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If y>0>x, and (3+5y)/(x−1) < −7, then which of the following  [#permalink]

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Question Stats: 70% (01:58) correct 30% (02:22) wrong based on 222 sessions

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If y>0>x, and (3+5y)/(x−1) < −7, then which of the following must be true?

A. 5y−7x+4 < 0
B. 5y+7x−4 > 0
C. 7x−5y−4 < 0
D. 4+5y+7x > 0
E. 7x−5y+4 > 0

P.S: Bunuel-although I got it right but I'm having confusion on the 'flip of sign' as we're to multiply this inequality with a negative quantity...Can you please explain in detail?

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Re: If y>0>x, and (3+5y)/(x−1) < −7, then which of the following  [#permalink]

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4
7
If y>0>x, and (3+5y)/(x−1) < −7, then which of the following must be true?

A. 5y−7x+4 < 0
B. 5y+7x−4 > 0
C. 7x−5y−4 < 0
D. 4+5y+7x > 0
E. 7x−5y+4 > 0

Since given that x is a negative number then x-1 is also some negative number. Multiply both sides of the inequality by x-1. When multiplying by negative flip the sig, so we'll have:

$$\frac{3+5y}{x-1} < -7$$ --> $$3+5y>-7(x-1)$$ --> $$3+5y>7-7x$$ --> $$5y+7x-4>0$$.

Hope it's clear.
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Retired Moderator B
Joined: 27 Aug 2012
Posts: 1046
Re: If y>0>x, and (3+5y)/(x−1) < −7, then which of the following  [#permalink]

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Bunuel wrote:
If y>0>x, and (3+5y)/(x−1) < −7, then which of the following must be true?

A. 5y−7x+4 < 0
B. 5y+7x−4 > 0
C. 7x−5y−4 < 0
D. 4+5y+7x > 0
E. 7x−5y+4 > 0

Since given that x is a negative number then x-1 is also some negative number. Multiply both sides of the inequality by x-1. When multiplying by negative flip the sig, so we'll have:

$$\frac{3+5y}{x-1} < -7$$ --> $$3+5y>-7(x-1)$$ --> $$3+5y>7-7x$$ --> $$5y+7x-4>0$$.

Hope it's clear.

There I'm having confusion...as x-1 is negative so multiplying with it both sides of the inequality should flip their signs.But for example, RHS remains with a negative sign!

Is it because of the fact that x-1 is itself negative so with -7 multiplied with x-1 will yield positive (negative * negative =positive)...Same logic goes for LHS of inequality which is primnarily negative because of its negative denominator, so LHS multiplied with (x-1) yields a positive. Let me know if this is correct...
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Joined: 02 Sep 2009
Posts: 59073
Re: If y>0>x, and (3+5y)/(x−1) < −7, then which of the following  [#permalink]

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debayan222 wrote:
Bunuel wrote:
If y>0>x, and (3+5y)/(x−1) < −7, then which of the following must be true?

A. 5y−7x+4 < 0
B. 5y+7x−4 > 0
C. 7x−5y−4 < 0
D. 4+5y+7x > 0
E. 7x−5y+4 > 0

Since given that x is a negative number then x-1 is also some negative number. Multiply both sides of the inequality by x-1. When multiplying by negative flip the sig, so we'll have:

$$\frac{3+5y}{x-1} < -7$$ --> $$3+5y>-7(x-1)$$ --> $$3+5y>7-7x$$ --> $$5y+7x-4>0$$.

Hope it's clear.

There I'm having confusion...as x-1 is negative so multiplying with it both sides of the inequality should flip their signs.But for example, RHS remains with a negative sign!

Is it because of the fact that x-1 is itself negative so with -7 multiplied with x-1 will yield positive (negative * negative =positive)...Same logic goes for LHS of inequality which is primnarily negative because of its negative denominator, so LHS multiplied with (x-1) yields a positive. Let me know if this is correct...

Flipping sign means flipping "<" sign to ">". Nothing else.
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Joined: 27 Aug 2012
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Re: If y>0>x, and (3+5y)/(x−1) < −7, then which of the following  [#permalink]

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Bunuel wrote:
debayan222 wrote:
Bunuel wrote:
If y>0>x, and (3+5y)/(x−1) < −7, then which of the following must be true?

A. 5y−7x+4 < 0
B. 5y+7x−4 > 0
C. 7x−5y−4 < 0
D. 4+5y+7x > 0
E. 7x−5y+4 > 0

Since given that x is a negative number then x-1 is also some negative number. Multiply both sides of the inequality by x-1. When multiplying by negative flip the sig, so we'll have:

$$\frac{3+5y}{x-1} < -7$$ --> $$3+5y>-7(x-1)$$ --> $$3+5y>7-7x$$ --> $$5y+7x-4>0$$.

Hope it's clear.

There I'm having confusion...as x-1 is negative so multiplying with it both sides of the inequality should flip their signs.But for example, RHS remains with a negative sign!

Is it because of the fact that x-1 is itself negative so with -7 multiplied with x-1 will yield positive (negative * negative =positive)...Same logic goes for LHS of inequality which is primnarily negative because of its negative denominator, so LHS multiplied with (x-1) yields a positive.Let me know if this is correct...

Flipping sign means flipping "<" sign to ">". Nothing else.

Ok!
But could you please let me know whether my reasoning above(as highlighted) is correct...?
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Math Expert V
Joined: 02 Sep 2009
Posts: 59073
Re: If y>0>x, and (3+5y)/(x−1) < −7, then which of the following  [#permalink]

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debayan222 wrote:
Bunuel wrote:
debayan222 wrote:
There I'm having confusion...as x-1 is negative so multiplying with it both sides of the inequality should flip their signs.But for example, RHS remains with a negative sign!

Is it because of the fact that x-1 is itself negative so with -7 multiplied with x-1 will yield positive (negative * negative =positive)...Same logic goes for LHS of inequality which is primnarily negative because of its negative denominator, so LHS multiplied with (x-1) yields a positive.Let me know if this is correct...

Flipping sign means flipping "<" sign to ">". Nothing else.

Ok!
But could you please let me know whether my reasoning above(as highlighted) is correct...?

Don't know why do you need this reasoning to solve this question, but anyway:

-7*(x-1) = negative*negative = positive;
(3+5y)/(x−1) = positive/negative = negative. When multiplied by (x-1)=negative, (x-1) reduces and we get (3+5y)=positive.

Hope it's clear.
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Re: If y>0>x, and (3+5y)/(x−1) < −7, then which of the following  [#permalink]

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Much Thanks Sir for the clarification... Well,it's just a self-satisfaction to know that I got the logic right for multiplying with variable...nothing else!
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Re: If y>0>x, and (3+5y)/(x−1) < −7, then which of the following  [#permalink]

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C. 7x−5y−4 < 0

why C is not always true, x is given negative, y is positive.

then 7(-ve) - 5 (+ve) -4 < 0

-ve - ve -ve < 0 this also negative always.
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Re: If y>0>x, and (3+5y)/(x−1) < −7, then which of the following  [#permalink]

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PiyushK wrote:
C. 7x−5y−4 < 0

why C is not always true, x is given negative, y is positive.

then 7(-ve) - 5 (+ve) -4 < 0

-ve - ve -ve < 0 this also negative always.

If y>0>x, and (3+5y)/(x−1) < −7, then which of the following must be true?

Here we have two equations. The 1st conforms (C) but the 2nd doesn't conforms (C). i mean we can't have (C) from the 2nd condition.
we have to select a combine effect of both as the result......
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Re: If y>0>x, and (3+5y)/(x−1) < −7, then which of the following  [#permalink]

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Asifpirlo wrote:
PiyushK wrote:
C. 7x−5y−4 < 0

why C is not always true, x is given negative, y is positive.

then 7(-ve) - 5 (+ve) -4 < 0

-ve - ve -ve < 0 this also negative always.

If y>0>x, and (3+5y)/(x−1) < −7, then which of the following must be true?

Here we have two equations. The 1st conforms (C) but the 2nd doesn't conforms (C). i mean we can't have (C) from the 2nd condition.
we have to select a combine effect of both as the result......

Ohhh IC, I got it.

Thanks.
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Re: If y>0>x, and (3+5y)/(x−1) < −7, then which of the following  [#permalink]

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Bunuel wrote:
If y>0>x, and (3+5y)/(x−1) < −7, then which of the following must be true?

A. 5y−7x+4 < 0
B. 5y+7x−4 > 0
C. 7x−5y−4 < 0
D. 4+5y+7x > 0
E. 7x−5y+4 > 0

Since given that x is a negative number then x-1 is also some negative number. Multiply both sides of the inequality by x-1. When multiplying by negative flip the sig, so we'll have:

$$\frac{3+5y}{x-1} < -7$$ --> $$3+5y>-7(x-1)$$ --> $$3+5y>7-7x$$ --> $$5y+7x-4>0$$.

Hope it's clear.

I tried plugging in sample values say y=2 x=-1 into 5y+7x-4

10-7-4 < 0
What am i missing here ?
Math Expert V
Joined: 02 Sep 2009
Posts: 59073
Re: If y>0>x, and (3+5y)/(x−1) < −7, then which of the following  [#permalink]

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goforgmat wrote:
Bunuel wrote:
If y>0>x, and (3+5y)/(x−1) < −7, then which of the following must be true?

A. 5y−7x+4 < 0
B. 5y+7x−4 > 0
C. 7x−5y−4 < 0
D. 4+5y+7x > 0
E. 7x−5y+4 > 0

Since given that x is a negative number then x-1 is also some negative number. Multiply both sides of the inequality by x-1. When multiplying by negative flip the sig, so we'll have:

$$\frac{3+5y}{x-1} < -7$$ --> $$3+5y>-7(x-1)$$ --> $$3+5y>7-7x$$ --> $$5y+7x-4>0$$.

Hope it's clear.

I tried plugging in sample values say y=2 x=-1 into 5y+7x-4

10-7-4 < 0
What am i missing here ?

Those values do not satisfy the given condition that (3+5y)/(x−1) < −7
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Re: If y>0>x, and (3+5y)/(x−1) < −7, then which of the following  [#permalink]

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