Correct option is (c) x = 2, y = –4
Given,
\(\frac{x + 2}{y + 2} + 2 =0\)
⇒ \(\frac{x + 2y + 6}{y + 2} = 0\)
⇒ x + 2y + 6 = 0 .......(1)
\(\frac{x - 4}{y-2} =\frac{x -1}{y + 7}\)
⇒ xy + 7x − 4y − 28 = xy − 2x − y + 2
⇒ 9x − 3y = 30
⇒ 3x − y = 10
⇒ y = 3x − 10 .....(2)
Putting the value of y in eq(1), we get
x + 2(3x − 10) + 6 = 0
⇒ 7x = 14
⇒ x = 2
Putting the value of x in eq(2), we get
y = 6 − 10 = −4
So, we get the value of x = 2 and y = −4.