Question

if x= -2 y=-1 then p^-q-q^p​

Answer 1

Step-by-step explanation:

Solution 12. From

p

y

+

q

x

= 1 we have py+qx = xy

and then xy qx py+pq = pq.

Factoring the left side of the last equation gives

(x

p)(y

q) = pq.

Because x and y are positive integers and p and q are prime we must

have

x

p = p and y

q = q

leading to (x, y) = (2p, 2q)

or

x

p = q

and y

q = p

leading to (x, y) = (p + q, p + q)

or

x

p = 1 and y

q = pq

leading to (x, y) = (p + 1, pq + q)

or

x

p = pq

and y

q = 1 leading to (x, y) = (pq + p, q + 1).

If p = q, then the first two solutions are the same and we have three

distinct solution sets. If p = q, then the four possibilities result in four

different solution sets.

1

Answer 2

Answer:

hope it is helpful

Step-by-step explanation:

From  p y + q x = 1 we have py+qx = xy   and then xy qx py+pq = pq.

Factoring the left side of the last equation gives    (x p)(y q) = pq.  

Because x and y are positive integers and p and q are prime we must

have  x p = p and y q = q

leading to (x, y) = (2p, 2q)   or   x p = q and y q = p

leading to (x, y) = (p + q, p + q)   or  x p = 1 and y q = pq

leading to (x, y) = (p + 1, pq + q)     or   x p = pq and y

q = 1 leading to (x, y) = (pq + p, q + 1).

If p = q, then the first two solutions are the same and we have three

distinct solution sets. If p = q, then the four possibilities result in four

different solution sets. 1