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

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

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New post 24 Apr 2018, 06:12
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A
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E

Difficulty:

  95% (hard)

Question Stats:

29% (01:58) correct 71% (01:50) wrong based on 83 sessions

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

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New post 24 Apr 2018, 10:06
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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
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Re: Given the graphs of y = (x^2 - 3x - 10)/(x + 2) and y= 3x - 1 in the x  [#permalink]

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New post 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? :-)
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Re: Given the graphs of y = (x^2 - 3x - 10)/(x + 2) and y= 3x - 1 in the x  [#permalink]

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New post 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.
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Re: Given the graphs of y = (x^2 - 3x - 10)/(x + 2) and y= 3x - 1 in the x  [#permalink]

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New post 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.
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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
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