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Bunuel
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Bunuel
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Hi Bunuel

I have a question for the number 3:
What if we select x= 10; y = -5 or x= -5; y= 10. For the first par we can say that x/y = -2, which is less than -1 or, on the other hand y=10 / x=-5 which is equal to -2 also.. Please help me understand. Thanks

Your pairs do not satisfy the condition x + y < 0; therefore, they are not valid.
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Sorry Bunuel , I meant (-10,5) or (5, -10) then, x/y would be -2 which is less than -1 or y/x = -2.. Thanks in advance!
Bunuel
Javmex
Hi Bunuel

I have a question for the number 3:
What if we select x= 10; y = -5 or x= -5; y= 10. For the first par we can say that x/y = -2, which is less than -1 or, on the other hand y=10 / x=-5 which is equal to -2 also.. Please help me understand. Thanks

Your pairs do not satisfy the condition x + y < 0; therefore, they are not valid.
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Javmex
Sorry Bunuel , I meant (-10,5) or (5, -10) then, x/y would be -2 which is less than -1 or y/x = -2.. Thanks in advance!
Bunuel
Javmex
Hi Bunuel

I have a question for the number 3:
What if we select x= 10; y = -5 or x= -5; y= 10. For the first par we can say that x/y = -2, which is less than -1 or, on the other hand y=10 / x=-5 which is equal to -2 also.. Please help me understand. Thanks

Your pairs do not satisfy the condition x + y < 0; therefore, they are not valid.

If x = -10 and y = 5, then y/x = -1/2, and in this case \(\frac{y}{x} > -1\) is true.

If x = 5 and y = -10, then x/y = -1/2, and in this case \(\frac{x}{y} > -1\) is true.

So, in any case either \(\frac{x}{y} > -1\) or \(\frac{y}{x} > -1\) is true, making III true.

Hope it's clear.
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Third inequality is tricky.

x/y>-1 or y/x>-1. Don't multiply each side to reach conclusion. Rather bring the terms on same side of inequality.

So lets say y was -ve and x was +ve from xy<0. and magnitude of y is greater than x as x+y<0,

therefore, x/y>-1 >>>>>>bringing to same side x/y+1>0 OR (x+y)/y>0. Since x+y<0 "given" the numerator is always -ve and denominator was also -ve (y we assumed to be negative) So (-ve)N/(-veD) becomes positive and hence rightly greater than 0.

Apply same to the viceversa case, x-ve, y+ve, Mag(x)>(Mag(y). Here, y/x>-1 inequaility holds true.
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