GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 20 Apr 2019, 09:31

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Given the graphs of y = (x^2 - 3x - 10)/(x + 2) and y= 3x - 1 in the x

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 54376
Given the graphs of y = (x^2 - 3x - 10)/(x + 2) and y= 3x - 1 in the x  [#permalink]

### Show Tags

24 Apr 2018, 06:12
00:00

Difficulty:

95% (hard)

Question Stats:

29% (01:52) correct 71% (01:50) wrong based on 76 sessions

### HideShow timer Statistics

Given the graphs of $$y = \frac{(x^2 - 3x - 10)}{(x + 2)}$$ and $$y= 3x - 1$$ in the xy-plane. Which of the following can be the number of points of intersection of the two graphs?

I. 0
II. 1
III. 2

(A) I only
(B) II only
(C) I and II only
(D) II and III only
(E) I, II, and III

_________________
Manager
Joined: 05 Dec 2017
Posts: 52
Location: India
WE: Analyst (Commercial Banking)
Re: Given the graphs of y = (x^2 - 3x - 10)/(x + 2) and y= 3x - 1 in the x  [#permalink]

### Show Tags

24 Apr 2018, 10:06
2
Bunuel wrote:
Given the graphs of $$y = \frac{(x^2 - 3x - 10)}{(x + 2)}$$ and $$y= 3x - 1$$ in the xy-plane. Which of the following can be the number of points of intersection of the two graphs?

I. 0
II. 1
III. 2

(A) I only
(B) II only
(C) I and II only
(D) II and III only
(E) I, II, and III

1st equation $$y = \frac{(x^2 - 3x - 10)}{(x + 2)}$$ can be written as $$y = \frac{(x+2)*(x-5)}{(x+2)}$$
or y = x-5, (provided $$x\neq{-2}$$)
2nd equation y = 3x-1

solving x = -2 and y = -7, but x=-2 is not possible

So 0 values..Hence Option A
Intern
Joined: 31 Jan 2018
Posts: 42
Location: Germany
GMAT 1: 610 Q44 V30
GPA: 3
WE: Supply Chain Management (Manufacturing)
Re: Given the graphs of y = (x^2 - 3x - 10)/(x + 2) and y= 3x - 1 in the x  [#permalink]

### Show Tags

24 Apr 2018, 11:11
rkgstyle wrote:
Bunuel wrote:
Given the graphs of $$y = \frac{(x^2 - 3x - 10)}{(x + 2)}$$ and $$y= 3x - 1$$ in the xy-plane. Which of the following can be the number of points of intersection of the two graphs?

I. 0
II. 1
III. 2

(A) I only
(B) II only
(C) I and II only
(D) II and III only
(E) I, II, and III

1st equation $$y = \frac{(x^2 - 3x - 10)}{(x + 2)}$$ can be written as $$y = \frac{(x+2)*(x-5)}{(x+2)}$$
or y = x-5, (provided $$x\neq{-2}$$)
2nd equation y = 3x-1

solving x = -2 and y = -7, but x=-2 is not possible

So 0 values..Hence Option A

Can you please explain why this is not possible?
Manager
Joined: 05 Dec 2017
Posts: 52
Location: India
WE: Analyst (Commercial Banking)
Re: Given the graphs of y = (x^2 - 3x - 10)/(x + 2) and y= 3x - 1 in the x  [#permalink]

### Show Tags

24 Apr 2018, 11:15
1
fallenx wrote:
rkgstyle wrote:
Bunuel wrote:
Given the graphs of $$y = \frac{(x^2 - 3x - 10)}{(x + 2)}$$ and $$y= 3x - 1$$ in the xy-plane. Which of the following can be the number of points of intersection of the two graphs?

I. 0
II. 1
III. 2

(A) I only
(B) II only
(C) I and II only
(D) II and III only
(E) I, II, and III

1st equation $$y = \frac{(x^2 - 3x - 10)}{(x + 2)}$$ can be written as $$y = \frac{(x+2)*(x-5)}{(x+2)}$$
or y = x-5, (provided $$x\neq{-2}$$)
2nd equation y = 3x-1

solving x = -2 and y = -7, but x=-2 is not possible

So 0 values..Hence Option A

Can you please explain why this is not possible?

Sure

When we plug x = -2, the 1st graph, $$y = \frac{(x^2 - 3x - 10)}{(x + 2)}$$, $$y = \frac{0}{0}$$, which is not defined, Hence x= -2 is not possible.

Please let me know, if there is still any doubt.
Manager
Status: Gathering chakra
Joined: 05 Feb 2018
Posts: 185
Re: Given the graphs of y = (x^2 - 3x - 10)/(x + 2) and y= 3x - 1 in the x  [#permalink]

### Show Tags

22 Mar 2019, 15:57
Tricky question...

(x²-3x-10)/(x+2) = 3x - 1
(x-5)(x+2)/(x+2) = 3x - 1
x-5 = 3x - 1
-4 = 2x
-2 = x

BUT we have division by a variable, if we plug -2 back into the first equation we have division by 0...which is undefined. So no solution.
Re: Given the graphs of y = (x^2 - 3x - 10)/(x + 2) and y= 3x - 1 in the x   [#permalink] 22 Mar 2019, 15:57
Display posts from previous: Sort by