Question

13 th text book in maths x^2y^2+3 xyp+2y^2=0

Answer 1

Let's solve for p.

x2y2+3xyp+2y2=0

Step 1: Add -x^2y^2 to both sides.

x2y2+3pxy+2y2+−x2y2=0+−x2y2

3pxy+2y2=−x2y2

Step 2: Add -2y^2 to both sides.

3pxy+2y2+−2y2=−x2y2+−2y2

3pxy=−x2y2−2y2

Step 3: Divide both sides by 3xy.

3pxy

3xy

=

−x2y2−2y2

3xy

p=

−x2y−2y

3x

Answer:

p=

−x2y−2y

3x