Question

y=5x^2-2
_______
3x^2+4. then[dy]
___
[DX]
X=1 is​

Answer 1

Answer:

Disclaimer: There is error in the Q. In (II) there should have been 2x - 3y = 12

5x + 3y = 9 -----(I)

2x - 3y = 12 ----- (II)

Add (I) and (II)

7x = 21

x = 3

Putting the value of x = 3 in (I) we get

5(3)+3y=9⇒15+3y=9⇒3y=9-15=-6⇒y=-2

Thus, (x, y) = (3, -6).