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# M19-29

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Math Expert
Joined: 02 Sep 2009
Posts: 56269

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16 Sep 2014, 01:06
1
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Difficulty:

75% (hard)

Question Stats:

55% (02:27) correct 45% (02:30) wrong based on 88 sessions

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If $$k$$ does not equal -1, 0 or 1, does the point of intersection of line $$y = kx + b$$ and line $$x = ky + b$$ have a negative x-coordinate?

(1) $$kb \gt 0$$

(2) $$k \gt 1$$

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Math Expert
Joined: 02 Sep 2009
Posts: 56269

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16 Sep 2014, 01:07
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Official Solution:

We have equations of two lines: $$y = kx + b$$ and $$y=\frac{x}{k}-\frac{b}{k}$$ (from $$x = ky + b$$). Equate to get the x-coordinate of the intersection point: $$kx + b=\frac{x}{k}-\frac{b}{k}$$, which gives $$x=\frac{b(k+1)}{1-k^2}=\frac{b(k+1)}{(1-k)(1+k)}=\frac{b}{1-k}$$.

So, the question basically asks whether $$x=\frac{b}{1-k}$$ is negative.

(1) $$kb \gt 0$$. This statement tells that $$k$$ and $$b$$ have the same sign. Now, if $$b \gt 0$$ and $$k=2$$ then the answer is YES but if $$b \gt 0$$ and $$k=\frac{1}{2}$$ then the answer is NO. Not sufficient.

(2) $$k \gt 1$$. So, the denominator of $$x=\frac{b}{1-k}$$ is negative, but we have no info about $$b$$. Not sufficient.

(1)+(2) Since from (2) $$k$$ is positive and from (1) $$k$$ and $$b$$ have the same sign, then $$b$$ is positive too. So, numerator ($$b$$) is positive and denominator ($$1-k$$) is negative, which means that $$x=\frac{b}{1-k}$$ is negative. Sufficient.

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Joined: 01 Aug 2013
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20 Jul 2016, 12:16
Hi Bunuel,

Is there a way to resolve this using graphical method? I tried but I am facing deadlock in statement 1.
Intern
Joined: 13 Jun 2016
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17 Aug 2016, 06:06
Bunuel,

how do you get from kx+b= (x - b) / k to x = b(k+1)/1−k2 ? the "1-k2" in the denominator i just cant figure out how you get that.
Math Expert
Joined: 02 Sep 2009
Posts: 56269

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17 Aug 2016, 07:05
1
lydennis8 wrote:
Bunuel,

how do you get from kx+b= (x - b) / k to x = b(k+1)/1−k2 ? the "1-k2" in the denominator i just cant figure out how you get that.

$$kx+b= \frac{(x - b)}{k}$$

Multiply by k: $$k^2x + bk = x - b$$

Rearrange: $$bk + b = x - k^2x$$

Factor out b and x: $$b(k + 1) = x(1 - k^2)$$

$$\frac{b(k + 1)}{(1 - k^2)} = x$$.

Hope it's clear.
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Joined: 07 Oct 2015
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24 Aug 2016, 06:08
I know this is a silly question: how did you get y=x/k-b/k?
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24 Aug 2016, 06:10
3ksnikhil wrote:
I know this is a silly question: how did you get y=x/k-b/k?

x = ky + b;

Re-arrange: x - b = ky;

Divide by k: x/k - b/y = y.
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Intern
Joined: 20 Aug 2017
Posts: 18

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13 Dec 2017, 11:45
hey!

would it be correct to solve this question this way?

ky+b=kx+b
x=0, y=b ; y=0 x=b
(1) both negative or both positive - not suf
(2) not suf
(1+2) if k is positive then according to (1) b also positive - then x positive.

am i right?

thanks!
Director
Joined: 14 Feb 2017
Posts: 691
Location: Australia
Concentration: Technology, Strategy
GMAT 1: 560 Q41 V26
GMAT 2: 550 Q43 V23
GMAT 3: 650 Q47 V33
GPA: 2.61
WE: Management Consulting (Consulting)

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16 Jun 2019, 18:01
Bunuel wrote:
lydennis8 wrote:
Bunuel,

how do you get from kx+b= (x - b) / k to x = b(k+1)/1−k2 ? the "1-k2" in the denominator i just cant figure out how you get that.

$$kx+b= \frac{(x - b)}{k}$$

Multiply by k: $$k^2x + bk = x - b$$

Rearrange: $$bk + b = x - k^2x$$

Factor out b and x: $$b(k + 1) = x(1 - k^2)$$

$$\frac{b(k + 1)}{(1 - k^2)} = x$$.

Hope it's clear.

The explanation is made much more clearer with this! Thanks
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Goal: Q49, V41

+1 Kudos if you like my post pls!
Re: M19-29   [#permalink] 16 Jun 2019, 18:01
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# M19-29

Moderators: chetan2u, Bunuel