If x < 0 and y > 0, determine the sign of the real number. (a) x y positive negative (b) xy2 positive negative (c) x − y xy positive negative (d) y(y − x) positive negative

From the question it is given that x<0 and y>0 or in other words x is negative and y is positive.

(a)x y : x is negative and y is positive

⇒negative x positive = negative

Therefore, xy is negative

(b)x :  x  is negative and is positive

⇒negative x positive = negative

Therefore, x is negative

(c) (x − y)xy : x  is negative and is positive

⇒(x − y) is negative and xy is negative {from (a)}

⇒negative x negative = positive

Therefore, (x-y)xy is negative

(d) y(y − x): x is negative and y is positive

here y - x is positive

⇒positive x positive = positive

Therefore, (y-x)y is negative