((t-8)(t+7) =0 \)
so \(t=8 \)
so \(x-y =8\)

Answer 2

\(y^2 - xy + y = 34\)
 \(x^2 - xy - x = 22\)
These two equations add up to 

\(x + y - (x - y)^2 = 56 \quad \text{or} \quad x + y - (y - x)^2 = 56\)
Given that we must determine whether \(x > y\), we should calculate \((x - y)^2 - x + y = 56\)
\((x - y)^2 - x + y = 56\)
\((x - y) (x - y - 1) = 56 \)
This reduces to \(n (n - 1) = 56\) if we choose \(x - y = n. \)
Consequently, \(x - y = 8. \)