Question

If y ≠ 0, then the number of order pair (x, y) such that

$\sf (x + y) + \frac{x}{y} = \frac{1}{2}$

and

$\sf (x + y)(\frac{x}{y}) = \frac{1}{2}$ is?
​

Answer 1

||✪✪ Correct Question ✪✪||

• (x + y) + xy = (1/2)
• (x + y) * (x/y) = (-1/2)

✯✯ To Find ✯✯

• the number of order pair (x, y) ?

|| ✰✰ ANSWER ✰✰ ||

Let ,

→ (x + y) = (1/2) - (x/y) -------------- Equation (1)

→ (x+y) * (x/y) = (-1/2) --------------- Equation (2)

Dividing Both Equations Now, we get :-

→ [ (x + y) / (x+y) * (x/y) ] = [ (1/2) - (x/y) / (-1/2) ]

→ [ 1/(x/y) ] = (-2)[ (1/2) - (x/y) ]

→ [ 1/(x/y) ] = (-1) + (2x/y)

Let us Assume That, (x/y) = t now,

So,

→ (1/t) = (-1) + 2t

→ 1 = (-t) + 2t²

→ 2t² - t - 1 = 0

Splitting The Middle Term now,

→ 2t² - 2t + t - 1 = 0

→ 2t(t - 1) + 1(t - 1) = 0

→ (2t + 1)(t - 1) = 0

Putting Both Equal to Zero now,

→ t = (-1/2) or, 1.

So,

→ (x/y) = (-1/2) or, 1.

_____________________

Case ❶ :- when (x/y) = 1 .

➼ (x/y) = 1

➼ x = y .

Putting This value in Equation (1) now, we get,

➼ (x + y) = (1/2) - (x/y)

➼ (x + x) = (1/2) - (x/x)

➼ 2x = (1/2) - 1

➼ 2x = (-1/2)

➼ x = (-1/4)

So,

☛ x = (-1/4) = y

____________________

Case ❷ :- when (x/y) = (-1/2) .

➪ (x/y) = (-1/2)

➪ (-2x) = y

Putting This value in Equation (1) now, we get ,

➪ (x + y) = (1/2) - (x/y)

➪ (x - 2x) = (1/2) - {x /(-2x)}

➪ (-x) = (1/2) - (-1/2)

➪ (-x) = 1/2 + 1/2

➪ (-x) = 1

➪ x = (-1)

So,

☛ y = (-2x) = (-2) * (-1) = 2.

____________________

Hence, The number of order pair (x, y) are [ (-1/4) , (-1/4) ] & [(-1) , 2 ] = Total 2 Pairs.

Answer 2

Refer The Attachment.......