Hi venmic!

Are you sure that C is correct?

problem:
if x > y, is zx > yx ?

(1) z > 0
(2) y < 0

Let as look at zx > yx
We can deduct yx on both sides of an inequality zx - yx > 0,
we can further reshape it x(z - y) > 0

(1) z > 0

at this point we can plugin numbers
z=5 x=8, y=7
z(x - y) = 8(5-7) = -16
thus x(z - y) < 0
let us try other numbers
z=5, x=4, y=2
z(x - y) = 4(5 - 2) = 12
thus x(z - y) > 0

It is insufficient

(2) y < 0

now let us try to plugin numbers
z=5 x=-2, y=-3
z(x - y) = -2(5+3) = -16
thus x(z - y) < 0

let us try other numbers
z=3, x=1, y=-2
z(x - y) = 1(3+2) = 5
thus x(z - y) > 0

it is insufficient

According to me correct answer is E

hope it helps!

Best

x>y
is zx>yx?

The question basically asks, is z>y (since the x's on either side cancel out).

C, together they are suff, since you one is greater than 0 and the other is less than 0.

we have to include the possibility that x can be positive or negative

if you multiply (or divide) both sides by a negative number "<" becomes ">"

I hope it is better explanation

Maciek wrote:we have to include the possibility that x can be positive or negative

if you multiply (or divide) both sides by a negative number "<" becomes ">"

I hope it is better explanation

ah, you're right.

if x > y, is zx > yx ?

(1) z > 0
(2) y < 0


(1)
First thing I did with (1) is suppose z=1 which tells me nothing about the equation. If z = 1 then zx>yx = x>yx

Three possibilities, x +ve, y +ve .. x +ve, y -ve .. x -ve, y -ve.

So is +ve greater than +ve x +ve, yea it can be. ( 1/2 > 1/2 x 1/4) but ( 4 < 4 x 3) so insuf

(2)

yx could be positive or negative and zx could be positive or negative independent of each other so insuf

(together)
Let's put z = 1 so it drops out, then I could have x is positive or negative while y is negative.

+ve > +ve x -ve
-ve < -ve x -ve
insuff